MTG Mathematics Eng

Grade Mat h e m a t i c s Study Guide 12

© Department of Basic Education 2015 This content may not be sold or used for commercial purposes. Curriculum and Assessment Policy Statement (CAPS) Grade 12 Mind the Gap study guide for Mathematics ISBN 978-1-4315-1935-4 This publication has a Creative Commons Attribution NonCommercial Sharealike license. You can use, modify, upload, download, and share content, but you must acknowledge the Department of Basic Education, the authors and contributors. If you make any changes to the content you must send the changes to the Department of Basic Education. This content may not be sold or used for commercial purposes. For more information about the terms of the license please see: http://creativecommons.org/licenses/by-nc-sa/3.0/. Copyright © Department of Basic Education 2015 222 Struben Street, Pretoria, South Africa Contact person: Dr Patricia Watson Email: watson.p@dbe.gov.za Tel: (012) 357 4502 http://www.education.gov.za Call Centre: 0800202933 The first edition, published in 2012, for the Revised National Curriculum Statement (RNCS) Grade 12 Mind the Gap study guides for Accounting, Economics, Geography and Life Sciences; the second edition, published in 2014 , aligned these titles to the Curriculum and Assessment Policy Statement (CAPS) and added more titles to the series in 2015, including the CAPS Grade 12 Mind the Gap study guide for Mathematics. ISBN 78-1-4315-1935-4 Mind the Gap team Series managing editor: Dr Patricia Watson Production co-ordinators: Lisa Treffry-Goatley and Radha Pillay Production assistants: Nomathamsanqa Hlatshwayo and Motshabi Mondlane Authors: Lynn Bowie, Ronald Peter Jacobs, Sue Jobson, Terrence Mongameli Mbusi, Sello Gert Motsoane, Nonhlanhla Rachel Mthembu, Ntshengedzeni Steven Muthige, Mpho Francis Phatlane, Josephina Mamaroke Phatlane, Peter Ketshepile Raadt, Percy Stephen Tebeila, Anita van Heerden and Carol Wyeth. Expert readers: Prof Bruce Watson, Leonard Mudau and Karen van Niekerk Proofreaders: John Ostrowick and Angela Thomas Designers: Sonja McGonigle and Erika van Rooyen Illustrators: Michele Dean, Vusi Malindi, Khosi Pholosa, John Ostrowick, Kenneth Tekane Cover illustration: Alastair Findlay Afrikaans Translation: Marga Vos Onsite writers’ workshop IT support: Wayne Cussons Special thanks to Winning Teams, CEO Denzil Hollis, for the organisation’s subject expertise and workshop support Winning Teams board game facilitators: Mantse Khoza and Sue Jobson

© Department of Basic Education 2015 Ministerial foreword The Department of Basic Education (DBE) has pleasure in releasing the second edition of the Mind the Gap study guides for Grade 12 learners. These study guides continue the innovative and committed attempt by the DBE to improve the academic performance of Grade 12 candidates in the National Senior Certificate (NSC) examination. The study guides have been written by teams of exerts comprising teachers, examiners, moderators, subject advisors and coordinators. Research, which began in 2012, has shown that the Mind the Gap series has, without doubt, had a positive impact on grades. It is my fervent wish that the Mind the Gap study guides take us all closer to ensuring that no learner is left behind, especially as we celebrate 20 years of democracy. Matsie Angelina Motshekga, MP The second edition of Mind the Gap is aligned to the 2014 Curriculum and Minister of Basic Education Assessment Policy Statement (CAPS). This means that the writers have considered the National Policy pertaining to the programme, promotion requirements and protocols for assessment of the National Curriculum Statement for Grade 12 in 2014. The CAPS aligned Mind the Gap study guides take their brief in part from the 2013 National Diagnostic report on learner performance and draw on the Grade 12 Examination Guidelines. Each of the Mind the Gap study guides defines key terminology and offers simple explanations and examples of the types of questions learners can expect to be asked in an exam. Marking memoranda are included to assist learners to build their understanding. Learners are also referred to specific questions from past national exam papers and examination memos that are available on the Department’s website – www.education.gov.za. The CAPS editions include Accounting, Economics, Geography, Life Sciences, Mathematics, Mathematical Literacy and Physical Sciences Part 1: Physics and Part 2: Chemistry. The series is produced in both English and Afrikaans. There are also nine English First Additional Language (EFAL) study guides. These include EFAL Paper 1 (Language in Context); EFAL Paper 3 (Writing) and a guide for each of the Grade 12 prescribed literature set works included in Paper 2. These are Short Stories, Poetry, To Kill a Mockingbird, A Grain of Wheat, Lord of the Flies, Nothing but the Truth and Romeo and Juliet. (Please remember when preparing for EFAL Paper 2 that you need only study the set works you did in your EFAL class at school.) The study guides have been designed to assist those learners who have been underperforming due to a lack of exposure to the content requirements of the curriculum and aim to mind-the-gap between failing and passing, by bridging the gap in learners’ understanding of commonly tested concepts, thus helping candidates to pass. All that is now required is for our Grade 12 learners to put in the hours required to prepare for the examinations. Learners, make us proud – study hard. We wish each and every one of you good luck for your Grade 12 examinations. __________________________________ Matsie Angelina Motshekga, MP Minister of Basic Education 2015

© Department of Basic Education 2015

© Department of Basic Education 2015 Table of contents Dear Grade 12 learner .... ............................................................ix How to use this study guide.........................................................xi Top 10 study tips..........................................................................xii Mnemonics. .................................................................................xiii Mind maps. .................................................................................. xiv On the day of the exam............................................................... xv Question words to help you answer questions. ...................... xvi Vocabulary................................................................................... xvii General terms............................................................................. xvii Technical terms........................................................................... xix The maths you need. .............................................................. xxviii Unit 1: Exponents and surds........................................................ 1 1.1 The number system. .................................................................... 1 1.2 Working with irrational numbers................................................. 3 1.3 Exponents..................................................................................... 6 1.4 Exponential equations. .............................................................. 12 1.5 Equations with rational exponents. .......................................... 14 1.6 Exam type examples. ................................................................. 17 Unit 2: Algebra.............................................................................19 2.1 Algebraic expressions. ............................................................... 19 2.2 Addition and subtraction. .......................................................... 19 2.3 Multiplication and division......................................................... 20 2.4 Factorising. ................................................................................. 21 2.5 Notes on factorising a trinomial................................................ 22 2.6 Quadratic equations. ................................................................. 24 2.7 Quadratic inequalities................................................................ 30 2.8 Simultaneous equations............................................................ 34 2.9 The nature of the roots. ............................................................. 37 Unit 3: Number patterns, sequences and series.................... 42 3.1 Number patterns........................................................................ 42 3.2 Arithmetic sequences. ............................................................... 43 3.3 Quadratic sequences................................................................. 45 3.4 Geometric sequences................................................................ 48 3.5 Arithmetic and geometric series. .............................................. 50 © Department of Basic Education 2015 CONTENTS v Mind the Gap Mathematics

© Department of Basic Education 2015 Unit 4: Funtions. ..........................................................................60 4.1 What is a function?. ................................................................... 60 4.2 Function notation. ...................................................................... 62 4.3 The basic functions, formulas and graphs............................... 63 4.4 Inverse functions........................................................................ 81 4.5 The logarithmic function............................................................ 84 Unit 5: Trig functions...................................................................88 5.1 Graphs of trigonometric functions............................................ 88 5.2 The effect of a on the shape of the graph: change in amplitude.................................................................................... 91 5.3 The effect of q on the shape of the graph: vertical shift. ........ 93 5.4 The effect of b on the shape of the graph: change in period.......................................................................................... 94 5.5 The effect of p on the shape of the graph: horizontal shift..... 95 Unit 6: Finance growth and decay. .........................................101 6.1 Revision: Simple and compound interest............................... 101 6.2 Calculating the value of P, i and n. .........................................104 6.3 Simple and compound decay formulae.................................. 107 6.4 Nominal and effective interest rates. .....................................109 6.5 Investments with time and interest rate changes. ................111 6.6 Annuities...................................................................................113 Unit 7: Calculus..........................................................................123 7.1 Average gradient. .....................................................................123 7.2 Average rate of change............................................................125 7.3 The derivative of a function at a point....................................126 7.4 Uses of the derivative. ............................................................. 131 7.5 Drawing the graph of a cubic polynomial. ..............................132 Unit 8: Probability......................................................................145 8.1 Revision. ...................................................................................145 8.2 Theoretical probability and relative frequency.......................146 8.3 Venn diagrams. ........................................................................ 147 8.4 Mutually exclusive events........................................................149 8.5 Complementary events............................................................150 8.6 Events which are not mutually exclusive................................152 8.7 Summary of symbols and sets used in probability................154 8.8 Tree diagrams and contingency tables...................................158 8.9 Contingency tables................................................................... 161 8.10 Counting principles. .................................................................164 8.11 Use of counting principles in probability. ............................... 170 vi CONTENTS  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 Unit 9: Analytical Geometry.....................................................172 9.1 Revise: Analytical Geometry.................................................... 172 9.2 The equation of a line.............................................................. 177 9.3 The inclination of a line. .......................................................... 179 9.4 Circles in analytical geometry. ................................................184 Unit 10: Trigonometry...............................................................191 10.1 Revise: Trig ratios. .................................................................... 191 10.2 Trig ratios in all the quadrants of the Cartesian plane..........194 10.3 Solving triangles with trig. .......................................................196 10.4 Using a calculator to find trig ratios. ....................................... 197 10.5 The trig ratios of special angles. .............................................198 10.6 Using reduction formulae. .......................................................201 10.7 Trigonometric identities. ..........................................................205 10.8 More trig identities................................................................... 207 10.9 Solving trigonometric equations. ............................................209 10.10 More solving trig equations using identities. ......................... 213 10.11 Compound and double angle identities. ................................ 215 10.12 Determining x for which the identity is undefined.................220 Unit 11: Trigonometry: Sine, cosine and area rules.............222 11.1 Right-angled triangles..............................................................222 11.2 Area rule. .................................................................................. 224 11.3 Sine rule....................................................................................226 11.4 Cosine rule................................................................................228 11.5 Problems in two and three dimensions..................................230 Unit 12: Euclidean Geometry. .................................................235 12.1 Revise: Proportion and area of trianges.................................235 12.2 Proportion theorems................................................................ 237 12.3 Similar polygons....................................................................... 240 Unit 13: Statistics......................................................................248 13.1 Bar graphs and frequency tables............................................ 249 13.2 Measures of central tendency.................................................250 13.3 Measures of dispersion (or spread)........................................254 13.4 Five number summary and box and whisker plot..................256 13.5 Histograms and frequency polygons.......................................260 13.6 Cumulative frequency tables and graphs (ogives).................263 13.7 Variance and standard deviation.............................................267 13.8 Bivariate data and the scatter plot (scatter graph)................ 271 13.9 The linear regression line (or the least squares regression line)......................................................................... 274 © Department of Basic Education 2015 CONTENTS vii Mind the Gap Mathematics

© Department of Basic Education 2015

© Department of Basic Education 2015 Dear Grade 12 learner We are confident that this This Mind the Gap study guide helps you to prepare for the end-of-year Mind the Gap study CAPS Grade 12 exam. guide can help you to prepare well so that you The study guide does NOT cover the entire curriculum, but it does focus on pass the end-of-year core content of each knowledge area and points out where you can earn exams. easy marks. You must work your way through this study guide to improve your understanding, identify your areas of weakness and correct your own mistakes. To ensure a good pass, you should also cover the remaining sections of the curriculum using other textbooks and your class notes. Overview of the Grade 12 exam The following topics make up each of the TWO exam papers that you write at the end of the year: Paper Topics Duration Total Date Marking Patterns and sequences 3 hours 150 October/ Externally Finance, growth and decay November Functions and graphs 1 Algebra, equations and inequalities Differential Calculus Probability Euclidean Geometry 3 hours 150 October/ Externally Analytical Geometry November 2 Statistics and regression Trigonometry © Department of Basic Education 2015 introduct ion ix Mind the Gap Mathematics

© Department of Basic Education 2015 Approximate Cognitive level Description of skills to be demonstrated Weighting number of marks in a 150-mark paper • Recall Knowledge • Identification of correct formula on the information 20% 30 marks sheet (no changing of the subject) • Use of mathematical facts • Appropriate use of mathematical vocabulary • Algorithms • Estimation and appropriate rounding of numbers • Proofs of prescribed theorems and derivation of Routine formulae 35% 52–53 marks Procedures • Perform well-known procedures • Simple applications and calculations which might involve few steps • Derivation from given information may be • involved • Identification and use (after changing the • subject) of correct formula • Generally similar to those encountered in class • Problems involve complex calculations and/or Complex • higher order reasoning 30% 45 marks Procedures • There is often not an obvious route to the • solution • Problems need not be based on a real world • context • Could involve making significant connections • between different representations • Require conceptual understanding • Learners are expected to solve problems by • integrating different topics. • Non-routine problems (which are not necessarily Problem Solving difficult) 15% 22–23 marks • Problems are mainly unfamiliar • Higher order reasoning and processes are involved • Might require the ability to break the problem down into its constituent parts • Interpreting and extrapolating from solutions obtained by solving problems based in unfamiliar contexts. x introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 How to use this study guide Look out for This study guide covers selected parts of the different topics of these icons in the the CAPS Grade 12 curriculum in the order they are usually taught study guide. during the year. The selected parts of each topic are presented in the following way: PAY SPECIAL ATTENTION • An explanation of terms and concepts; NB NB PAY SPECIAL ATTENTION • NB Worked examples to explain and demonstrate; NB • Activities with questions for you to answer; and PAY SPECIAL ATTENTION • Answers for you to use to check your NBown work. NB HINT! HINT! hint hint Hints to help you remember a concept NB PAY SPECIAL ATTENTION Pay special attention NBhint HINT! orEG - guide you in e.g. EG - worked examples Worked examples e.g. worked examples solving problems exams HINT! Step-by-step e.g. exams EG - worked examples Refers you to the Activities with hint questions for you instructions exemplar paper to answer exams e.g. EG - worked examples activity • The activities are based on exam-type questions. activity Cover the answers ACTIVITIES boy and girl-left and provided and do each activity on your own. Then check your answers. ACTIVITIES boy right of page exams and girl-left and Reward yourself for things you get right. If you get any right incorrect of page activity answers, make sure you understand where activity you went wrong before activity ACTIVITIES boy moving on to the next section. and girl-left and right of page • In these introduction pages, we will go through the mathematics that you activity need to know, in particular, algebra activity and graphs. These are Step by step comment comment Step by step comment comment crucial skills that you willACTIVITIES need boy for any subject that makes use of mathematics. Make sureright ofyou pageunderstand these pages before you go and girl-left and any further. activity Step by step comment comment • Go to www.education.gov.za to download past exam papers for you to practice. Step by step comment comment Use this study guide as a workbook. Make notes, draw pictures and highlight important concepts. © Department of Basic Education 2015 introduct ion xi Mind the Gap Mathematics

© Department of Basic Education 2015 Top 10 study tips 1. Have all your materials ready before you begin studying – pencils, pens, highlighters, paper, etc. Try these study tips to make learning easier. 2. Be positive. Make sure your brain holds on to the information you are learning by reminding yourself how important it is to remember the work and get the marks. 3. Take a walk outside. A change of scenery will stimulate your learning. You’ll be surprised at how much more you take in after being outside in the fresh air. 4. Break up your learning sections into manageable parts. Trying to learn too much at one time will only result in a tired, unfocused and anxious brain. 5. Keep your study sessions short but effective and reward yourself with short, constructive breaks. 6. Teach your concepts to anyone who will listen. It might feel strange at first, but it is definitely worth reading your revision notes aloud. 7. Your brain learns well with colours and pictures. Try to use them whenever you can. 8. Be confident with the learning areas you know well and focus your brain energy on the sections that you find more difficult to take in. 9. Repetition is the key to retaining information you have to learn. Keep going – don’t give up! 10. Sleeping at least 8 hours every night, eating properly and drinking plenty of water are all important things you need to do for your brain. Studying for exams is like strenuous exercise, so you must be physically prepared. If you can’t explain it simply, you don’t understand it well enough. Albert Einstein xii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 Mnemonics A mnemonic code is a useful technique for learning information that is difficult to remember. Here’s the most useful mnemonic for Mathematics, Mathematical Literacy and Physical Science: BODMAS: Mnemonics encode information and make it easier to remember B – Brackets O – Of or Orders: powers, roots, etc. D – Division M– Multiplication A – Addition S – Subtraction Throughout the book you will be given other mnemonics to help you remember information. The more creative you are and the more you link your ‘codes’ to familiar things, the more helpful your mnemonics will be. Education helps one cease being intimidated by strange situations. Maya Angelou © Department of Basic Education 2015 introduction xiii Mind the Gap Mathematics

© Department of Basic Education 2015 Mind maps There are several mind maps included in the Mind the Gaps guides, summarising some of the sections. Mind maps work because they show information that we have to learn in the same way that our brains ‘see’ information. As you study the mind maps in the guide, add pictures to each of the branches to help you remember the content. You can make your own mind maps as you finish each section. How to make your own mind maps: 1. Turn your paper sideways so your brain has space to spread out in all directions. 2. Decide on a name for your mind map that summarises the information you are going to put on it. 3. Write the name in the middle and draw a circle, bubble or picture Mind around it. mapping 4. Write only key words on your branches, not whole sentences. Keep it your notes makes short and simple. them more interesting and easier to 5. Each branch should show a different idea. Use a different colour for remember. each idea. Connect the information that belongs together. This will help build your understanding of the learning areas. 6. Have fun adding pictures wherever you can. It does not matter if you can’t draw well. xiv introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 On the day of the exam 1. Make sure you have all the necessary stationery for your exam, i.e. pens, pencils, eraser, protractor, compass, calculator (with new batteries). Make sure you bring your ID document and examination admission letter. 2. Arrive on time, at least one hour before the start of the exam. 3. Go to the toilet before entering the exam room. You don’t want to waste valuable time going to the toilet during the exam. 4. Use the 10 minutes reading time to read the instructions carefully. This helps to ‘open’ the information in your brain. Start with the question you think is the easiest to get the flow going. 5. Break the questions down to make sure you understand what is being asked. If you don’t answer the question properly you won’t get any marks for it. Look for the key words in the question to know how to answer it. Lists of difficult words (vocabulary) is given a bit later on in this introduction. 6. Try all the questions. Each question has some easy marks in it so make sure that you do all the questions in the exam. 7. Never panic, even if the question seems difficult at first. It will be linked with something you have covered. Find the connection. 8. Manage your time properly. Don’t waste time on questions you are unsure of. Move on and come back if time allows. Do the questions that you know the answers for, first. GOOD LUCK! 9. Write big and bold and clearly. You will get more marks if the marker can read your answer clearly. 10. Check weighting – how many marks have been allocated for your answer? Take note of the ticks in this study guide as examples of marks allocated. Do not give more or less information than is required. © Department of Basic Education 2015 introduct ion xv Mind the Gap Mathematics

© Department of Basic Education 2015 Question words to help you answer questions It is important to look for the question words (the words that tell you what to do) to correctly understand what the examiner is asking. Use the words in the table below as a guide when answering questions. Question word/phrase What is required of you Analyse Separate, examine and interpret Calculate This means a numerical answer is required – in general, you should show your working, especially where two or more steps are involved Classify Group things based on common characteristics Compare Point out or show both similarities and differences between things, concepts or phenomena Define Give a clear meaning Describe State in words (using diagrams where appropriate) the main points of a structure/process/ phenomenon/investigation Determine To calculate something, or to discover the answer by examining evidence Differentiate Use differences to qualify categories Discuss Consider all information and reach a conclusion Explain Make clear; interpret and spell out Identify Name the essential characteristics PAY SPECIAL ATTENTION Label Identify on a diagram or drawing List Write a list of items, with no additional detail Mention Refer to relevant points Name Give the name (proper noun) of something State Write down information without discussion Suggest Offer an explanation or a solution Tabulate Draw a table and indicate the answers as direct In every exam pairs question, put a CIRCLE around the question word and underline any other important key words. These words tell you exactly what is being asked. xvi introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 Vocabulary C The following vocabulary consists of all the category (n). Class or group of things. difficult words used in Mind the Gap Mathematics, complex (adj). Consisting of many different Mathematical Literacy, and Physical Science. We parts; not easy to understand suggest that you read over the list below a few (n). a group or system of things times and make sure that you understand each connected in a complicated way. term. Tick next to each term once you understand component (n). A part. it so you can see easily where the gaps are in your compose (v). To make up from parts. knowledge. composite (n). Something made up of parts; KEY (adj). made up of several parts. Abbreviation Meaning conjunction (n). When two or more things come together at the same point; (v) verb: doing-word or action word, in grammar, a part of speech such as “walk” that connects words, sentences, (n) noun: naming word, such as phrases or clauses, e.g.: “and” “person” consecutive (adj). One after another without any (adj) adjective: describing word such as gaps or breaks. “big” consider (v). think about. (adv) adverb: describing word for verbs, contrast (v). Show the difference between; such as “fast” (n). something that is very different (prep) preposition: a word describing a from what it is being compared position, such as “on”, “at” with. (sing) singular: one of conversely (adv). The opposite of. (pl) plural: more than one of (abbr) abbreviation D General terms data (pl), (n). Information given or found. datum (sing) Term Meaning deduce (v). To work something out by A reasoning. deduction (n). Conclusion or idea that abbreviate (v). Make shorter. someone has worked out. account for (v). Explain why. define (v). Give the meaning of a word or adjacent (adj). Next to something. words. analyse (v). Examine something in detail. definition (n). The meaning of a word or words. annotated (adj). Something that has comments or explanations, usually denote (v). To refer to or mean something. written, added to it. descending (adj). Going down. apply (v). Make a formal application; be determine (v). Work out, usually by experiment relevant to; work hard; place on. or calculation. approximate (v. & adj.). Come close to (v); discreet (adj). Careful, polite. roughly, almost, not perfectly discrete (adj). Single, separate, distinct, a accurate, close but not exact. The part. verb is pronounced “approxi-mayt” and the adjective is pronounced “approxi-mitt”. E ascending (adj). Going up. arbitrary (adj). Based on random choice; establish (v). Show or prove, set up or create. unrestrained and autocratic. exceed (v). Go beyond. excess (n). More than necessary. © Department of Basic Education 2015 introduction xvii Mind the Gap Mathematics

© Department of Basic Education 2015 excluding (prep). Not including. inter- (adj). Can be swapped or exclusive (adj). Excluding or not admitting changeable exchanged for each other. other things; reserved for one investigate (v). Carry out research or a study. particular group or person. exemplar (n). A good or typical example. M exempt (v). To free from a duty. exempt (adj). Be freed from a duty. magnitude (adj). Size. exemption (n). Being freed from an obligation. manipulate (v). Handle or control (a thing or a exhibit (v). To show or display. person). exhibit (n). A part of an exhibition. motivate (v). Give someone a reason for extent (n). The area covered by something. doing something. Limit. multiple (adj). Many. F N factor (n). A circumstance, fact or negligible (adj). Small and insignificant; can influence that contributes to be ignored. From “neglect” (ignore). a result; a component or part. numerical (adj). Relating to or expressed as a A number that is divisible into number or numbers. another number without a numerous (adj). Many. remainder. factory (n). A place where goods are made or put together from parts. O find (v). Discover or locate. find (n). Results of a search or discovery. obtain (v). Get. finding (n). Information discovered as the optimal (adj). Best; most favourable. result of an inquiry. optimum (adj). Best; (n) the most favourable format (n). Layout or pattern; the way situation for growth or success. something is laid out. P H provide (v). Make available for use; supply. horizontal (adj). Across, from left to right or right to left. (From “horizon”, the R line dividing the earth and the sky). hypothesis (n). A theory or proposed explanation. reciprocal (adj). Given or done in return. hypothetical (adj). Theoretical or tentative; record (v). Make a note of something in waiting for further evidence. order to refer to it later (pronounced ree-cord). record (n). A note made in order to refer I to it later; evidence of something; a copy of something (pronounced identify (v). Recognise or point out. rec-cord. illustrate (v). Give an example to show what relative (adj). Considered in relation to is meant; draw. something else; compared to. imply (v). Suggest without directly saying relative (n). A family member. what is meant. represent (v). Be appointed to act or speak for indicate (v). Point out or show. someone; amount to. initial (n). First. resolve (v). Finalise something or make it clear; bring something to a insufficient (adj). Not enough. conclusion. xviii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 respect (v). Admire something or someone; angel (n). In Abrahamic religions, a consider the needs or feelings of messenger from God. Note the another person. spelling. respectively (adj). In regards to each other, in angle (n). The difference in position relation to items listed in the same between two straight lines which order. meet at a point, measured in degrees. Note the spelling. S annual (adj). Once every year. (E.g. “Christmas is an annual holiday”). annum, per (adv). For the entire year. (E.g. “You simultane- (adv). At the same time. should pay R 100 per annum”). ously annuity (n). A fixed sum of money paid suffice (v). Be enough. to someone each year, typically surplus (adj). More than is needed. for the rest of their life, as an survey (n). A general view, examination, insurance policy. See policy. or description of someone or apex (n). The tip of a triangle or two lines something. meeting. survey (v). Look closely at or examine; approach (v). To approximate or come close consider a wide range of opinions to in value. or options. area (n). Length x breadth (width). In common usage: a place. T asymptote (n). A line that continually approaches a given curve but does not meet it at any finite distance. tendency (n). An inclination to do something in a particular way; a habit. average (n). Mathematics: The sum of parts divided by the quantity of parts. In transverse (adj). Extending across something. common use: neither very good, strong, etc., but also neither very V weak, bad, etc; the middle. If you are asked to find the average, you always have to calculate it using verify (v). Show to be true; check for truth; the information you have. For confirm. example, the average of (1;2;3) is vice versa (adv). The other way round. 2, because (1+2+3)/3 = 2. See versus (prep). Against. Abbreviated “vs” also mean, median and mode. and sometimes “v”. axiom A basic truth of mathematics. vertical (adj). Upright; straight up; standing. axis (sing), (n). A line along which points axes (pl, can be plotted (placed), showing Technical terms pronounced “akseez”) how far they are from a central point, called the origin. See origin. “Vertical axis” or “y-axis” refers A to how high up a point is above the origin (or how far below). abscissa (n). The distance from a point to “Horizontal axis” or “x-axis” refers the vertical or y-axis, measured to how far left or right a point is parallel to the horizontal or x-axis; away from the origin. the x-coordinate. See ordinate. acute (adj). Having an angle less than B 90˚. algebra (n). A mathematical system base (n). The horizontal lowest line where unknown quantities are on a diagram of a geometrical represented by letters, which shape, usually of a triangle. Or: can be used to perform complex a number used as the basis of a calculations through certain rules. numeration scale. Or: a number in altitude (n). Height. terms of which other numbers are expressed as logarithms. © Department of Basic Education 2015 introduction xix Mind the Gap Mathematics

© Department of Basic Education 2015 bias (n). To be inclined against consecutive (adj). Following on from one something or usually unfairly another. opposed to something; to not continuous (adj). Mathematics: having no accurately report on something; to breaks between mathematical favour something excessively. points; an unbroken graph or curve binomial (n). An algebraic expression of the represents a continuous function. sum or the difference of two terms. See function. bisect (v). To cut into two. control (n. and v.). To ensure something bivariate (adj). Depending on two variables. does not change without being allowed to do so (v); breadth (n). How wide something is. From an experimental situation to the word “broad”. which nothing is done, in order to compare to a separate C experimental situation, called the ‘experiment’, in which a change is attempted. The control is then calculus (n). A branch of mathematics compared to the experiment to see that deals with the finding and if a change happened. properties of derivatives and control (n). A variable that is held constant integrals of functions, by methods variable in order to discover the relationship originally based on the summation between two other variables. of infinitesimal (infinitely small) “Control variable” must not be differences. The two main types confused with “controlled variable” are differential calculus and (see independent variable). integral calculus. coordinate (n). The x or y location of a point on cancel (v). To remove a factor by dividing a Cartesian graph, given as an x or by the factor. y value. Coordinates (pl) are given chance (n). The same as possibility or as an ordered pair (x, y). likelihood; that something might correlate (v). To see or observe a happen but that it is hard to relationship between two things, predict whether it will. without showing that one causes chart (v). To draw a diagram comparing the other. values on Cartesian axes. correlation (n). That there is a relationship chord (n). A line cutting across a circle between two things, without or arc at a position other than the showing that one causes the other. diameter. Note spelling. correspond (v). To pair things off in a circum- (n). The distance around the outer correlational relationship. For two ference rim of a circle. things to agree or match. E.g. A coefficient (n). A constant value placed corresponds to 1, B corresponds to next to an algebraic symbol as a 2, C corresponds to 3, etc. multiplier. Same as constant (see cubed (adj). The power of three; below). Or: a multiplier or factor multiplied by itself three times. that measures a property, e.g. cubic (adj). Shaped like a cube; having coefficient of friction. been multiplied by itself three complement (n).Geometry: the amount in times. degrees by which a given angle is cut (n). A subdivision of a line or less than 90°. point where one line crosses over Mathematics: the members of a another. set or class that are not members of a given subset. Do not confuse cyclic (adj). Pertaining to a circle. with compliment (praise). cylinder (n). A tall shape with parallel sides composite (adj). Made of parts. and a circular cross-section – think of a log of wood, for example, or a compound (n). Interest charged on an amount tube. See parallel. The formula for interest due, but including interest charges the volume of a cylinder is πr2h. to date. Compare to simple interest. constant (n). See coefficient. Means “unchanging”. xx introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 D divisor (n). The number below the line in a fraction; the number that is dividing the other number above dependent (adj/n). A variable whose value the fraction line. See numerator, (variable) depends on another; the thing denominator. that comes out of an experiment, the effect; the results. See also domain (n). The possible range of x-values independent variable and control for a graph of a function. See variable. The dependent variable range. has values that depend on the independent variable, and we plot E it on the vertical axis. derivation (n). Mathematics: to show the working of your arithmetic or element (n). Mathematics: part of a set of answer or solution; the process of numbers. Popular use: part of. finding a derivative. eliminate (v). To remove from an equation. derivative (n). Mathematics: The rate of See cancel. change of a function with respect equiangular (adj). Having the same angle. to an independent variable. See equidistant (adj). Having the same distance or independent variable. In common length. use: something that comes from equilateral (adj). Having sides of the same length. something else. estimate (n., v.). To give an approximate determine(s) (v). To cause; to ensure that; to set value close to an actual value; an (causation) up so that; to find out the cause of. imprecise calculation. deviation (n). A variation from a statistical Euclidean (adj). Pertaining to geometry of norm; not as far out as an outlier. straight lines on flat planes. An amount by which a single measurement differs from a fixed even (adj). Divisible by two without a value such as the mean. A significant remainder. deviation from the average value. exponent (n). When a number is raised to di- (prefix). Two. a power, i.e. multiplied by itself as many times as shown in the diagonal (adj. & n.). A line joining two power (the small number up opposite corners of an angular above the base number). So, 23 shape. means 2 x 2 x 2. See also cubed. diameter (n). The line passing through the exponential (adj). To multiply something many centre of a shape from one side times; a curve representing an of the shape to the other, esp. a exponent. circle. Formula: d = 2r. See radius, radii, circumference. expression (n). A formula or equation. difference (n). Mathematics: subtraction. extrapolation (n). To extend the line of a graph Informally: a dissimilarity. How further, into values not empirically things are not the same. documented, to project a future event or result. In plain language: digit (n). A number represented in to say what is going to happen writing. based on past results which were dimension (n). The measurable size or extent obtained (gotten) by experiment of usually a geometric shape and and measurement. If you have often on a Cartesian Coordinate a graph and have documented system, e.g. the x-dimension certain results (e.g. change vs (breadth). time), and you draw the line discriminant (n). A function of the coefficients of further in the same curve, to say a polynomial equation whose value what future results you will get, gives information about the roots that is called ‘extrapolation’. See of the polynomial. predict. Mathematics: to project distribution (n). How something is spread another iteration, value or solution, out. Mathematics: the range and based on a formula that covers or variety of numbers as shown on a formulates a previous solution. graph. © Department of Basic Education 2015 introduction xxi Mind the Gap Mathematics

© Department of Basic Education 2015 F homologous (n). Belonging to the same group of things; analogous. factorial (n). The product of an integer hyperbola (n). Mathematics: a graph of a and all the integers below it; e.g. section of a cone with ends going factorial four (4!) is equal to 24. off the graph; a symmetrical (both sides the same) open curve. factorise (v). To break up into factors. hypotenuse (n). The longest side of a right- formula (n). See expression. angled triangle. fraction (n). Mathematics: Not a whole number; a representation of a division. A part. E.g. the third I 2 fraction of two is 0,666 or ​ __ 3​ meaning two divided into three illuminate (v). To explain or light up. parts. incline (n. & v.). Slope. See gradient (n); to frequency (n). How often. Usually represented lean (v). 12 1 as a fraction, e.g. ___ ​ 48 ​ = ​ __ 4 ​or 0,25. independent (n). The things that act as input function (n). Mathematics: when two (variable) to the experiment, the potential attributes or quantities correlate. causes. Also called the controlled If y changes as x changes, then variable. The independent variable y = f(x). See correlate, graph, is not changed by other factors, Cartesian, axis, coordinate. Also: and we plot it on the horizontal a relation with more than one axis. See control, dependent variable (mathematics). variable. inequality (n). A relation between two G expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than’, or < geometric (adj). Progressing or growing in a ‘less than’. regular ratio. inflation (n). That prices increase over time; geometry (n). The mathematics of shape. that the value of money decreases gradient (n). A slope. An increase or over time. General use: the action decrease in a property or of getting bigger. measurement. Also the rate of such insufficient (adj). Not enough. a change. In the formula for a line integer (n.) a whole number not a fraction, graph, y=mx+c, m is the gradient. can be negative. gradually (adv). To change or move slowly. intercept (n.) Where a line cuts an axis on a graph (n). A diagram representing graph. See cut. experimental or mathematical interest (n). Finance: money paid regularly values or results. Cartesian at a particular rate for the use or Coordinates. loan of money. It can be paid by graphic (n., adj.). A diagram or graph a finance organisation or bank (n). Popular use: vivid or clear or to you (in the case of savings), remarkable (adj.). or it may be payable by you to a graphically (adv). Using a diagram or graph. finance organisation on money you Popular use: to explain very clearly. borrowed from the organisation. See compound interest and simple interest, see also borrow. H interquartile (adj). Between quartiles. See quartile. histogram (n). A bar graph that represents intersection (n). Where two groups overlap on a continuous (unbroken) data (i.e. Venn Diagram. data with no gaps). There are no spaces between the bars. A interval (n). Gap. A difference between two histogram shows the frequency, or measurements. the number of times, something inverse (n). The opposite of. Mathematics: 1 happens within a specific interval or one divided by. E.g. ​ __ 2 ​is the “group” or “batch” of information. inverse of 2. xxii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 imaginary (n). I; a number which is a multiple minimise (v). To make as small as possible. numbers of the square root of (–1). The minimum (n). The smallest amount possible. opposite of real numbers. Not examinable/advanced. modal (adj). Pertaining to the mode, or method. Can mean: about the irrational (n). Fractions which recur, or which mathematical mode or about the numbers cannot be expressed as a ratio of method used. See mode. whole numbers. Decimals. mode (n). the most common number in a isosceles (n., adj.). A triangle with two sides series of numbers. See also mean, (triangle) of equal length. median. model (n). A general or simplified way L to describe an ideal situation, in science, a mathematical description that covers all cases of law (n). A formula or statement, the type of thing being observed. A deduced (discovered) from prior representation. axioms (truths), used to predict a result. mutually (adj). In respect to each other, affecting each other. likely (adj). To be probable; something that might well happen. linear (adj). In a line. Mathematics: in a N direct relationship, which, when graphed with Cartesian coordinates, natural (n). Any number which is not a turns out to be a straight line. numbers fraction and greater than –1 logarithm (n). A quantity representing the (includes zero). Positive whole power by which a fixed number (the numbers. base) must be raised to produce negative (adj). Below zero. a given number. The base of a common logarithm is 10, and normal (n., adj.). Mathematics and that of a natural logarithm is the Physics: a force, vector or line that number e (2,7183…). A log graph acts at right angles to another can turn a geometric or exponential force, vector or line or object. (n). relationship, which is normally Common use: Regular or standard curved, into a straight line. (adj). numerator (n). The opposite of a denominator; the number on top in a fraction. M O magnitude (n). Size. manipulate (v). To change, or rearrange something. Usually in Mathematics obtuse (adj). Having an angle greater than it means to rearrange a formula to 90˚ but less than 180˚. solve for (to get) an answer. odd (adj). Not divisible by two without a mean (n). See average. remainder. median (n). Mathematics: the number in ogive (adj). A pointed arch shape; a the middle of a range of numbers cumulative frequency graph. written out in a line or sequence. optimal (adj). Best, most. metric (adj). A measurement system, ordinate (n). A straight line from any point using a base of 10 (i.e. all the drawn parallel to one coordinate units are divisible by 10). The USA axis and meeting the other, uses something known as the especially a coordinate measured Imperial system, which is not used parallel to the vertical. See in science. The Imperial system is abscissa. based on 12. Examples: 2,54 cm origin (n). Mathematics: the centre of (metric) = 1 inch (imperial). 1 foot a Cartesian coordinate system. = 12 inches = approx. 30 cm; General use: the source of 1 metre = 100 cm. 1 Fl.Oz (fluid anything, where it comes from. ounce) = approx 30 mℓ. © Department of Basic Education 2015 introduction xxiii Mind the Gap Mathematics

© Department of Basic Education 2015 outlier (n). Statistics: a data point which pi (n). π, the Greek letter p, the lies well outside the range of ratio of the circumference of a related or nearby data points. circle to its diameter. A constant without units, value approximately 3,14159. P plan (n). Architecture: a diagram representing the layout and parallel (adj). Keeping an equal distance structure of a building, specifically along a length to another item as viewed from above. More (line, object, figure). Mathematics: general use: any design or two lines running alongside each diagram, or any intended sequence other which always keep an equal of actions, intended to achieve a distance between them. goal. parallelogram (n). Any four-sided figure with two plane (n). A flat surface. sides parallel. Abbr.: parm. plot (v). To place points on a Cartesian parameter (n). A value or algebraic symbol in coordinate system; to draw a a formula. Statistics: a numerical graph. characteristic of a population, poly- (prefix). Many. as distinct from a statistic of a sample. polygon (n). Any shape with many (at least three) equal sides and angles. A quantity whose value is selected for the particular circumstances polyhedron (n). A three-dimensional shape and in relation to which other with many usually identical flat variable quantities may be sides. expressed. polynomial (n). An expression of more than particular (adj). A specific thing being pointed two algebraic terms, especially the out or discussed; to single out or sum of several terms that contain point out a member of a group. different powers of the same variable(s). pent- (prefix). Five. population (n). Statistics: the larger body from pentagon (n). A five sided figure with all sides which the statistical sample is equal in length. taken. per (prep). For every, in accordance positive (adj) Above zero. with. predict (v). General use: to foresee. per annum (adv). Once per year; for each year. Physical Science: to state what will percent (adv). For every part in 100. The happen, based on a law. See law. rate per hundred. prime number (n). Any number divisible only by percentile (n). A division of percentages one and itself. into subsections, e.g. if the scale probability (n). How likely something is. See is divided into four, the fourth likely. Probability is generally a percentile is anything between 75 mathematical measure given and 100%. as a decimal, e.g. [0] means perimeter (n). The length of the outer edge; unlikely, but [1,0] means certain, the outer edge of a shape. and [0,5] means just as likely period (n). The time gap between events; versus unlikely. [0,3] is unlikely, a section of time. and [0,7] is quite likely. The most common way to express periodic (adj). Regular; happening regularly. probability is as a frequency, or permutation (n). The action of changing the how often something comes up. 1 arrangement, especially the linear E.g. an Ace is ___ ​ 13 ​or 0,077 likely, order, of a set of items. in a deck of cards, because perpendicular (adj). Normal; at right angles to there are 4 of them in a set of (90˚). 52 cards. xxiv introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 product (n). Mathematics: the result of quartile (n). A quarter of a body of data multiplying two numbers. represented as a percentage. This is the division of data into project (n. & v.). A project (n., pronounced 4 equal parts of 25% each. To PRODJ-ekt) is a plan of action determine the quartiles, first divide or long-term activity intended the information into two equal to produce something or parts to determine the median reach a goal. To project (v., (Q2), then divide the first half into pronounced prodj-EKT), is to two equal parts, the median of throw something, or to guess or the first half is the lower quartile predict (a projection). To project (Q1), then divide the second half a result means to predict a into two equal parts, and the result. See extrapolate. median of the second half is the proportion (n). To relate to something else upper quartile (Q3). Data can be in a regular way, to be a part of summarised using five values, something in relation to its volume, called the five number summary, size, etc; to change as something i.e. the minimum value, lower else changes. See correlate and quartile, median, upper quartile, respectively. and maximum value. pyramid (n). A polyhedron of which one quotient (n). A ratio. face is a polygon of any number of sides, and the other faces are triangles with a common vertex. R Pythagoras’s (n). The square on the hypotenuse Theorem is equal to the sum of the squares radical (n). A root. on the other two sides of a right- radius (sing), (n). The distance between the angled triangle. Where h is the radii (plur) centre of an object, usually a hypotenuse, a is the side adjacent circle, and its circumference or to the right angle, and b is the outer edge. Plural is pronounced other side: h2 = a2 + b2. “ray-dee-eye.” random (n). Unpredictable, having no cause Q or no known cause. Done without planning. quadrilateral (n). A shape with four sides. range (n). The set of values that can be supplied to a function. The set of qualitative (adj). Relating to the quality possible y-values in a graph. See or properties of something. domain. A qualitative analysis looks at changes in properties like rate (n). How often per second (or per colour, that can’t be put into any other time period). Physics: numbers. Often contrasted with number of events per second; see quantitative. frequency. Finance: the exchange rate or value of one currency when quantitative (adj). Relating to, or by comparison exchanged for another currency; to, quantities. Often contrasted how many units of one currency with qualitative. A quantitative it takes to buy a unit of another analysis is one in which you currency. Also “interest rate”, or compare numbers, values and what percentage of a loan consists measurements. of interest charges or fees. quantity (n). Amount; how much. ratio (n). A fraction; how one number relates to another number; exact proportion. If there are five women for every four men, the ratio of women to men is 5:4, written with a colon (:). This ratio can be 5 1 represented as the fraction __ ​ 4 ​ or 1​ __ 4​ or 1,25; or we can say that there are 25% more women than men. © Department of Basic Education 2015 introduct ion xxv Mind the Gap Mathematics

© Department of Basic Education 2015 rational (n). A fraction which can be sphere (n). A perfectly round three- numbers expressed as a ratio of whole dimensional shape. A ball. numbers. See irrational numbers. square (n). Mathematics: a shape or figure ray (n). A line from amongst a set of with four equal sides and only right lines passing through the same angles; the exponent 2 (e.g. the central point. See radius. square of 4 is 42 = 16). reciprocal (n). A complement of a number squared (adj). Having been multiplied by which when added to the number itself, put to the exponent 2. See yields 1.0. square. rectangle (n). A parallelogram with only right statistics (n). The mathematics of chance angles (90˚). and probability. real number (n). Any non-imaginary number, steep (adj). Having a large gradient. that is, a number not a multiple subscript (n). A number placed below the of the square root of (-1). Includes rest of the line, e.g. CO2. rational and irrational numbers, substitute (v). To replace. integers. substitution (n). The process of substituting. remainder (n). Leftovers. Mathematics: an Mathematics: to replace an amount left over after division algebraic symbol in a formula with which cannot be divided further a known value or another formula, unless one wishes to have a so as to simplify the calculation. decimal or fraction as a result, i.e. See simplify. where the divisor does not exactly divide the numerator by an integer subtotal (n). Finance: the total amount due (whole number). on a statement or invoice, usually without VAT (tax) charges given. Or: rhombus (n). A quadrilateral (four sided) a total for a section of an invoice figure (diagram or shape) which or statement or series of accounts, has equal sides, but no right- but not the total of the whole angles (90˚ angles). invoice, statement or account. right angle (n). An angle of 90˚. successive (adj). One after the other. round (v). To approximate, especially an sum (n., v.). To add things up. irrational number, to a shorter Represented by Greek Sigma (s): series of decimals. ∑ or the plus sign (+). superscript (n). A number placed above the S rest of the line, e.g. πr2. surd (n). An irrational root (e.g. √2). scale (n). A system of measurement, with regular intervals or gaps between T units (subdivisions) of the scale. scalene (adj). A triangle with unequal sides. tally (n). A total count; to count in fives simple (n). Interest charged on the original by drawing four vertical lines then interest amount due only, resulting in the crossing through them with the same fee every time. fifth line. simplify (v). To make simpler. Mathematics: tangent (n). A straight line touching a curve to divide throughout by a common only at one point, indicating the factor (number or algebraic letter) slope of the curve at that point; the that will make the equation easier trigonometric function of the ratio to read and calculate. of the opposite side of a triangle to solution (n). Mathematics: the step-by-step the adjacent side of a triangle in a displaying of calculations to arrive right-angled triangle; a curve that at answers. Common use: the goes off the chart. answer to a problem, in the sense tax (n). A compulsory levy imposed on of dissolving (removing) a problem. citizens’ earnings or purchases to solve (v). To come up with a solution fund the activities of government. (answer). Show your working. xxvi introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 taxable (adj). A service, purchase or item V or earning that has a tax applied to it. variable (n., adj.). A letter used to represent tetra- (prefix). Four. Same as quad. an unknown quantity in algebra theory (n). A usually mathematical (n); a quantity that changes (n); representation of an explanation subject to change (adj). for something in the sciences, Venn Diagram (n). A diagram representing sets which does not depend on the (classes of objects) as circles. thing being explained. vertex (n). The angular point(s) of a theorem (n). A general proposition not polygon; apex. Plural: vertices. self-evident but proved by a chain of reasoning; a truth established volume (n). A measure of the space by means of accepted truths. occupied by an object, equal to Compare to theory. length x breadth x height. trapezium (n). A quadrilateral with one pair of sides parallel (and the other sides W usually having complementary angles). whole number (n). Any number not a fraction or treble (adj). Triple. decimal, greater than zero. Natural trends (n). Regular patterns within data. numbers and zero. tri- (prefix). Three. trigonometry (n). The relationship and ratios Y between sides and angles within a right-angled triangle. yard (n). Old Imperial measurement of length, approximately equal to a U metre (1,09 m). yield (n., v.). An answer or solution. unit (n). A subdivision of a scale. See scale. union (adj). When two sets in a Venn Diagram overlap into one set. © Department of Basic Education 2015 introduct ion xxvii Mind the Gap Mathematics

© Department of Basic Education 2015 The maths you need This section gives you the basic mathematical skills that you need to pass any subject that makes use of mathematics. Do not go any further in this book until you have mastered this section. 1. Basic Pointers • If a formula does not have a multiplication (×) sign or a dot-product (·), and yet two symbols are next to each other, it means “times”. So, m1m2 means mass 1 times mass 2. You can also write it as m1 × m2, or m1·m2 • Comma means the same as decimal point on your calculator (i.e. 4,5 = 4.5). Do not confuse the decimal point with dot product (multiply): 4.5 = 4½ but 4·5 = 20. Rather avoid using the dot product for this reason. • A variable is something that varies (means: changes). So, for example, the weather is a variable in deciding whether to go to the shops. Variables in science and mathematics are represented with letters, sometimes called algebraic variables. The most common you see in maths is x, probably followed by y, z. 2. Subject of Formula or Solving For Very often in mathematics you have to “make something the subject of a formula” or “solve for something”. This refers to finding the value of an unknown quantity if you have been given other quantities and a formula that shows the relationship between them. The word ‘formula’ means a rule for working something out. We work with formulas to draw graphs and also to calculate values such as area, perimeter and volume. You are usually given the formulas in an exam question, so you don’t have to remember them, but you do need to select the right numbers to put into the formula (substitute). For example, the formula for the area of a triangle is 1 Area = ​ __ 2 ​ base × height. perpendicular height base In this formula: • the word Area stands for the size of the area of a triangle (the whole surface that the triangle covers) • the word base stands for the length of the base of the triangle • the word height stands for the length of the perpendicular height of the triangle. A formula can be written in letters rather than words, for example: 1 A = ​ __ 2 ​ b × h. The quantity on its own on the left is called the subject of the formula. xxviii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 e.g. 1 EG - worked examples If John has 5 apples, and he gives some to Joanna, and he has two apples exams left, how many did he give to Joanna? Well, the formula would be something like this: 5–x=2 To solve for x, we simply have to swap the x and the 2. What we’re actually doing is adding “x” to both sides: 5–x+x=2+x this becomes: 5 = 2 + x thenNB activity we subtract PAY SPECIAL ATTENTION 2 from both sides to move theNB 2 over: 5 – 2 = 2 – 2 + x ACTIVITIES boy and girl-left and 5–2=x right of page 3 = x ... so John gave Joanna three apples. activity The same procedures apply no matter how complex the formula looks. Just HINT! hint either add, subtract, square, square root, multiply or divide throughout to move the items around. Step by step comment comment e.g. 2 EG - worked examples Let’s take an example from Physical Science: V = IR. This means, the exams voltage in a circuit is equal to the current in the circuit times the resistance. Suppose we know the voltage is 12 V, and the resistance is 3 Ω. What is the current? V = IR 12 = 3 × PAYISPECIAL ATTENTION NB divide throughout by 3 so as to isolate the I NB activity 12 ​ __ (12 ) __ 3 ​= ​​  3 ​  ​I ACTIVITIES boy remember that anything divided by itself is 1, so: and girl-left and 12 12 ​ __ 3 ​= (1) × I ... and __ HINT! ​  3 ​right = 4of...page so hint activity 4 = I or I = 4 A ... The circuit has a current of 4 amperes. e.g. 3 EG - worked examples Step by step comment comment Here’s a more tricky example from Physical Science. Given exams Kc = 4,5 [SO3] = 1,5 mol/dm3 [SO2] = 0,5 mol/dm3 (x – 48) [O2] = ​ ______ 64 ​ mol/dm 3 solve for x. [SO3]2 (1,5)2 Kc = ________ activity ​  [SO ]2[O ] ​ ∴ 4,5 = __________ ​  (x – 48) ​ 2 2 (0,5)2 ______ ​  64 ​ ACTIVITIES boy ∴ x = 176 g and girl-left and right of page How did we get that answer? activity © Department of Basic Education 2015 introduction xxix Mind the Gap Mathematics

© Department of Basic Education 2015 Step by Step Let’s see how it works. First, solve for the exponents (powers): 2,25 4,5 = __________ ​  (x – 48) ​ (0,25) ______ ​  64 ​ Now, we can see that 2,25 and 0,25 are similar numbers (multiples of five), so let’s divide them as shown. 2,25 x – 48 4,5 = ____ ​  0,25 ​× _____ ​  64 ​ That leaves us with (x – 48) 4,5 = 9 × ______ ​  64 ​ But if we’re dividing a divisor, that second divisor can come up to the top row. Here’s a simple example: 1 1 ÷ (2 ÷ 3) = __ ​  2__ ​ ​  3 ​ 1×3 = ____ ​  2 ​ 3 = __ ​  2 ​= 1,5 If you doubt this, try it quickly on your calculator: 1 ÷ (2 ÷ 3) ... this means, one, divided by two-thirds. Well, two-thirds is 0,6667, which is almost one. So how many “two-thirds” do you need to really make up one? The answer is one and a half “two-thirds”... i.e. 0,6667 + (0,6667 ÷ 2) = 1. Hence the answer is 1,5. So, back to the original problem, we can bring the 64 up to the top line and multiply it by nine: 4,5 = 9 × ​_____(x – 48 ) ​  64 ​  ​ 9 × 64 4,5 = _____ ​  x – 48 ​ 576 4,5 = _____ ​  x – 48 ​ Now we can inverse the entire equation to get the x onto the top: ___ 1 x – 48 ​  4,5 ​= _____ ​  576 ​ Now we multiply both sides by 576 to remove the 576 from the bottom row (x – 48) 576 ___ ​ = _________ ​  576 ​  576 ​ 4,5 and we cancel the 576’s on the right hand side as shown above. Now, if 576 ÷ 4,5 = 128, then 128 = x – 48 Now we add 48 to both sides to move the 48 across 128 + 48 = x – 48 + 48 ... hence, 128 + 48 = x = 176. xxx introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 e.g. 4 EG - worked examples A triangle has a base of 6 cm and a perpendicular height of 2 cm. exams its area. Determine Step 1: Write down the value that you need to find. Need to find: Area Step 2: Write down the information that you have. base = 6 cm Write down the numbers and the units. height = 2 cm 1 Step 3: Write down the formula that you are going Area = ​ __2 ​base × height to use. activity 1 Area = __ PAY SPECIAL ATTENTION Step 4: Write down the formula again, butboywrite ​  2 ​× 6 cm × 2 cm NB the numbers that you know instead ACTIVITIES andof girl-left theand words NB right of page or letters. NB activity We PAY SPECIAL ATTENTION call this process NB substituting. Step 5: Now calculate. HINT! = 3 cm × 2 cm hint HINT! hint Step 6: Write your answer with the correct units. = 6 cm2 e.g. EG - worked examples comment Step by step comment e.g. 5 exams EG - worked examples 1,2 m 163 cm activity Calculate the area and the perimeter of the triangle alongside. ACTIVITIES boy exams and girl-left and right of page This looks like an easy problem, but you need to stay on your toes. As you activity follow the steps you will see why. 110 cm comment Step 1: Write down what you need to Need to find: Area and Perimeter. Step by step comment find. Let’s start with area. Step 2: Write down the information From the diagram: that you have. activity base = 110 cm height = 1,2 m ACTIVITIES boy The sides of the triangle are at right and girl-left and angles to each other, so one side is right of page the perpendicular height. activity NB PAY SPECIAL ATTENTION NB The units of the two lengths are not the same. Always HINT! write the values with the same units. hint height = 1,2 m = 120 cm Step by step e.g. comment (because 100 comment EG - worked examples cm = 1 m) Step 3: Write down the formula. Area = base = ​ _1 ​ × height exams 2 __ 1 Step 4: Write down the formula Area = ​  2 ​× 110 cm × 120 cm again, but write the numbers that you know instead of the words. activity ACTIVITIES boy Step 5: Now calculate. and girl-left and right of page = 55 cm × 120 cm activity Step 6: Write your answer with the = 6 600 cm2 correct units. Step by step comment comment Step 7: Calculate the perimeter. Perimeter = 120 + 163 + 110 = 393 cm © Department of Basic Education 2015 introduction xxxi Mind the Gap Mathematics

HINT! © Department of Basic Education 2015 hint e.g. 6 EG - worked examples In the United States, people use degrees Fahrenheit to measure temperature. exams67 °F into degrees Celsius (°C). Round off your answer to two Convert decimal places. The formula to use is °C = (°F – 32°) ÷ 1,8. Need to find: Temperature in degree Notes Celsius. activity Information we have: Temperature in degrees Fahrenheit = 67 °F. ACTIVITIES boy °C = (67° – 32°) ÷ 1,8 Replace °F with 67° in the formula: and girl-left and right of page °C = (°F – 32°) ÷ 1,8. activity °C = 35° ÷ 1,8 Remember the order of operations: PAY SPECIAL ATTENTION Calculate the brackets first and then NB NB do the division. °C = 19,444 …° Round off to two decimal places. Temperature in Step degrees by stepCelsius comment Look at comment the number in the third = 19,44 °C HINT! decimal place. It is less than 5, so round the second decimal place hint down. e.g. 7 EG - worked examples A circular piece of land has a diameter of 40 m. What is the area of the land? d = 40m exams Use the formula: A = πr2 for the area of a circle and use the value of 3,142 for π. Need to find: Area Notes Information we have: diameter = 40 m; π = 3,142 PAY SPECIAL ATTENTION NB NB Butwe need the radius, which is half of the diameter, activity so r = 20 m. Always make sure you use the quantity that is HINT! ACTIVITIES boy hint and girl-left and written in the formula – right of page radius, not diameter. e.g. EG - worked examples A = πr activity 2 exams A = 3,142 × (20)2 A = πr2 means Area = pi times the radius squared. A = 3,142 × (20 × 20) A = 3,142 × 400 activity comment ACTIVITIES boy A = 1 256,8 m 2 Step by step Are the units right? Yes, the diameter was givenrightin and girl-left and comment of page metres, so the area will be in square activity metres (m2 ). Step by step comment comment xxxii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 e.g. 8 EG - worked examples When we work with a formula, we want the quantity that we are calculating on its examsown on one side of the formula, so that it is the subject of the formula. We can work out area easily if the formula is Area = length × breadth. Now let’s use the same formula to find the length. 1. Look at the formula. Which is Area = length × breadth the quantity that you want to calculate? 2 activity What do you need to do to get length on its own? ACTIVITIES Length is multiplied by breadth. We boyneed to divide by breadth to leave and girl-left and length on its own. right of page You can only do something to a formula if you do the same activity PAY SPECIAL ATTENTION NB NB to both sides! HINT! 3. Divide both sides Area ÷ breadth = length × breadth ÷ breadth hint by breadth: e.g. 4. EG - worked examples Now simplify the formula: PAY SPECIAL ATTENTION = length NBarea ÷ breadth Step by step comment NB ÷ breadth = 1) comment (because: breadth exams 5. Length = area ÷ breadth 6. Use the formula to solve activity the problem by substituting HINT! the values for area and hint ACTIVITIES boy and girl-left and breadth. right of page activity Step by step comment comment e.g. 9 EG - worked examples To calculate the profit made from selling an item, we use the formula: exams profit = selling price – cost price. But what if we already know the profit and the cost price, but we need to calculate the selling price? An example: It costs R121 to buy a necklace at cost price, and Thabo wants to make R65 profit. How much must he sell it for? (What is the selling price?) activity Selling price 1. Look at the formula. WhichACTIVITIES is profit = selling price – cost price boy and girl-left and the quantity that you want to right of pageP = SP – CP calculate? 2. activity Substitute the values thatyou R65 = SP – R121 have i.e. profit and cost price. 3. Add cost price to both sides. R65 + R121 = SP – R121+ R121 4. Now simplify. Step by step comment R186 = SP comment (because cost price – cost price = 0) © Department of Basic Education 2015 introduct ion xxxiii Mind the Gap Mathematics

HINT! © Department of Basic Education 2015 hint e.g. 10 EG - worked examples This example has a fraction in it. See what you need to do in that case to make a quantity exams the subject of the formula. 5 miles is approximately the same as 8 kilometres. The formula to convert kilometres to miles is: 5 number of miles = __ ​  8 ​× number of kilometres. Gavin has cycled 30 miles and he wants to know what this is in kilometres. The formula must start with “number of kilometres = … ” activity Rearrange the formula. Then work out how many kilometres he has cycled. ACTIVITIES boy and girl-left and 5 1. Look at the formula. number of miles = ​ __8 ​× number of kilometres right of page Which is the quantity thatactivity you want to calculate? 5 8 2. Number of kilometres is multiplied by __ ​  8 ​. So we need to multiply by __ ​  5 ​ 5 __ __ 8 because ​  8 ​× ​  5 ​ = 1. 8 5 3. Multiply both sides Step by step number of miles × ​ __ comment comment __ 5 ​= ​  8 ​× number of 8 by ​ __5 ​. 8 __ kilometres × ​  5 ​ 8 5 8 4. Now simplify the number of miles × ​ __5 ​= __ ​  8 ​× __ ​  5 ​× number of formula: kilometres 8 Move the “× __ ​  5 ​”. 5 8 8 Cancel out: __ ​  8 ​× __ ​  5 ​= 1. number of miles × ​ __5 ​= number of kilometres 8 5. Now we have number of kilometres = number of miles × __ ​  5 ​ 8 6. Use the formula to number of kilometres = number of miles × ​ __5 ​ solve the problem. 8 number of kilometres = 30 × ​ __5 ​= 48 km PAY SPECIAL ATTENTION NB You can do this in your Gavin cycled 48NB km. head: 30 × 8 = 240 240 ÷ 5 = 48 Or use a calculator: HINT! 30 hint [×] 8 [÷] 5 [=] e.g. 11 EG - worked examples Thami needs to make a circle with an area of 40 cm2. What should the radius ofexams the circle be? Round off your answer to two decimal places. The formula for the area of a circle is A= πr 2. Use the value of 3,142 for π. activity ACTIVITIES boy and girl-left and xxxiv introduction  right of page © Department of Basic Education 2015  Mind the Gap Mathematics activity

© Department of Basic Education 2015 1. Look at the formula. Which is A = πr2 the quantity that you want to calculate? 2. What do you need to do to get radius on its own on one side of the equation? There are two things: • radius is first squared • then it is multiplied by pi (π) 3. Divide both sides by π Area ÷ π = πr2 ÷ π 4. When then have Area ÷ π = r2 area ____ Which we write as: ​  π ​= r2 ____ __ Now take the square root of both sides. √ Area ​ ____ ​  π ​ ​ = ​√r2 ​ ____ 5. Now we have r = ​ ____ √ Area ​  π ​ ​ ____ 6. Use the formula to solve the problem, by substituting the given √_____ Area r = ​ ____ ​  π ​ ​ values. √ 40 r = ​ _____ ​  3,142 ​ ​ = 3,568 ... To do this on your calculator: first enter 40 ÷ 3,142 = r = 3,57 cm __ Then press ​√0 ​ She needs to make the circle with a radius of 3,57 cm. Round off to two decimal places 3. Statistics You should at least know the following terminology: Dependent variable: The thing that comes out of an experiment, the effect; the results. Independent variable(s): The things that act as input to the experiment, the potential causes. Also called the controlled variable. Control variable: A variable that is held constant in order to discover the relationship between two other variables. “Control variable” must not be confused with “controlled variable”. Correlation does not mean causation. That is, if two variables seem to relate to each other (they seem to co-relate), it doesn’t mean that one causes the other. A variable only causes another variable if one of the variables is a function f(x) of the other. We will see more about this when we look at graphs, below. Mean: The average. In the series 1, 3, 5, 7, 9, the mean is 1 + 3 + 5 + 7 + 9 divided by 5, since there are 5 bits of data. The mean in this case is 5. Median: The datum (single bit of data) in the precise middle of a range of data. In the series 1, 3, 5, 7, 9, the median value is 5. Mode: The most common piece of data. In the series 1, 1, 2, 2, 3, 3, 3, 4, 5, the mode is 3. © Department of Basic Education 2015 introduct ion xxxv Mind the Gap Mathematics

© Department of Basic Education 2015 4. Triangles b The area of a triangle is half the base times the height: a = ​ __ 2 ​(h). A triangle with a base of 5 cm and a height of 3 cm will have an area of 2,5 × 3 = 7,5 cm2. A = 7,5 b Base 5 a hb c hb Height 3 PAY SPECIAL ATTENTION γ NB NB b Lengths of Triangle Sides You can calculate the lengths of sides of right-angled triangles using Pythagoras’ Theorem. HINT! The square of the hypotenuse is equal to the sum of hint the squares of the other two sides: In this diagram, b = base, hb = height, and c = the hypotenuse: c2 = hb2 + b2. e.g. 12 EG - worked examples In the triangle shown, the hypotenuse, marked “?”, can be obtained by examsboth sides, adding them, and then square-rooting them for the squaring ? 3cm length of the hypotenuse. That is: 32 + 52 = 9 + 25 = 34. Since in this case 34 = hyp2 it follows that the square root of 34 gives the value of “?”, the hypotenuse. That is, 5,83cm. 5cm 5. Trigonometry activity You can use trigonometry to calculate triangle sides’ sizes if you do not have enough information, e.g. you do boy ACTIVITIES not have the sizes of at least two and girl-left and sides (but do have the angle). right of page sin = opposite / hypotenuse sin = O/H cos activity = adjacent / hypotenuse cos = A/H tan = opposite / adjacent tan = O/A It’s probably easiest to remember this as SOHCAHTOA (soak a toe or soccer toe). Hypotenuse is the longest side next Step by step to o the comment comment angle, usually represented with a theta (θ). “Opposite” means the side of the triangle directly opposite the angle. “Adjacent” means the side adjacent to (next to) the angle, that is not the hypotenuse. xxxvi introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 e.g. 13 EG - worked examples In this triangle, the side opposite the angle θ is 3cm long. The side adjacent to hyp theexams angle θ, and the hypotenuse, are unknown. Theta, the angle, is 30 degrees. oten opposite use 3cm How do we calculate the hypotenuse? Well, O sin θ = __ ​  H ​= 3 cm ÷ H. θ = 30° adjacent sin 30° = 0,5 (you can get this from your calculator, or memorise it). thus 3 0,5 = __ ​  H ​ activity solving for H, we multiply throughout by H to make H the subject of ACTIVITIES boy the formula: and girl-left and right of page H × 0,5 = 3 × H ÷ H H × 0,5 = 3 activity now divide throughout by 0,5 to isolate the H: H × 0,5 ÷ 0,5 = 3 ÷ 0,5 H = 3 ÷ 0,5 ∴ H = 6 cm Let’s try work out what the adjacent side is equal to, assuming we don’t know the hypotenuse. Step by step comment comment O tan θ = __ ​  A ​ tan 30° = 3 cm ÷ A 0,57735 = 3 ÷ A A × 0,57735 = 3 × A ÷ A A × 0,57735 = 3 A = 3 ÷ 0,57735 A = 5,196 cm ≈ 5,2 cm. Let’s check this with Pythagoras. Suppose we want to prove that the opposite side is 3 cm. We have H = 6 and A = 5,2. So, Pythagoras tells us that A2 + O2 = H2. So, 5,22 + O2 = 62 O2 = 62 – 5,22 O2 = 36 – 27 O2 = 9 So the square root of O2 will give us O .... namely, O = 3 cm. The trigonometric calculation is correct. Lastly, there are three other operations you can do in trigonometry, but they’re just inverses of the first three: cosecant, secant and cotangent. Cosec, sometimes abbreviated csc, is the reciprocal (inverse) of sine. Sec is the inverse of cosine. And cot is the inverse of tangent. So this means if sin = O/H, then cosec = H/O, and so on. © Department of Basic Education 2015 introduct ion xxxvii Mind the Gap Mathematics

© Department of Basic Education 2015 HINT! hint e.g. 14 EG - worked examples unknown Earth orbits the sun at a distance of 149 597 870 700 metres or 149exams 597 870,7 km (one hundred and forty nine million km). This distance is called the AU, or astronomical unit. The flat disc that corresponds θ to earth’s orbit is called the ‘ecliptic’. Suppose that on 21 December, an sun unknown object is observed at an angle of 88˚ to the ecliptic, and that 2 AU earth on June 21 the same object is observed at 92˚. How far is the unknown ecliptic object in AU? activity ACTIVITIES boy Step by Step and girl-left and right of page Solutionactivity Step 1. Ignore extra information. Since the earth orbits the sun, the angle to the unknown object relative to the earth is the same in both cases; it’s just that on one date, the earth is on one side of the unknown object, and on the other date, it is on the other side. Step by step comment comment From the angles given, you can tell that the unknown object is at 90° to the sun relative to the earth. earth θ Step 2. We know the angle to the unknown object, and the distance to ? the sun. So, if we draw a triangle where the sun is the right-angle, the earth is at the tip of the hypotenuse, and the distant unknown object is opposite the sun, we get the following triangle. sun ? unknown So, we want the hypotenuse. We know that triangles add up to 180°, so the difference between θ and the given angle of 88° is 2°. That means that the angle the unknown object makes in reference to earth is 2°. Thus: O sin = __ ​  H ​ sin 2° = 1 AU ÷ H = 149 597 870,7 km ÷ H 0,035 = 149 597 870,7 km ÷ H H = 149 597 870,7 km ÷ 0,035 H = 4 286 533 756,4964 km = 28,6 AU This means that the unknown object is 4,2 billion km away, or 28,6 AU away. 6. Graphs It’s probably best to start from scratch with Cartesian Coordinates. “Coordinates” are numbers that refer to the distance of a point along a line, or on a surface, or in space, from a central point called the “origin”. Graphs that you will use have only two dimensions (directions). The positions of points on these graphs are described using two coordinates: how far across (left-to-right) the point is, called the x-coordinate, and how far up-or-down on the page the point is, called the y-coordinate. xxxviii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 y 3 e.g. 15 EG - worked examples 2 Consider the following graph. It shows six points in a straight 1 line. exams The coordinates shown can be described using what are called –x x “ordered pairs”. For example, the furthest point in this graph -3 -2 -1 1 2 3 is 3 units across on the “x-axis” or horizontal line. Likewise, it is also 3 units up on the y-axis, or vertical (up and down) line. -1 So, its coordinates are (3;3). The point just below the midpoint or “origin”, is one unit down of the x-axis, and one unit left of -2 activity the y-axis. So its coordinates are (–1;–1). Note that anything to the left or below of the origin (theACTIVITIES circle boyin the middle), takes a –y -3 minus sign. and girl-left and right of page This series of dots look like they’re related to each other, activity because they’re falling on a straight line. If you see a result like this in an experimental situation, it usually means that you can predict what the next dot will be, namely, (4;4). This kind of prediction is called “extrapolation”. If you carry out the experiment, and find that the result is (4;4), and then PAY SPECIAL ATTENTION (5;5),NB NB or correlation. you’ve established that there is a strong relation Step by step comment comment Now, another way of saying that x relates to y, or x is proportional to y, is to say that y is a function of x. This is written y = f(x). So, in the example given above, voltage is a function of resistance. But how is y related to x in thisHINT! graph? Well, it seems to be in a 1 to 1 ratio: y = x. So hint the formula for this graph is y = x. In this case, we’re only dealing with two factors; y = x and y. e.g. 16 EG - worked examples y 3 2 Now, let’s take a slightly more complex case, illustrated exams next to this paragraph. 1 In this next graph,PAYwe see ATTENTION SPECIAL that wherever x is equal to something, y is one more. So, trace your finger NB NB from the bottom left dot upwards. It meets the x-axis at the point –x x -3 -2 -1 1 2 3 –3. Do the same for the same point towards the y-axis. You’ll see it meets the y-axis at –2. You’ll see the next -1 coordinates activity are HINT! (–2;–1), then (–1;0), then (0;1), (1;2), and hint finally (2;3). From this we can see that whatever x is, y is one more. So, y = x + 1 is the formula ACTIVITIES boy forandthis line. -2 and girl-left right of page activity –y -3 e.g. 17 EG - worked examples Let’s take another case. In this next case, we see the following values: where exams x has a certain value, Step byy has double that step comment value. Let’s comment tabulate it. When x is 1,5, y is 3, when x is 1, y is 2. Thus, the formula for this line is: y = 2x. This value next to x is called the “gradient” or “slope” of the line. The larger the value next to x is, i.e. the larger the gradient, the steeper the slope. The gradient is usually abbreviated as “m” when it is unknown. activity © Department of Basic Education 2015 introduct ion xxxix Mind the Gap Mathematics ACTIVITIES boy and girl-left and

© Department of Basic Education 2015 y 3 2 x y 1,5 3 1 1 2 –x x 0,5 1 -3 -2 -1 1 2 3 0 0 -1 –0,5 –1 –1 –2 -2 –1,5 –3 –y -3 Let’s do one more case. In this case, we can see that y is a function of x, since it’s a straight line graph. However, it’s not that easy to see the relationship between x and y. We can see that the slope is the same as the previous graph, so it must be something like y = 2x. However, it doesn’t quite make sense, since 2(–1,5) is not –2. We see that where x is zero (at the origin), y is at 1. But the slope is the same, so it must be y = 2(0) + 1. So the formula is: y = 2x + 1. y 3 As you can see, this is where mathematicians get the general equation for a 2 straight line of y = mx + c. (“c” stands for “constant”). 1 –x x -3 -2 -1 1 2 3 -1 -2 –y -3 x y 2x + 1 –1,5 –2 2(–1,5)+1 = –3+1 = –2 –1 –1 2(–1)+1= –2+1 = –1 –0,5 0 2(–0,5)+1 = –1+1 = 0 0 1 2(0)+1 = 0+1 = 1 0,5 2 2(0,5)+1 = 1+1 = 2 1 3 2(1)+1 = 2+1 = 3 xl introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 7. Circles nce • Diameter is the width of a circle (2r); radius is half the diameter (d/2). fere The edge of a circle is called the “circumference”. “Diameter” means c um to “measure across”. Compare “diagonal” which means an angle Cir across a square or rectangle, so “dia-” means “across” (Greek). “Circumference” means to “carry in a circle” (Latin); think of how Diameter the earth carries us in a circle or orbit around the sun. To remember the difference between these things, just remember that the sun’s rays radiate out from the sun in every direction, so the radius is the dius Ra distance from the centre of a circle, e.g. the sun, to the outer edge of a circle surrounding it, e.g. earth’s orbit (the circumference). • Area of a circle = π r2 • Circumference = 2 π r • You can use the above to solve for radius or diameter. 8. Reading Tables PAY SPECIAL ATTENTION 8.1 Reading Tables NB A table is a way of showing information in rows and columns. A row goes right across. A column in a building is row upright. HINT! hint column e.g. EG - worked examples Getting information from tables Reading a table means finding information in the cells. Each block exams in a table is called a cell. Reading a table is like reading a grid. A B Look at the table on the right. 1. A and B are the column headings. 2. 1, 2, 3, 4, and 5 are the row PAY SPECIAL headings. ATTENTION NB NB • What is in A2? Go across to column A and read down to row 2. 3. A bell. activity • What is in B3? A hand. 4. ACTIVITIE and girl-le • Give the row and column for the star. Row 4 and column A. You 5. right of p HINT! can also write A4. •hintGive the row and column for the clock. Row 5 and column B. activity You can also write B5. e.g. 18 EG - worked examples Step by step com Look at the table below. In a question, you might have to find information inexams the table and write it down, or you might have to use the information in the table to do a calculation. The table below shows the average maximum and minimum temperatures (highs and lows) in Mauritius (measured in degrees Celsius) each month. activity © Department of Basic Education 2015 introduct ion xli Mind the Gap Mathematics ACTIVITIES boy and girl-left and

© Department of Basic Education 2015 Average monthly maximum and minimum temperatures in Mauritius AL ATTENTION NB Average maximum Average minimum Month of the year temperature °C temperature °C This shows that in April January 35 24 the average maximum February 30 22 temperature was 29 °C. This shows that in November March 30 21 the average minimum April 29 21 temperature was 22 °C. May 25 19 June 24 17 d examples July 26 18 August PAY SPECIAL ATTENTION 27 19 NB NB September 29 20 The table doesn’t have units October 32 22 in the cells, but we know the November 32 22 units because they are inHINT! the December 34 24 hint headings for each column. Always give the units in your Look at the table above to answer these questions. answer. 1. Which month of the year had the highest average maximum temperature in Mauritius? ACTIVITIES boy e.g. EG - worked examples AL ATTENTION and girl-left and 2. Which month had the lowest average maximum temperature? right of page NB 3. What is the difference between theaverage maximum temperature in December and the average minimum temperature in December? exams Be careful! Here we are still working with the average maximum temperature Solution column. 1. Reading down the average maximum temperature column, you can PAY SPECIAL ATTENTION p comment comment NB NB see that January has a temperature of 35 °C, and none of the other temperatures are higher. activity 2. The lowest maximum temperature is 24 °C in June. PAY SPECIAL ATTENTION d examples NB NB 3. Here you will need to find the row for December and look across to ACTIVITIES boy and girl-left and get the lowestHINT! right of pageand highest temperatures for that month, then subtract The difference between the hint the lowest temperature from the highest temperature to find the activity lowest and highest numbers difference: 34 – 24 = 10°C. is called the range. HINT! hint e.g. 19 EG - worked examples Step by step comment comment e.g. EG - worked examples The average monthly increases in the cost of electricity (excluding VAT) exams2011 and 2012 between ACTIVITIES boy and girl-left and right of page exams Electricity consumption in kWh 50 150 600 1 000 Amount payable in 2011 R27,35 R85,83 R393,67 R728,63 Amount payable in 2012 activityR28,83 R94,99 R467,43 R888,83 p comment Increase comment activity between 2011 and 2012 R1,48 ACTIVITIES boy R9,16 and girl-left and R73,67 R160,20 right of page Percentage increase between 2011 andACTIVITIES 2012 boy 5,39% 10,67% 18,74% 21,99% activity and girl-left and right of page activity xlii introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

© Department of Basic Education 2015 Read from the table to answer the questions. 1. If a household used 600 kWh of electricity in 2011, what would they have paid? 2. How much more would you pay for 1 000 kWh of electricity in 2012 compared to 2011? 3. What was the percentage increase for 150 kWh of electricity between 2011 and 2012? 4. Was the percentage increase higher for lower electricity consumption, or for higher electricity consumption? Solution When you answer a question like this, take a few minutes to look at the table and write down some notes about what it shows. Don’t get too detailed, just to understand what the table is showing. The columns show 4 different amounts of electricity consumption. The unit is kWh. Electricity consumption in kWh 50 150 600 1 000 First row shows the cost Amount payable for 2011 and 2nd row R27,35 R85,85 R393,67 R728,63 in 2011 shows 2012. Amount payable This is what the table is in 2012 R28,83 R94,99 R467,43 R8883,83 comparing. Increase between R1,48 R9,16 R73,67 R160,20 2011 and 2012 These amounts are Percentage increase calculated for us! between 2011 and 5,39% 10,67% 18,74% 21,99% These are differences 2012 between 2011 and 2012: Amount and Percentage. Notice that there is an increase in costs this way. 1. Read off the 2011 row showing the amount, and the 600PAYkWh SPECIALcolumn: ATTENTION R393,67. NB NB 2. You don’t have to calculate; this difference is given in the third row. The question is asking for 3. The percentage increase is given in the last row. So look at the last row the increase in the amount of and second column (for 150 kWh): 10,67%. money. So we are interested HINT! 4. In the fourth row, there is a steady increase in the percentages from hint lower to higher electricity consumption. So the percentage increase is in the third row. The consumption is 1 000 kWh, bigger for higher consumption. so look at the 4th column and third row: R160,20. 8.2 Reading Two-Way Tables e.g. EG - worked examples Two-way tables are a useful way to display information, and they help you to work out missing information. exams for the same sample. These tables show the numbers of two categories PAY SPECIAL ATTENTION One category is shown in rows, and the other category is shown in columns. NB For example, the table below shows how many Grade 12 learners in a These numbers are for the school own a cell phone or not and how many of the same learners own a same group of learners. music player or not. HINT! activity hint ACTIVITIES boy and girl-left and right of page © Department of Basic Education 2015 activity e.g. EG - worked examples introduction xliii Mind the Gap Mathematics

© Department of Basic Education 2015 Own an MP3 player Do not own an MP3 player Own a cell phone 57 21 Do not own a cell phone 13 9 What’s interesting about this table is that the totals of both columns and the totals of both rows are the PAY SPECIAL ATTENTION NB same. We can see that the sample was of 100 learners. NB Own an MP3 player Do not own an MP3 player Total Own a cell phone 57 21 78 HINT! Do not own a cell phone 13 9 22 Total 70 hint 30 100 e.g. 20 EG - worked examples During one month, 75 of the 180 babies born in a hospital were boys, exams and 40 of the babies weighed 4 kg or more. There were 26 baby boys who weighed 4 kg or more. 1. Put this information in a two-way table and fill in the missing numbers. 2. What percentage of girl babies weighed 4 kg or more? Solution 1. activity First draw up the grid and fill in the information given. (It doesn’t matter whether you put theACTIVITIES weights boy or the gender in the columns or rows.) and girl-left and right of page activity Boys Girls Total Weighed less than 4 kg Weighed 4 kg or more 26 0 40 Total 75 180 Step by step comment comment When you’ve got the table in this form, you can find the missing information. Work back from the totals. For example, if 26 of the baby boys weighed 4 kg or more, then 75 – 26 = 49 of them weighed less than 4 kg. Boys Girls Total Weighed less than 4 kg 49 91 140 Weighed 4 kg or more 26 14 40 Total 75 105 180 2. There were 14 girl babies who weighed 4 kg or more, out of a total of 105 girl babies. ___ 14 ​  105​× 100% = 13,33% xliv introduction  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint © Department of Basic Education 2015 e.g. 21 EG - worked examples One hundred passengers on a bus trip were asked whether they wanted examsor beef and whether they preferred rice or potato for their meals. chicken Out of 30 passengers who liked rice, 20 liked chicken. There were 60 passengers who chose chicken. 1. Put this information in a two-way table and fill in the missing numbers. 2. How many meals with beef and potato should the bus company produce? activity Solution ACTIVITIES boy 1. Here is the information we and aregirl-left given: right of page and activity Chicken Beef Total Rice 20 Potato Total 60 Step by step comment comment Here are the rest of the information: Chicken Beef Total Rice 20 10 30 Potato 40 30 70 Total 60 40 100 © Department of Basic Education 2015 introduction xlv Mind the Gap Mathematics

© Department of Basic Education 2015

Unit 1 Exponents and surds To understand exponents and surds, you need to revise the number system thoroughly. 1.1 The number system All terminating, recurring decimals are rational numbers examples:   •   • •    1.1.1 Real numbers 0.​ 3​; 2,​ 7​​ 1​; 5,321784571 The numbers that we work with every day are called real numbers. The set of real numbers has subsets shown in the Venn diagram: 핉 핈 핈´ 핑 핅0 핅 PAY SPECIAL ATTENTION NB NB 1. Natural Numbers ℕ = {1; 2; 3; …} (positive whole numbers) Pi (π)isHINT! an interesting irrational number. 2. Whole Numbers hint It is the ratio of the circumference to the ℕ0 = {0; 1; 2; …} (Natural numbers and 0) diameter of any circle: circumference ______________of circle 3. Integers π = ​      ​ diameter of circle ℤ = {…NB ;–3; –2; –1; 0; 1; 2; 3; …} NB PAY SPECIAL ATTENTION e.g. = 3,141592653 … EG - worked examples 4. Rational Numbers a • A rational number is a Real number which can be written in the form __ ​ b ​, exams HINT! where hint a, b ∈ 핑 and b ≠ 0. The rational numbers include all the integers. ___ __    • __      •• NOTE: e.g. 1 5; ​√16 ​; ​√8 ​; __​ 37​; ​ ___ 3 EG - worked examples –13 132 ___ __ 22 –16 ___ 3 269 ___ 9 ​; ​  1 ​; ​  7 ​; ​  4 ​; 3,14; 0,​ 3​= ​  10 ​; 2,​ 7​​ 1​= ​  99 ​ ​  22 __ 7 ​and 3,14 are approximate exams 5. Irrational Numbers activity rational numbers that • Irrational numbers are numbers that cannot be written have aACTIVITIES value close to π. boy • as fractions. 22 ​≠ So ​ __ π and 3,14 ≠ π and girl-left and 7 right of page • All decimals that do not terminate or recur are irrational. activity activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 1 comment Step by step comment Mind the Gap Mathematicss Step by step comment comment

HINT! 1 Unit hint __ e.g. 2 ​√5 ​= 2,23606… pi (π) = 3,141592… EG - worked examples • They have decimals that continue indefinitely with no pattern. exams • Look at these numbers on a calculator. • The calculator will round them off. However, they continue indefinitely without a pattern. • The symbol for the irrational numbers is ℚ′, which means the Some quadratic activity complement PAY SPECIAL ATTENTIONof ℚ or not ℚ. NB NB equations do not have ACTIVITIES boy and girl-left and real roots, but 6. Real Numbers right of page some do activity The set of real numbers, ℝ, is the set of all rational and irrational numbers together. HINT! hint We can also write ℝ = ℚ ⋃ ℚ′ Step by step comment comment __ __ e.g. NB 3 –3; –​√7 ​; –1​ 1__4 ​; –1;0; ​ NB PAY- worked EG SPECIAL examples ATTENTION __ 1 2 ​; 1; ​√ 2 ​; 2; 3; π 1.1.2 exams Non-real Numbers HINT! hint The square root (or any even root) of a negative number, is a non-real number. ____ NB 4 ​√ −25 ​is a non-realNB PAY SPECIAL ATTENTION e.g. number. EG - worked examples activity_____ 4 ​√−100 ​ isanda non-real number ACTIVITIES boy girl-left and 6 _____ right of page exams ​√−120 ​is a non-real number HINT! hint activity • The calculator will show an error. e.g. PAY SPECIAL ATTENTION 5 x2 + 5x + 9 = 0 EG - worked examples Step by step comment comment NB activity NB ACTIVITIES boy Use the quadratic formula exams to find the values of x: and girl-left and right of page ______ ​  number ​= undefined Any _____ −b ± ​√b2−4ac ​ 0 x = ___________ ​  activity 2a ​ The calculator will also show __________ an error. HINT! −5 ± ​√​5​2​– 4​(1 )​​(9 )​ ​ =______________ ​    ​ hint (1 )​ 2​_______ activity −5 ± ​√ step ​ 25Step–by36 comment = ___________ comment ​   2____ ​ ACTIVITIES boy and girl-left and − 5 ± ​√−11 ​ = ​ _________ right of page ​ e.g. EG - worked examples ____ 2 activity ​√−11 ​ is a non-real number so the value of x is non-real. There are no real roots for the equation, so the graph of the function y = x2 + 5x + 9 has no intercepts Step with the x-axis. by step comment comment exams y activity y = x2 + 5x + 9 ACTIVITIES boy and girl-left and right of page x activity You will learn more about the nature of roots in Unit 2. 2 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015 Step by step comment comment  Mind the Gap Mathematics

Unit 1 1.2 Working with irrational numbers 1.2.1 Surds All square PAY roots; cube SPECIAL ATTENTIONroots, etc. that they are not rational are called surds. __ NB__ __ __ __ __ NB ​√2 ​; ​√3 ​; ​√5 ​; ​√6 ​; ​√7 ​; ​√8 ​; are all surds. Surds are real numbers which when expressed as decimals are non- HINT! recurring and non-terminating. hint We can work out where a surd lies between two integers on a number line. __ __ __ e.g. 6 ​√1 ​= 1 and ​√4 ​= 2, so ​√2 ​lies somewhere between 1 and 2. EG - worked examples 3 ___ 3 ____ 3 ____ ​√64 ​= 4 and ​√125 ​= 5, so ​√102 ​lies between 4 and 5. exams WeNB can show their approximateNB PAY SPECIAL ATTENTION positions on the number line: __ ____ 1 HINT! ​√2 ​ 2 4 3 ​√102 ​ 5 hint activity Some roots PAY orSPECIAL radical numbers are rational and are not surds: ACTIVITIES boy ATTENTIONand girl-left and NB right of page NB 7 Examples of roots that are not surds include: e.g. activity EG - worked examples __ __ __ __ ___ 3 4 ​√1 ​= 1; ​√4 ​= 2; ​√9 ​= 3; ​√8 ​= 2; ​√81 ​= 3 HINT! exams hint 1.2.2 Simplifying surds Step by step comment comment e.g. 8 EG - worked examples __activity __ ___ 1.exams ​√5 ​× ​√3 ​= ​√15 ​ 3ACTIVITIES boy (1) __ __ __ and girl-left and 2. (​​ ​√5 ​  )​​ ​= ​√5 ​× ​√5 ​= 5 3 2 right of page (1)   ___ activity 3 __ √______​√27 ​ 8 ​√8 ​ 2 ​ ​  27 ​ ​ = ​ ____ ​= ​ __3 ​ 33 3 __ 3. 3 ___ (2) 4. ​ a2activity √ − b2 ​ cannot be simplified comment 3 ___ ____ Step by step 3 ___ comment 3ACTIVITIES boy 12 __ 5. ​√274 ​= ​√​(33 )4​ ​= ​√and ​312 ​girl-left ​ ​ 3and = ​3​​  3 ​​= ​3​4​= 81 3 (2) PAY SPECIAL ATTENTION ______ ___ NB right of page 6. + 16 ​= ​√25 ​= 5 3 ​√9 activity (1) __ ___ 7. √​ 9 ​+ ​√16 ​= 3 + 4=7 3 (1) Step by step comment comment HINT! ______ hint √ ​ 9 + 16 ​  ≠ 3 + 4 e.g. EG - worked examples exams © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 3 Mind the Gap Mathematics activity

1 Unit Activity 1 Write in simplest form without using a calculator (show all working). __ __ 3 __ 3 __ 1. ​√8 ​×___ ​√2 ​ 2. ​√4 ​× ​√2 ​ 9 + ​√45 ​ __ __ 3. ​ ______ 3 ​ 4. ​(2 + ​√5 ​  )(​​ 2 − ​√5 ​  )​ [10] Solutions __ __ ____ ___ 1. ​√8 ​× ​√8 ​= ​√8×2 ​= ​√16 ​= 4 3 (1) 3 __ 3 __ 3 ____ 3 __ 2. ​√4 ​× ​√2 ​= ​√4×2 ​3 = ​√8 ​= 2 3 (2) ___ __ __ 9+​√45 ​ 9+3​√5 ​ 3​(3+​√5 ​  )​ __ 3. _____ ​  3 ​= _____ ​  3 ​3 = _______ ​  3 ​3 = 3 + ​√5 ​3 (3) __ __ 4. ​(2 + ​√5 ​  )​​(2 − ​√5 ​  )​ __ __ = 2 × 2 – ​√5 ​× ​√5 ​3 = 4 – 5 = –13 (2) Or multiply out the brackets: __ __ __ __ __ __ (​2 + ​√ 5 ​  )(​​ 2 − ​√ 5 ​  )​= 4 + 2​√ 5 ​– 2​√ 5 ​– ​√ 5 ​.​√ 5 ​3 = 4 – 5 = –1 3 (2) [10] 1.2.3 Rationalising a denominator When a fraction contains a surd in the denominator, you can change the denominator to a rational number. This is called ‘rationalising the denominator’. PAY SPECIAL ATTENTION NB NB If you multiply the numerator and the denominator by the same surd, you are not __ changing the value of the number. You are multiplying by 1 ​√2 ​ (i.e ​ ___ __ ​= 1) to HINT!change what the number looks like, not its value. Doing this ​√2 ​ hint can give you a rational denominator. e.g. 9 EG - worked examples __ ​√3 ​ Rationalise the denominator of ___ ​  __ ​. ​ √ 2​ __exams NB __ PAY SPECIAL ATTENTION __ __ __ __ NB ​√__ 3 ​ ___ ​√3 ​ ​√__ 2 ​ ______​√3 ​× ​√2 ​ ​√6 ​ ​ ___ ​= ​  __ ​× ​ ___ ​ = ​  2 ​= ​ ___ 2 ​3 (1) ​√2 ​ ​√2 ​ ​√2 ​ PAY SPECIAL ATTENTION NB Now the NB denominator is a__ rational HINT! __ value. hint ​√3 ​ ​√6 ​ on a calculator: ___ Check activity ​  __ ​= ___ ​  2 ​= 1,2247… ​ √ 2​ ACTIVITIES boy and girl-left and e.g. 10 right of page EG - worked examples HINT! activity hint 3 ​ _____ __ ​√3 ​− exams ​ 1 (has an irrational number in __ denominator) If the denominator is √ ​ 3 ​ − 1, __ Step by step comment comment __ __ multiply 3 ​√3 ​+ 1 ​√3 ​+ 1 ​√3 ​+ 1 __ = _____ ​  __ ​× _____ ​  __ ​ 3 (multiply by _____ ​  __ ​, since 1= ​ _____ __ ​) e.g. ​√3 ​ + 1 by ​ _____ __ ​√3 ​ + 1 EG - worked examples ​. This will give us the ​√3 ​− 1 __ ​√3 ​+ 1 __ ​√3 ​+ 1 ​√3 ​+ 1 3(​√3 ​+ 1) ____________ √3 3​__ ​+ 3 ____________ difference of two squares. = ​     __ activity __ ​= ​     __ ​ (notice how the surd terms cancel) (​√3 ​− 1)(​√3 ​+ 1) 3 + ​√3 ​− ​√3 ​− 1 ACTIVITIES boy __ and girl-left and 3​√3 ​+ 3 right of page exams = ​ ______ ​3 (2) 2 activity (now the denominator is rational) Step by step comment comment 4 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics activity

Unit 1 Activity 2 Interpret a graph 1. Complete the table for each number by marking the correct columns. Non- Real Rational Irrational Integer Whole Natural real number number number ℤ number number number ℝ ℚ ℚ′ ℕ0 ℕ a) 13 b) 5,121212… ___ c) √ ​ –6 ​ d) 3π 0 e) ​ __9 ​= 0 ___ f) ​√17 ​ 3 ___ g) ​√64 ​= 4 22 h) ​ __ 7​ (23) 2. Which of the following numbers are rational and which __ are irrational? ___ ___ __ a) ​√16  ​ b) ​√8 ​ ___ 22 √ c) ​ __ 9 ​  4 ​ ​ √1 d) ​ 6​ __4 ​  ​ e) ​√47 ​ f) ​ __ 7 ​ g) 0,347347… h) π − ​(−2 )​ __ i) 2 + ​√2 ​ j) 1,121221222… (10) [33] Solutions 1. Non-real Real Rational Irrational Integer Whole Natural number number number number ℤ number number ℝ ℚ ℚ′ ℕ0 ℕ a) 13 3 3 3 3 3 (5) b) 5,121212… 3 3 (2) ___ c) ​√−6 ​ 3 (1) d) 3π 3 3 (2) 0 e) __ 9​  ​= 0 3 3 3 3 (4) ___ f) ​√17 ​ 3 3 (2) 3 ___ g) ​√64 ​= 4 3 3 3 3 3 (5) 22 h) __ ​  7 ​ 3 3 (2) ___ __ 2. a) ​√16 ​= 4 (rational) 3 (1) b) ​√___ 8 ​(irrational) 3 (1) __ ___ √ ___ 3 ​  94 ​ ​ = ​ __2 ​(rational) 3 c) ​ __ (1) √1 √ d) ​ 6​ __4 ​ ​ = ​ ​  25 5 __​ ​ = ​ __​(rational) 3 4 2 (1) 22 e) ​√47 ​(irrational) 3 (1) f) ​ __ 7 ​(rational) 3 (1) g) 0,347347… (rational, because it is a recurring decimal) 3 (1) h) π −  ​(−2 )​(irrational, because π is irrational) 3 (1) __ __ i) 2 + ​√2 ​(irrational, because ​√2 ​is irrational) 3 (1) j) 1,121221222… (irrational, because it is a non-recurring and non-terminating decimal) 3 (1) [33] © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 5 Mind the Gap Mathematics

1 Unit 1.3 Exponents The exponent exponent PAY of SPECIALa number tells us how ATTENTION NB NB many times to multiply the number (the power base) by itself. So a2 = a × aHINT! a3 = a × a × a base hint a = a × a × a × … n times n e.g. 11 EG - worked examples 34 read as: three to the power of 4, or 3 exponent 4 which is equal to 3 exams × 3 × 3 × 3 = 81 1.3.1 Rules of exponents These rules work for exponents that are integers, rational numbers or activity irrational numbers. ACTIVITIES boy and girl-left and 1. ​a​ ​× ​a​ ​= a m n m+n right of page ​a​5​× ​a​3​ = ​a​5 + 3 ​= ​a​8​ activity To multiply two powers with same bases, add their exponents. ​3​5​× ​3​3​ = ​3​5 + 3 ​= ​3​8​ 2. ​a​m​÷ ​a​n​= am – n ​a​8​÷ ​a​2​= ​a​8 – 2 ​= ​a​6​ To divide two powers with same bases, subtract their exponents. Step by step comment comment 3. (am​)n​ ​= amn (​a4​ ​​)3​ ​= ​a​4 × 3​ = a​​12​ To raise a power to an exponent, multiply the exponents. (​a​2​× ​b​3​​)5​ ​= ​a​2 × 5​. ​b​3 × 5​= ​a​10​. ​b​15​ (ab) m = ​(​​am​ ​bm​ ​  )​ ​a​5×3​ ​a​15​ a m ​am​ ​ ​​(​a5​ ​/​b2​ ​  )3​​ ​= ​ ___ ​= ​ ___​ ​b​2×3​ ​b6​ ​ () ​​ __ ​  b ​  ​ ​= ​ __ ​ ​bm​ ​ 4. a0 = 1, (b​)​0​ = 1 ; ​​(3 )​​ 0​= 1 ; (5​a​2​​b3​ ​​)0​ ​= 1 Any base raised to 0 is 1 1 1 5. ​ __ ​a​n​​= ​a​ ​ −n ​ ​ = ​ __ ​b–3 ​ ​b3​ ​ A positive exponent in the denominator is the same as a negative ​b​3​= ​ ___ ​ 1 exponent in the numerator. ​b−3 ​ ​ ​  b ​  )​ ​= ​​(__ ​  a ​  )​ ​ a −3 b 3 ​  b ​  )​ ​= ​​(__ ​​(__ a −m ​  a ​  )​ ​ b m ​​(__ (2 ) −3 (5 ) ​  5 ​  ​​ ​= ​​ __ ​​ __ 3 125 ​  2 ​  ​​ ​= ​ ___ 8 ​ n ___ 1 __ m __ __ 2 __ 1 __ 1 __ 6. ​√​a​m​ ​ =​​( ​a​m​  )​ ​n​​= ​a​ ​n ​​ (n ≥ 2). ​√2 ​= ​√​2​1​ ​ = ​​(​2​1​  )​ ​​2​​= ​2​ ​2​​ __ To find the root of a power, divide the exponents. __ 2 1 __ 1 __ ​√a ​= ​√​a1​ ​ ​ = ​​(​a1​ ​  )​ ​​2​​= ​a​​ 2​​ 3 __ 1 __ 2 __ ​√​a2​ ​ ​ = ​​(​a2​ ​  )​ ​​3​​= ​a​​ 3​​ 6 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 1 1.3.2 Algebraic expressions with exponents Remember to work in this order: signs → values → variables Values Variables Answer a) –3​a​3​​b2​ ​× –4​a​4​​b4​ ​ –3 × –4 = 12 ​a​3​​b2​ ​× ​a​4​​b4​ ​= ​a​7​​b6​ ​ = +12​a7​ ​​b6​ ​ b) 12​x​5​​y8​ ​÷  –4​x2​ ​​y4​ ​ 12 ÷ –4 = –3 ​x5​ ​​y8​ ​÷ ​x2​ ​​y4​ ​= ​x3​ ​.​y4​ ​ = − 3​x​3​​y4​ ​ c) ( −3​a​3​​b​2​​)​3​ (–3​)​3​= –27 (​a​3​​b​2​​)​3​= ​a​9​​b​6​ = − 27​a​9​​b​6​ _____ __ ___ ​a​ ​ ​ = ​a​4​ 4 4 4 16 d) ​√16​a16 ​ ​ ​ ​ √ 16 ​ = 2 ​ √ = 2​a​4​ (​24​ ​= 16) Where necessary, we work out the inside set of brackets first and follow the order of operations: BODMAS Brackets/Of, Division/Multiplication/Addition/Subtraction Activity 3 Calculate ____ ____ __1 a) −3 ​(​​(−2​a3​ ​  )2​​ ​+ ​√9​a12 ​ ​ ​  )​ ​  ​ ​√9​a12 ​ ​ ​ = ​​(​32​ ​​a12 ​ ​  )​​2​ 5​​(2​a4​ ​  )3​​ ​ b) ___________ ​  ​​(−5​a3​ ​  )2​​ ​− 5​a​6​​ [5] Solutions ____ a) –3​(​​(–2​a3​ ​  )2​​ ​+ ​√9​a12 ​ ​ ​  )​ simplify exponents inside the brackets and the square root = −3(4​a​ ​3+ 3​a​ ​3) 6 6 add like terms inside the bracket = –3 ​(7​a6​ ​  )​= –21​a6​ ​3 simplify (3) 5​​(2​a​4​  )​​3​ b) __________ ​  ​​(−5​a3​ ​  )2​​ ​− 5​a​6​​ simplify brackets at the top and the bottom first 5​(8​a12 ​ ​  )​ 40​a​12​ = __________ ​  + 25​a​6​ − ​5a​6​​  3  = ​ ____ 2​0a​6​ ​ = 2​a​6​ 3 (2) [5] © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 7 Mind the Gap Mathematics

1 Unit 1.3.3 Prime factors When the bases are different, we can write each base as a product of its prime factors. Remember: A prime number has only two different factors. A composite number has more than two factors. The number 1 is neither a prime number nor a composite number. PAY SPECIAL ATTENTION NB NB PAY SPECIAL ATTENTION NB NB 2; 3; 5; 7; 11; 13 … Prime numbers: Every composite HINT! number can be written as the product of prime numbers. hint This helps us to factorise and to simplify. HINT! hint e.g. 12 EG - worked examples 4 = 22; 6 = 2 × 3; 8 = 23; 9 = 32; 10 = 2 × 5; 12 = 22 × 3 exams 24 = 8 × 3 = 23 × 3 e.g. EG - worked examples NOTE: To find the factors of 2 700, divide by the lowest prime that is a factor eg 2; then move activity PAY SPECIAL ATTENTION onto NB3; then 5 etc. NB exams ACTIVITIES boy 2 700 = 22 × 33 × 52 3 and girl-left and right of page PAY SPECIAL ATTENTION Find outactivity how your scientific calculator can calculate the prime factors of a number for NB you. NBHINT! hint Step by step comment comment activity e.g. 13 EG - worked examples HINT! hint ACTIVITIES boy Express 7​2x−2 ​ ​in prime factors and girl-left and exams right of page 2 72 7​2​x−2 ​= ​(2​3​.​32​ ​3​)x−2 ​ ​ activity 2 3 2 36 = ​2​3(x−2)​.​32(x−2) ​ ​ e.g. 2 EG - worked examples 18 activity = ​2​ ​.​3​ ​ 3 3x−6 2x−4 3 9 32 ACTIVITIES boy 3 3 1.3.4 Working comment and with negative exponents girl-left and Step by step comment right of page exams 1 activity It is easier to write answers with positive exponents, so we use the exponent rule: 1 __ 1 ​ ​ andStep by​ ___ ​  ​an​ ​​= ​a−n ​a−n​step​​= ​a​ ​ comment comment n a −2 b )2 This also means that (​​__ ​   ​  )​​ ​= ​​(__ ​  a ​  ​​ ​ b activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 8 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 1 Activity 4 1 ​ + ​ __ ​ __ 1 ​ ≠ ​ ______ 1 Simplify the following. Write answers with positive exponents where x y x + y ​ necessary. ​a–3 ​​ 4​a7​ ​​b–4 ​ ​​c–1 ​​ 1. ___ ​  ​b​–2​​ 2. ______ ​  ​​d​–2​e​5​ ​ 3. ​x–1 ​ ​ + ​y–1 ​​ [5] Solutions ​a​−3​ __ ​b​2​ 4 ​a​7​​b​−4​​c​−1​ _____ 4 a7d 2 1. ___ ​  ​b−2​ ​​= ​  ​a3​ ​​3 2. _______ ​  ​d−2 ​ ​​e5​ ​ ​= ​  b4c1e5 ​33 1 1 y+x 3. x–1 + y –1 = __ ​  x ​ + __ ​  y ​ 3 = ____ ​  xy ​ 3 [5] NOTE: A surd is also called a radical. 1.3.5 Working___with surd (root) signs n m The exponential rule ​√​am​ ​ ​ = ​a​ ​n ​ (a > 0; n ≥ 2), can be used to simplify certain __ expressions. We can find a root Activity 5 of a negative number if the root is odd, but not the root of a negative number 1. Rewrite __ these expressions___ without surd signs and____ simplify if possible. if the root is even. 3 4 3 a) ​√5 ​ b) ​√16 ​ c) ​√–32 ​ PAY SPECIAL ATTENTION [3] NB NB Solution __ __ 3 1 __ 4 1 __ 1​ ​ __ a) ​√5 ​= ​5​​ 3​​3 b) ​√1 ​6 = 1​6​​ 4​​= ​​(​24​ ​  )​​4​= 2 3 5 _____ HINT! ​  1 __​ ​  1 __​ hint c) ​√ − 32 ​= ​​(−32 )​​5​= ​​[ ​​(  2 )5​​ ​  ]​​5​=  −23 [3] e.g. 14 EG - worked examples 3 _____ 3 _____ ​√ exams −27 ​= − 3 because (–3)3 = –27 ∴ ​√ −27 ​is real 4 _____ ​√ −16 ​ is non-real 1.3.6 Watch out for these common activity mistakes! ACTIVITIES boy and girl-left and Correct Warning right of page activity 1.    ​2​ ​. ​3​ ​ = ​6​ ​ n n n ​2.3​ ​≠ ​6​ ​ n n 2.    ​34​ ​× ​35​ ​ = ​3​9​ ​34​ ​× ​35​ ​≠ ​9​9​ 3.    ​4​10​÷ ​4​5Step​=by ​4 ​5​ step comment comment ​4​10​÷ ​4​5​≠ ​4​2​ ​410 ​ ​÷ ​4​5​≠ ​1​5​ ​410 ​ ​÷ ​4​5​≠ ​1​2​ ​ ​ = ​3​n – 1 ​bn​ ​– 1 4.    ​​(3 b )​n−1 (​​ 3 b )n−1 ​​ ​ ≠ 3. ​b​n–1​ © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 9 Mind the Gap Mathematics

1 Unit 5.    ​​(a + b )2​​ ​= ​a​2​ + 2ab + ​b ​2​ ​​(a + b )2​​ ​ ≠ ​a​2​+ ​b​2​ ______ ______ 6.    ​√16 ​×16 ​ ​ ​ = 4​x​8​ [​√16 ​× 16 ​ ​ ​ ≠ 4 ​x​4​ ] ______ ______ 1 __ 7.    ​√​a​ ​+ ​b​ ​ ​ = ​​(​a2​ ​+ ​b​2​  )​ ​​2​​≠ a + b 2 2 ​√​a2​ ​+ ​b​2​ ​ ≠ a + b ______ e.g. ​√​52​ ​− ​3​2​ ​ ≠ 5 − 3 = 2 ______ ______ ___ because ​√​52​ ​− ​3​2​ ​ = ​√25 − 9 ​= ​√16 ​= 4 3 1 ​ ​= ​ __ 8.   3​×−3 ​ ​x3​ ​ 3​x−3​≠ ​ ___ ​ 3​x3​ ​ 1 ​​ ​= ​ ______ 9.    ​​(x + y )−2 ​ ​​(x + y )2​ ​ ​​(x + y )−2 ​​ ​≠ ​x​−2​+ ​y​−2​ 1.3.7 Simplification of exponential expressions Activity 6 Simplify the following and leave answers with positive exponents where necessary: (​​ ​a​4​  )n–1 ​​ ​ . (​​ ​a​2​b )–3n ​​ ​ ​ ___________ ​​(ab )​​–2n​. ​b​–n​ ​ [4] Solution ​​(​a​4​  )​​n−1​ . ​​(​a​2​b )​​−3n​ ___________ ​a​4n−4​ . ​a​–6n​ . ​b​−3n​ 3 ​  ​​(ab )−2n    ​​ ​. ​b​–n​ ​ = ​ ____________   ​    ​a​−2n​ . ​b​−2n​ . ​b–n​​ 3 = ​a​4n–4 – ​6​​+ ​2​​​. ​b​−3n + 2n + n​ n n = ​a​−4​. ​b0​ ​ 3 1 1 = ​ __4 ​. 1 = ​ __4 ​ 3 ​a​ ​ ​a​ ​ [4] PAY SPECIAL ATTENTION NB NB 1.3.8 Algebraic fractions with exponents 1. Expressions with only products of terms • Factorise HINT! the terms using prime factors. hint • Use laws of exponents. e.g. 15 EG - worked examples ​ ​. ​92n ​52n _______ ​ – 3​ ​ ​. (​​ ​32​ ​  )2n–3 ​52n ​​ ​ ​  ​15​2n exams​. ​34n−1 ​ ​ ​ = _________ (​​ ​  5.3 )2n ​​ ​. ​32n−1 ​ ​ ​ 3     (use prime number bases) ​ ​. ​34n ​52n ​ – 6​ 3 = _________ ​  ​5​2n​. ​32n​ ​. ​3​2n−1​​ 3   (to remove brackets, × exponents) = ​5​2n–2n​. ​3​4n–6–2n–(2n–1)​3 (same bases × / ÷ , activity add/subtract exponents ) ACTIVITIES boy and girl-left and = ​5​0 ​. ​3​4n−6−2n−2n+1 right of page​ activity = 1 . ​3−5 ​ ​= 1 × __ 1 ​  ​   (write negative exponent as positive ​3​5​ exponent) 1 =Step ___ ​  by243 step ​3 comment comment 10 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

PAY SPECIAL ATTENTION Unit 1 NB NB 2. Expressions with terms added or subtracted • First try to factorise both the numerator and denominator. • Use laws of exponents. HINT! hint • Cancel any common factors. e.g. 16 EG - worked examples (​​ ​3​n​  )2​ ​ + ​3​2n – 1​ __________ ​ ​.​3−1 ​32n ​ ​ ​  exams ​9​n​ ​ ​ + ​ _____ ​ = ​32n ​​ ​32n ​              ​ ​ ​(1 + ​3​−1​  )​ ​32n = __________ ​  ​​ ​32n ​ 33       (factorise: take out HCF) 1 4 = 1 __ ​  3 ​= ​ __3 ​ 3 activity ACTIVITIES boy Activity 7 and girl-left and right of page activity Simplify the following and leave answers with positive exponents where necessary: ​27​3−2x​. ​9​x−1 ​ by step 1. ​ ________ comment 8​1​2−x​ ​ Step comment ​ ​− 2.​5​x+2​ 6.​5x+1 2. __________ ​  ​5​x+3​ ​ ​22009 ​ ​− ​2​2012​ 3. ________ ​  ​2​2010​ ​ [13] Solutions ​​ ​. ​​(​3​2​  )x−1 ​​(​3​3​  )3−2x 2​7​3−2x​. ​9​x−1​ ___________ ​3​9−6x​.​3​2x−2​ 3 ​​ ​ _______ 1. _________ ​  8​12−x ​ ​ ​ = ​  ​​(​34​ ​  )2−x ​​ ​ ​= ​  ​3​8−4x​ ​ 3                                                         = ​3​9−6x+2x−2−8+4x​ 3 1 = ​3​−1​= ​ __3 ​ 3 (4) 6.​5​x+1​− 2.​5​x+2​ 6.​5x​ ​.​51​ ​− 2.​5​x​.​5​2​ 2. __________ ​  ​5x+3 ​ ​ ​ = ____________ ​   ​ ​5​x​.​53​ ​ ​5​x​​(6 × 5 − 2 × ​5​2​  )​       = _____________    ​ 33    ​  ​5x​ ​.​53​ ​ 30 − 50 − 20 4 (4) = ​ ______ 3 ____ __ 3 125 ​ =  ​  125 ​=  − ​  25 ​ 33 ​22009 ​ ​− ​2​2012​ ​ ​​(1 ​22009 − ​2​3​  )​ ​(​22009 ​ ​ 1 − 8 )​ 3. ​ ________ ​2​ ​ 2010 ​ = _________ ​  ​2​ ​ 2010 ​ = ________ ​  ​22010 ​ ​ ​ ​22009 ​ ​​(− 7 )​                = ________ ​  ​2​2010​ ​    = ​22009−2010 ​ ​3 ×− 7 1 7               = ​2​−1​×  − 7 3 = __ ​  2 ​×  − 7 =  −​ __ 2​ 3 (5) [13] © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 11 Mind the Gap Mathematics

1 Unit 1.4 Exponential equations NB equations where x is part PAY SPECIAL ATTENTION Solving NB of the exponent: • Write the powers as products of prime factors. • Aim to get HINT! ONE power with the same base on each side of the equation. •hint Equate the exponents. • Solve for x. e.g. 17 EG - worked examples 1. Solve for x exams ​ 2x​ ​ = 8 write 8 as a power of 2 ​ 2x​ ​ = ​23​ ​ get the same base on each side ∴ x = 3 3 equate the exponents ​​ activity 2. ​​5​2x+1 ​− 12​​5​2x−3 ​​ ​= 0 equate the two powers ACTIVITIES boy ​​5​ ​​ ​= 12​​52​ x−3 2x+ 1 ​​ ​ and girl-left and right of page ​​5​2x+ ​​ ​ 3 ​​ 1​= ​​​(​53​ ​  )2​​ x−3 write with prime bases activity 6x−9 3 ​​5​ ​​ ​= ​​5​ ​​ ​ 2x+ 1 ∴ 2x + 1 = 6x –9 ∴ –4x =Step–10 by step comment comment ∴ x = ​  5__​ 333 2 3. ​ 2​x​= ​5​x​ () x ​​ ∴ ​  ​2__ ∴ ​​ ​  2__5 ​  ​​ ​= 1 3 x ​5​x​​= 1 ∴ (​​ __ ​  5 ​  )​​ ​ = ​​(__ ​  5 ​  )​​ ​  3 x 2 2 0 ∴ x =03 4. ​ ​− ​3​x−1​ = 216 ​3x+1 ∴ ​3​x​.​31​ ​− ​3​x​.​3​−1​ = 216 ∴ ​3​x​​(3 − ​3​−1​  )​= 216 33 ( ) ∴ ​3x​ ​​ 3 − ​ __13 ​  ​ = 216 (3 ) ∴ ​3x​ ​​ ​  8__​  ​ = 216 ∴ ​3​ ​ = 216 × ​  3__8 ​3 x PAY SPECIAL ATTENTION NB NB∴ ​3x​ ​ = 81 ∴ ​3x​ ​= ​3​4​3 x = 43 5. ​3​2x​− 12.​3​x​+ 27 = 0 HINT! ∴ ​3x​ ​.​3x​ ​− 12.​3​x​+ 27 = 0 hint Remember: ​ 3x​.​​3x​​= ​32x ​​ Method 1:     Method 2: ∴ ​3​x​.​3x​ ​− 12.​3​x​+ 27 = 0       ∴ ​3x​ ​.​3x​ ​− 12.​3​x​+ 27 = 0 ​(​3x​ ​− 9 )​​(​3x​ ​− 3 )​= 0 33          let ​3​x​= k ∴ k.k − 12k + 27 = 0         ​3​x​= 9  or  ​3​x​= 3   3                e.g. EG - worked examples ∴ ​k2​ ​− 12k + 27 = 0   3        ​3​ ​= ​3​ ​  or  ​3​ ​= ​3​ ​                            ​(k − 9 )​​(k − 3 )​= 0 3 x 2 x 1       ∴ x = 2 3 or x = 1 3                     ∴ k = 9 or k = 3 3 but ​3​x​= k  ∴ ​3​x​= 9   or ​3​x​= 3    exams ∴ x = 2 3 or  x = 1   3 [24] 12 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 1 Activity 8 Solve for x: ​ ​  )​= ​27​2x–1​ 1. 3​(​9x+3 2. ​ ​= 1 ​32x–12 3. ​2x​ ​= 0,125 4. ​10​x​(x+1 )​​= 100 5. ​5​x​+ ​5​x+1​= 30 6. ​ ​– ​5x​ ​= ​5x​ ​.23 + 1 ​52+x 2 __ _ 1 7. ​5x​ ​ + 15.​5−x ​ ​= 2 8. ​x​​ 3​​– ​x​ ​3​​– 12 = 0 [31] Solutions Remember: When adding or subtracting terms, you need to factorise first. 1. 3 (​​9​x+ ​​ 3​) = 2​​7​2x– ​​ 1​ 2. ​​32​ x​​ −12​ = 1 3​1​​​(​32​ ​  )x+ ​ ​​ 3​= ​​(​33​ ​  )2x–1 ​​ ​ prime bases ​​​3​2x​​ ​​ – 12​ = ​3​0​ 3 make same bases by putting 1 = 30 ​​3​1+2x+6 ​​ ​ = ​​3​6x–3 ​​ ​ 3 same bases ∴ 2x – 12 = 0 3 equate exponents ∴ 7 + 2x = 6x – 3 3 equate exponents 2x = 12 –4x = – 3 – 7 x=63 (3) x = ​  −____ 10 ​ = ​  5_ ​ 3 (3) −4 2 3. ​   2​x​= 0,125 convert to a common fraction 4. 1​0x(x+1) ​ ​= 100 ​2​x​= ​  125 ____ ​= ​  1 _ ​= ​  1 __ ​ 3 simplify 1​0x(x+ ​ 1)​ = 1​0​2​ same bases 1 000 8 3 ​2​ ​ ​​   2​ ​​ ​= ​2​ ​ 3 x −3 same bases ∴ x (x + 1) = 2 3 equate exponents ∴ x = –3 3 equate exponents (3) x2+x–2=0 set quadratic equation = 0 (x + 2)(x – 1) = 0 3 factorise the trinomial PAY SPECIAL ATTENTION x + 2 = 0 or xNB – 1 = 0 make each factor = 0 NB x = –2 3 x=1 (4) 5. ​​5​2 +x​​ ​–​5x​ ​ = ​5​x​  · 23 + 1 3 ​​5​2 +x​​ ​− ​5​x​– ​5x​ ​  · 23 = 1 like terms 2 +x HINT! ​​5​ ​​ ​– 24 · ​5​ ​ = 1 3x ​52​ ​. ​5x​ ​– 24 · ​5​x​ = 1 factorise PAY SPECIAL ATTENTION NB(Common Factor) hint NB When adding or subtracting terms, you need to factorise first. 5​ ​​ ​5​ ​– 24 )​ = 1 33 ​ x( 2 PAY SPECIAL ATTENTION ​          5​ ​​ 1 )​ = 1 x( NB NB 5​x​= ​5​0​ ∴ x = 03 ​ (4) HINT! e.g. EG - worked examples hint ​5​ ​= ​5​x.​ 5 x+1 6. ​5​x​+ ​5​x+1​ = 30 HINT! ​5​ ​+ ​5​x​. ​51​ ​ = 30 x factorise hint exams When adding or subtracting terms, you need to factorise first. ​5x​ ​​(1 + ​5​1​  )​ = 30 33 common factor 5x ​5x​ ​​(6 )​ = 30 3 e.g. divide 30 by 6 EG - worked examples ​5​x​= 5 same bases e.g. EG - worked examples ∴x=13 equate exponents (4) exams activity exams ACTIVITIES boy and girl-left and right of page activity activity © Department of Basic Education 2015 ACTIVITIES boy and girl-left and activity UNIT 1 EXPONENTS AND SURDS 13 right of page Mind the Gap Mathematics Step by step ACTIVITIES boy comment comment activity and girl-left and right of page

PAY SPECIAL ATTENTION NB NB 1 Unit PAY SPECIAL ATTENTION 7. ​ 5x​ ​ +   15.​5−x ​ ​= 2 NB HINT! NB 15 __ ∴ ​5​ ​+ ​  ​5​x​​= 2 x hint x ​ ​= ​​ _15 ​​ ​  ∴ 15.​5−x ​5−x ​ ​= 15 × ​  ​51 15 __​= ​ __ ​x​ ​5​x​ ​ 15    × ​5x​ ​∴ ​5​x​.​5x​ ​+ ​5​x​.__ ​  ​5x​ ​​= 2.​5​x ​3     ∴ ​5​x​.​5​x​+ 15 = 2.​5​x​ HINT!     ∴ ​5​x​.​5​x​− 2.​5​x​+ 15 = 0 3 e.g. hint NB Factorise - trinomial EG SPECIAL PAY - workedATTENTION examples NB     ∴ (​​5​x​− 5 )​​(​5​x​+ 3 )​= 0 33     ∴ ​5x​ ​= 5  or  ​5​x​=  − 3   (no solution)     ∴ x = 1   3    exams (5) e.g. EG - worked examples HINT! 2 __ 1 __ 8. ​x​ ​3​​ – ​x​ ​3​​– 12 = 0 hint ​x​ ​3​​. ​x​ ​3​​= ​x​ ​3​​ 1 _ 1 _ 2 _ ∴ ​(​x​ ​3​​– 4 )​​(​x​ ​3​​+ 3 )​33 = 0 1 __ __1 1 __ ∴ ​x​ ​3​​= 4 or ​x​ ​3​​= –3 3 1 __ exams ∴ x = 64 3 or x = –267 3 e.g. activity EG - worked examples (5) [31] ACTIVITIES boy and girl-left and right of page If the numerator 1.5 Equations exams with rational activity activity of the exponent is odd, then we will always have exponents ACTIVITIES boy one and only one and girl-left and solution 1.5.1 Tips right of page • NBWhen working activity with equations, PAY SPECIAL ATTENTION NB you Step bymust step do the same commentoperation comment to both sides of the equation. activity • Get the variable with the fraction exponent onACTIVITIES one side boy by itself. • Get x byHINT! itself by changing the fraction exponent to an and girl-left and exponent of 1. right of page m so that ​x​ ​ncomment ​ __ •hint Do this by choosing an exponent for both sides, comment Step by step  ​ becomes ​x​ ​. 1 activity e.g. 18 EG - worked examples 1 __ 1.exams ​x​ ​2​​ = 3 (Raise both sides to the power of 3) Step by step comment comment ​​(​x​ ​  )​ ​= (​​3 )2​​ ​ 3 ​  2 ​ 2 _ 1 ∴ ​x​1​= 9 3 (Raise both sides to the power of 2) 1    ​x​__​=   − 3 ​ __ 2​ ​x​ ​2​activity 1 __ ∴ ​√x ​ = − 3      2. ​=  −3 3 The square root of a x = (​​ ​x​ ​2​​  )​ ​= 9 1 _1 2 ACTIVITIES boy and girl-left and negative is non-real right of page 4 activity (Raise both sides to the power of __ ​  3 ​) 3 ​ __​ 3. ​x​ ​= 8 4 (​​ ​x__​ ​43​​  )__​ ​​34​​=3 ​8​__​ 34​​ 4 __ 3 If the numerator 3 )​  3by​ step 4 comment    ​x1​ ​= ​​(​2Step ​ ​  ​​ ​ = ​2​ ​= 16comment 3   of the exponent is even, then we get a quadratic 2 __ equation with two possible answers. 4. ​x​​ 3​​= 4 2 __ ​x​ ​3 ​ = 4     ​​(​x​ ​3​​  )​​2 ​= ± ​4​​ 2​​3 2 __ __ 3 3 __ ​  ​ 2 __ 3 ​ __​ ∴ ​x​ ​3 ​ − 4 = 0    x = ± ​​(​22​ ​  )​​2​=  ± ​​(2 )3​​ ​ 3 1 1 ∴ (​​x​ ​3 ​​− 2 )​​(​x​ ​3 ​​+ 2 )​= 0 __ __    ∴ x = −8 3  or  x = 8   3 1 1 ​ __ __ ∴ ​x​ ​= 2   or ​x​ ​3 ​​=   − 2 3 ​ 1 3 1 3 ∴ ​​(​x​ ​3 ​​  )​​ ​= ​2​3​   or ​​(​x​ ​3 ​​  )​​ ​= ​​( − 2 )3​​ ​ __ __ ∴ x = 8  or x =   − 8 14 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 1 Activity 9 Solve for x: 3 _ 5 __ 1. ​x–​ ​ 2​​= 8 2. ​√​x4​ ​ ​ = 256 [7] Solutions 3 __ 1. ​x–​ ​ 2​​ = 8 2 ​​(​x−​ ​ 2​​  )−​ ​​ 3​​ = ​​(​2​3​  )−​​ ​ 3​​ 33 (raise both sides to the power of − __ 3 __ 2 __ 2 __ ​  3 ​) ​ ​ = ​2​–2 3 ​ 2 __ ​x+1 )3 ×−​  3 ​ ​​(2 ​​ ​ The numerator 1 of the exponent is x = ​ __4 ​    (3) even, therefore two solutions are possible. __ ​√​x4​ ​ ​ = 256 5 2. (change radical to fractional exponent form) 5 ​  54 ​ (raise both sides of equation to the power of __ __ ​x​ ​=  ± ​​(2 )8​​ ​ 3 ​  4 ​) ​​(​x​ ​5​​  )​​4​=  ± ​​(​28​ ​  )​​4​3 __ ​  5 4 __​ ​  5 __​ NB∴ x =  ± ​​(2 )​​ ​ 3 =  ± 1024 NB PAY SPECIAL 10ATTENTION 3 (4) [7] HINT! 1.5.2 hint Exponential equations with surds e.g. 19 EG - worked examples Solve for x: exams _____ 3​√x + 2 ​ + x = 2  [7] Solution _____ activity 3​√x + 2 ​+ x = 2 _____ ∴ 3​√x + 2 ​= 2 − x ACTIVITIES boy and girl-left and _____ right of page ∴ ​​(3​√x + 2 ​  )2​​ ​= ​​(2 − x )2​​ ​ 3 activity ∴ 9(x + 2)  =   (​2 − x )​​(2 − x )​ 3 ∴ 9x + 18 = 4 − 4x + ​x​2​ 3 ∴ 0 = ​x​2​−Step 13x by step− 14 3 comment comment ∴ 0 = ​(x − 14 )​​(x + 1 )​ 3 ∴ x = 14 or x =  − 1 3 Check: ______ ___ x = 14    LHS = 3​√ 14 + 2 ​ + 14 = 3​√ 16 ​ + 14 = 3 × 4 + 14 = 26         RHS = 2 ∴ x = 14 is not a solution 3 _______ __ x =  − 1   LHS = 3​√ − 1 + 2 ​ + ( − 1) = 3​√ 1 ​ − 1 = 3 × 1 − 1 = 2         RHS = 2 ∴ x =  − 1 is a solution    [7] © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 15 Mind the Gap Mathematics

1 Unit Activity 10 Solve these equations and check your solutions. ______ 1. ​√3x + 4 ​− 5 = 0 (3) ______ 2. ​√3x − 5 ​− x = 5 (5) [8] Solutions _______ 1. √ ​ 3 x + 4​ − 5 = 0 Check: _______ _______ ​√3 x + 4 ​ = 5 ​(isolate the radical )​ LHS: ​√3(7) + 4 ​− 5 _______ ______ (​​ √ ​ 3 x + 4 ​  )​​ ​= ​5​2​ 3 2 (square both sides of = ​√21 + 4 ​− 5 ___ the equation) = ​√25 ​− 5 3x + 4 = 25 3 =0 3x = 21 = RHS x=7 3 ∴ x = 7 is a solution (3) ______ 2. ​√3x − 5 ​ − x = 5 ______ ​√3x − 5 ​= x − 5 (always isolate the radical first) ______ (​​ √ ​ 3x − 5 ​  )​​ ​3= ​​(x − 5 )2​​ ​ 2 (square both sides) 3x–5 = ​x​ ​ –10x + 25 3    2 Remember: (x– 5​)​2​ ≠ ​x​2​+ 25 0 = ​x2​ ​ – 13x + 30 3 (quadratic equation, set = 0) 0 = (x – 10)(x – 3 ) 3 (factorise the trinomial and make each factor = 0) x = 10 or x = 3 Check your answer: If x = 10 If x = 3 LHS: LHS ________ _______ ​√3(10) − 5 ​− 10 ​√3(3) − 5 ​− 3 ___ __ = ​√25 ​− 10 = ​√4 ​− 3 =  −5 = RHS =  −1 ≠   RHS (5) ∴ x ≠ 3 and only x = 10 is a solution. 3 [8] 16 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 1 1.6 Exam type examples Activity 11 1. Simplify the following: 2. Solve for x: ​6​ ​.​9​ ​ 6x 3x a) ​3x​ ​− ​3​x−1​= 6 a)  ​ ________ 1 2−x ​ __ () ​ ​ ​.​​ ​  4 ​  ​​ 5​44x b) ​4(x+1)(x−3) ​ ​= ​8​−x​ c) ​2​2x​− 3 × ​2x​ ​− 4 = 0 ​22x+2 ​ ​− ​2​2x−1​ b) _________ ​  ​4x​ ​+ 8.​2​2x−4​​ [22] Solutions 6x ( 2 )3x 3 ​66x ​ ​.​9​3x​ ​​(2 × 3 ) ____________ ​​ ​.​​ ​3​ ​  ​​ ​ ​ ​× ​3 ​26x ____________ ​ ​× ​36x 6x ​​ a) ​ ________    2−x ​ = (​     4x( −2 )2−x ​ =   ​     ​2​ ​× ​3​ ​x ​2–4+2x ​ ​ ​ 3 5​4​4x​.​​(​ __4 ​  )​​ ​ ​​ 2 × ​3​ ​  ​​ ​​​ ​2​ ​  ​​ ​3 1 3 ) 4x 12x ​ ​× ​3​6x​× ​3​6x​ ​26x = ​ ____________   ​ 3    ​ ​× ​​3​12x​​ ​× ​2​−4+2x​ ​24x = ​2​4​× ​3​0​= 163 (5) ​22x+2 ​ ​− ​2​2x−1​ ​ ​.​22​ ​− ​2​2x​.​2−1 ​22x ​ ​ b) _________ ​  ​4​x​+ 8.​2​2x−4​​= ​ ___________ ​ ​2​2x​+ ​2​3​.​2​2x​.​2​−4​ ​ ​(​22​ ​− ​2​−1​) ​22x = __________ ​  ​22x ​3 ​ ​(1 + ​2​3​.​2​−4​) 3 ​ ​​(​2​2​– __ ​  2 ​  )​ 1 ​22x = ________ ​  ​ ( __​3​ ​2 ​2​ ​​ 1 + ​  ​2​4​​  ​ 2x ) 1 4 − ​ __2 ​ = ​  1__ ​3= ​_____ ____ 1 + ​  2 ​ ​  8−1 2 ( 2+1 ​  ​÷​ ​ ____ 2 ​  ​ ) ( ) ​  2 ​  )​÷ ​(__ = ​(__ ​  2 ​  )​= ​ __2 ​× ​ __3 ​= ​ __3 ​3 7 3 7 2 7 (4) 2. a) ​  3​x​− ​3​x−1 ​= 6 b) ​4(x+1)(x−3) ​ ​= ​8​–x​ ​3x​ ​− ​3​x​.​3−1      ​4​​x​​−2x−3 ​ = ​​(​23​ ​  )​​ ​ –x ​ ​= 6 2 ​3x​ (​​ 1 − ​3​−1​  )​= 6 33    ​​(​22​ ​  )​x​​ ​ ​−2x−3​ = ​2​−3x​ 3 2 ​3x​ ​​(1 − ​ _​  )​= 6 3 1 ​ ​    ​ 2​2​x​ ​−4x−6​ = ​2−3x 2 3 ​        3​ ​​(_​  ​  )​= 6 x 2 3 ∴ 2​x​ ​− 4x − 6 = −3x 3 2 ​3​ ​= 6 × ​ _2 ​ 3 x ∴ 2​x​2​− x − 6 = 0 3 ​3x​ ​= 9 ∴ (​2x + 3 )(​​ x − 2 )​= 0 3 ​3x​ ​ = ​3​2​∴ x = 2 3  (4) ∴ x =  − ​  3 __​  or x = 2 3 (5) 2 ​ ​− 3 × ​2x​ ​− 4 = 0 c)     ​22x         ​(​2x​ ​− 4 )(​​ ​2x​ ​+ 1 )​ 33 = 0          ∴ ​2x​ ​= 4   or ​2​x​= −1 (no solution) 3          ∴ x = 2 3 (4) [22] © Department of Basic Education 2015 UNIT 1 EXPONENTS AND SURDS 17 Mind the Gap Mathematics

1 Unit What you need to be able to do: • Use the laws of exponents to simplify expressions • Calculate with negative powers • Mmultiply and divide powers • Add and subtract powers • Solve exponential equations, including those with rational exponents • Simplify surds and do operations with surds • Rationalise the denominator if necessary • Solve equations involving surds. Feb/March 2014 Q 1.1.3 Nov 2013 Q 1.3 Feb/March 2013 Q 1.1.3 Feb/March 2011 Q 1.3 Nov 2010 Q1.3 Feb/March 2010 Q1.4 Keep going! 18 UNIT 1 EXPONENTS AND SURDS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 Algebra 2.1 Algebraic expressions PAY SPECIAL ATTENTION NB Algebraic expressions are madeNB up of constants, variables and number operations (add, subtract, divide and multiply). The variables are shown with letters such as x, y, a, b, p, m, n, etc. HINT! hint The terms in an algebraic expression are separated by a plus or a minus sign. e.g. 1 EG - worked examples • 2x + 3y is an algebraic expression with two terms which are 2x and 3y. • exams 2x(3y) is only one term. • (2x + 3y)(2x – 3y) is also only one term because it is two expressions in brackets ______ multiplied together. The brackets are not separated by + or –. • ​√2x − 3 ​is also an algebraic expression ______ with one term because square roots 1 __ can be written as exponents. ​√2x − 3 ​= ​​(2x − 3 )​ ​​2​​ activity ACTIVITIES boy and girl-left and 2.2activity Addition and subtraction right of page Check NB that you know these NB facts: PAY SPECIAL ATTENTION • We canStepadd by step or subtract comment like terms. comment • If the terms are like, we add or subtract the coefficients. • Like terms HINT! have the same variables (letters) and the variables must have the hint same exponents. e.g. 2 EG - worked examples 3x + 5x = 8x exams –3a + 10a = 7a 6​x2​ ​y + 3x − 10​x2​ ​y = −4​x​2​y + 3x a c ad + cb ​ __b ​+ __ ​  d ​= ​  ______ bd ​ activity We cannot add or subtract ACTIVITIES boy unlike terms. and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 2 ALGEBRA 19 Mind the Gap Mathematics

2 Unit 2.3 Multiplication and division c Dividing by ​ __ ​is d Check that you know these facts: the same as multiplying d NB number × positive number positive NB = positive answer. 3x × 5y2 = 15xy2 PAY SPECIAL ATTENTION by __ ​ c . ​ positive number × negative number = negative answer. 3x × –5y2 = –15xy2 negative number × positive number = negative answer. –3x × 5y2 = –15xy2 negative number × negative number = positive answer. –3x PAY SPECIAL × –5y2 = 15xy2 ATTENTION NB NB HINT! __ hint a c ac ​  b ​× __ ​  d ​= ​ __ bd ​ PAY SPECIAL ATTENTION NB NB e.g. 3 EG - worked examples HINT! __ 6x ​× __ 3 18x ​  5z ​= ​ ____ ​  7yexams 35yz ​ HINT! hint Multiply numerators and multiply hint denominators. Simplify if possible. PAY SPECIAL ATTENTION NB e.g. 4 e.g. EG - worked examples EG - worked examples activity6x3z + 3(2y) ________18xz + 6y 6x 3 ___ __________ ​ __ ​+ ​  8yexams 12z ​= ​  24yz ​and = ​  boy ACTIVITIES 24yz ​ girl-left and HINT! hint Find the lowest right of page a c a d ad 3 ​ __b ​ ÷ ​ __d​ = ​ __b ​ × __ activity ​  c ​ = ​ __ bc ​ exams common multiple PAY SPECIAL ATTENTION NB NB or denominator (LCM) The distributive law: first. 8 and 12 have an activity Step by step comment ACTIVITIES boy comment e.g. LCM of 24. EG - worked examples c(a + b) = c × a + c × b = ac + bc HINT! hint and girl-left and right of page activity activity exams e.g. 5 EG - worked examples ACTIVITIES boy and girl-left and right of page Step by step comment comment –3x(5x – 6y) = –15x2 + 18xy exams activity PAY SPECIAL ATTENTION NB L NB F (x + y)(a + b) = ax + bx + ay + by activity I O activity HINT! Step by step comment comment ACTIVITIES boy hint ACTIVITIES boy and girl-left and and girl-left and right of page right of page activity activity e.g. 6 EG - worked examples (2x + y)(3x – 2y) = 6x2 – 4xy + 3xy – 2y2 = 6x2 – xy – 2y2 exams Step by step comment comment Step by step comment activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 20 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 PAY SPECIAL ATTENTION NB NB 2.4 Factorising It is useful to use ‘FOIL’ What does itHINT!mean to ‘factorise an expression’? (multiply FIRST terms, OUTER hint terms, INNER terms and It means to write the expression as a product of its factors. LAST terms). e.g. 7 EG - worked examples Here are ways to factorise an expression: exams 1. Find the common factor: 9​ x2 ​ ​– 6xy2 = 3x(3x – 2y2) 2. Factorise by grouping in pairs and then finding a common factor: 3xy – 2x + 3y – 2 activity = 3xy + 3y – 2x – 2 ACTIVITIES boy and girl-left and = 3y(x + 1) – 2(x + 1)page right of When you take out a negative factor, signs activity in bracket change. = (x + 1)(3y – 2) 3. Factorise a difference of two squares: 16x2 – yStep2 by=step(4x – y)(4xcomment + y)comment PAY SPECIAL ATTENTION 4. Factorise a difference of two cubes: NB 8x3 – y3 = (2x – y)(4x2 + 2xy + y2) 5. Factorise a sum of two cubes: 27a3 + 64b3 = (3a + 4b)(9a2 – 12ab + 16b2) 6. Factorise a trinomial: HINT! 9x2 + 5x – 4 = (9x – 4)(x + 1) hint When factorising, always take out a common factor e.g. first, if possible. EG Then lookexamples - worked factorise the difference of two to squares or the sum/difference of two cubes or a trinomial. exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment © Department of Basic Education 2015 UNIT 2 ALGEBRA 21 Mind the Gap Mathematics

2 Unit 2.5 Notes onNBfactorising a PAY SPECIAL ATTENTION NB trinomial HINT! hint The following steps will explain how to factorise a trinomial: PAY SPECIAL ATTENTION NB e.g. 8 NB EG - worked examples Factorise 3x2 + 11x + 6 Step 1: Multiply the coefficient of x2 and the constant value exams (+3 × +6 = 18). HINT! Step 2: Write down all the products of 18: 10 × 1 hint 9×2 6×3 We write the 9x first and the activity 2x second because: There Step 3: We will use 9ACTIVITIES × 2,boybecause 9 + 2 = 11, the middle term. e.g. is a common factor between and girl-left and EG - worked examples Step 4: We write theright middle of page term (11x) as 9x + 2x 3x and 9x. 2 activity ∴ 3x2 + 11x + 6 There is a common factor = 3x 2 + 9x + 2x + 6......we wrote the 9x first, followed by the 2x PAY SPECIAL ATTENTION between 2x and 6. NB NB exams Step 5: We now group the four terms and factorise by taking out a comment common Step by step factor. comment 3x2HINT! + 9x + 2x +6 hint = 3x(x + 3) + 2(x + 3) = (x + 3)(3x + 2) NB activity PAY SPECIAL ATTENTION e.g. 9NB EG - worked examples Factorise 4x2 + 9x – 13 Step 1: Multiply the coefficient of x2 and the constant value ACTIVITIES boy exams and girl-left and (+4 × –13 = –52). right of page Step 2: Write down all the products of 52: 52 × 1 activity HINT! hint 26 × 2 13 × 4 We write the –4x first and the activity Step 3: We will use 13 × 4, because 13 – 4 = 9, the middle term. 13x second because: ACTIVITIES boy and girl-left and e.g. Step 4: We write theright middle term (9x) as –4x +13x of page EG - worked examples There is a common factor Step by step comment comment between 4x2 and –4x. There activity ∴ 4x + 9x – 13 2 is a common factor between = 4x2 – 4x + 13x – 13......we write the –4x first, followed 13x and –13 . by the 13x PAY SPECIAL ATTENTION exams NB5: WeStep by Step now step groupcommentthe fourNBterms and factorise by taking out a comment common factor. 4x2 – 4x + 13x – 13 HINT! hint = 4x(x – 1) + 13(x – 1) = (x – 1)(4x + 13) activity e.g. 10 Factorise 8x2 – 18x + 9 EG - worked examples ACTIVITIES boy Step 1: Multiply the coefficient of x2 and the constant value and girl-left and right ofexams page (+8 × +9 = 72). activity Step 2: Write down all the products of 72: 72 × 1 36 × 2 24 × 3 activity 18 × 4 ACTIVITIES boy 12 × 6 Step by step comment comment and girl-left and right of page 9×8 activity 22 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics Step by step comment comment

NB N Unit 2 HINT! Step 3: We will use 12 × 6, because –12 – 6 = –18, the middle term. hint Step 4: We write the middle term (–18x) as –12x – 6x or –6x – 12x ∴ 8​x​2​– 18x + 9 In this example, we can write = 8x2 – 12x – 6x + 9......we write the –4x first followed by the 13x –12x first and then –6x or –6x Step 5: We now group the four terms and factorise by taking out a e.g. EG - worked examples first and then –12x. We have a common factor between 8x2 common factor and –12x and between –12x 8x2 – 12x – 6x + 9 and 9. We have a common = 4x(2x – 3) – 3(2x – 3) exams factor between 8x2 and –6x and = (2x –3)(4x – 3) between –6x and 9. Activity 1 Factorise each of the following completely: activity 1. 12x + 17x + 6 2 2. 5x – 23x – 10 2 3. 9x + 5x – 4 2 ACTIVITIES boy and girl-left and 4. 12x2 – 11x + 2 5. 5x2 – 45 6. 2x3 + 16 right of page 7. 6x3 – 13x2 + 5x [16] activity Solutions 1. 12x2 + 17x + 6 Step by step comment co = 12x2 + 9x + 8x +6...........12 × 6 = 72 and 72 = 9 × 8 (9 + 8 = 17) = 3x(4x + 3) + 2(4x + 3) = (4x + 3)(3x + 2)33 (2) 2. 5x2 – 23x – 10 = 5x2 – 25x + 2x – 10............5 × –10 = –50 and 50 = 25 × 2 (–25 + 2 = –23) = 5x(x – 5) + 2(x – 5) = (x – 5)(5x + 2)33 (2) 3. 9x2 + 5x – 4 = 9x2 + 9x – 4x – 4..............9 × –4 = –36 and 36 = 9 × 4 (9 – 4 = 5) = 9x(x + 1) – 4(x + 1) = (x + 1)(9x – 4)33 (2) 4. 12x2 – 11x + 2 = 12x2 – 3x – 8x + 2..............12 × 2 = 24 and 24 = 8 × 3 (–8 – 3 = –11) = 3x(4x – 1) – 2(4x – 1) = (4x – 1)(3x – 2)33 (2) 5. 5x2 – 45............................Common factor of 5 = 5(x2 – 9) ............................Difference of two squares = 5(x – 3)(x + 3)33 (2) 6. 2x3 + 16............................Common factor of 2 = 2(x3 + 8) 3............................Sum of two cubes = 2(x + 2)(​x2​ ​– 2x + 4) 33 (3) 7. 6x3 – 13x2 + 5x............................Common factor of x = x(6x2 – 13x + 5) 3............................Trinomial factorising = x(6x2 – 3x – 10x + 5) .............. 6 × 5 = 30 and 30 = 10 × 3 (–3 – 10 = –13) = x[3x(2x – 1) – 5(2x –1)] = x[(2x – 1)(3x – 5)] 33 (3) [16] © Department of Basic Education 2015 UNIT 2 ALGEBRA 23 Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB 2 Unit HINT! 2.6 Quadratic equations hint e.g. 11 EG - worked examples Here are some quadratic equations: 1.exams x2 + 5x PAY+SPECIAL NB 6 =ATTENTION 0 NB 2. 3x – 7x = 12 2 3. 3x(x – 9) + 2 = 5x 3x × x = 3x2 so the equation has x2 as its highest power of x HINT! hint activity Quadratic equations can be put into the standard form: ​ax​2​ + bx + c = 0 ACTIVITIES boy and girl-left and right of page e.g. activity 12 EG - worked examples 1. x2 + 5x + 6 = 0 So a = 1, b = 5 and c = 6 exams2 2. 3x – 4x = 12 comment 3x2 – 4xStep–by step 12 = 0 comment So a = 3, b = –4 and c = –12 3. 3x(x – 9) + 2 = 5x 3x2 – 27x + 2 – 5x = 0 activity 3x2 – 32x + 2 = 0 So a = 3, b = –32 and c = 2 ACTIVITIES boy and girl-left and right of page 2.6.1activity Solving a quadratic equation by If (A) × (B) = 0, then either A = 0 factorising: PAY SPECIAL ATTENTION NB NB OR B = 0. What does it mean to ‘solve a quadratic equation’? Step by step comment comment It means to find the unknown value(s) of x in a quadratic equation. The x-values in aHINT!quadratic equation are also called the roots of the equation hintthe equation is equal to zero. when e.g. 13 EG - worked examples Solve for x: examsx2 – 7x = –10 x2 – 7x + 10 = 0 Write in standard form and equal to 0 x2 – 5x – 2x + 10 = 0 Factorise the trinomial x(x–5) –2(x–5) = 0 (x – 5)(x – 2) = 0 activity ∴x – 5 = 0 or x –ACTIVITIES and 2 =and0 boy girl-left right of page x=5 ∴x = 2 activity Activity 2 Step by step comment comment Solve for x: 1. x(x + 3) = 0 2. x(2x – 5) = 12 3. 2​x2​+ x − 6 = 0 _____ 1 4. 2​x​2​= 32 5. 3x + __ ​  x ​= 4, x ≠ 0 6. 2​√x − 3 ​= x − 3 [22] 24 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 Solutions 1. x(x + 3) = 0 We have a product = 0. Therefore put each factor = 0 x = 0 or x + 3 = 0 x = 0 3 or x = –3 3 (2) 2. x(2x – 5) = 12 We need a product = 0. Therefore, multiply out brackets and write in standard form with all the terms on one side and equal to 0 (2x + 3)(x – 4) = 0 Factorise 2x + 3 = 0 or x – 4 = 0 Put each factor = 0 2x = –3 or x = 4 3 x = –​ __2 ​ 3 x = 4 3 (2) 3. 2​x2​+ x − 6 = 0 (2x − 3)(x + 2) = 0 33 ∴ 2x = 3 or x = − 2 Find the solutions by putting each factor equal to zero ∴ x = ​  3_ ​  or  x =  −2 33 (4) 2 4. 2x2 = 32 2x2 – 32 = 0 Write in standard form with all the terms on one side and equal to 0 x2 – 16 = 0 3 Divide every term on both sides by 2 (x + 4)(x – 4) = 0 3 Factorise (the difference of two squares) ∴ x + 4 = 0 or x – 4 = 0 ∴ x = –4 3 or ∴ x = 4 3 (4) 1 5. 3x + __ ​  x ​= 4, x ≠ 0 Multiply through by x to get rid of the denominator 3x + 1 = 4x Write in standard form with all the terms on one side and 2 equal to 0 3x2 – 4x + 1 = 0 3 Factorise (the trinomial) (3x – 1)(x – 1) = 0 33 ∴ 3x – 1 = 0 or x – 1 = 0 ∴ 3x = 1 or x = 1 ∴ x = ​  1_ ​ 3 or ∴ x = 1 3 (5) 3 _____ 6. 2​√x_____ − 3 ​= x − 3 Square both sides ​​(2​√x − 3 ​  )​​ ​= ​​(x − 3 )2​​ ​ 2 4​(x − 3 )​= ​(x − 3 )(​​ x − 3 )​    3 4x − 12 = ​x2​ ​− 6x + 9           0 = ​x2​ ​− 10x + 21 3           0 = (​x − 7 )(​​ x − 3 )​3 ∴ x − 7 = 0    or  x − 3 = 0 ∴ x = 7   3   or x = 3 3   (5) [22] Check your answers: x=7 _____ __ LHS = 2​√7 − 3 ​= 2​√4 ​= 2(2) = 4   RHS = 7 − 3 = 4    ∴ x = 7 is a solution x=3 _____ __ LHS = 2​√3 − 3 ​= 2​√0 ​= 0         RHS = 3 − 3 = 0    ∴ x = 3 is a solution If a quadratic equation cannot be factorised, there are other ways to find the roots or solutions. Sometimes the solutions do not exist! © Department of Basic Education 2015 UNIT 2 ALGEBRA 25 Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB 2 Unit HINT! hint 2.6.2 Completing the square e.g. 14 EG - worked examples Write y = 3x2 + 12x + 9 in the form y = a(x + p)2 + q. exams To do this, we can follow a few steps: y = 3x2 + 12x + 9 To complete the square, the coefficient of x2 must be one (1). activity We take 3 out as a factor so that ACTIVITIES boy and girl-left and the coefficient of x2 is one. right of page y = 3[x2 + 4x + 3] Take (half of the coefficient of x) activity and square the number. Add and subtract this answer to keep the equation balanced. Step by step comment comment The coefficient of x is +4. Halving 4 = 2. (+2)2 =4. y = 3[x2 + 4x + (+2)2 + 3 – (+2)2] Thus, add 4 and subtract 4. y = 3[x2 + 4x + 4 + 3 – 4] = 3[x2 + 4x + (+2)2 + 3 – 4] Now we can complete the square { (x + 2)2 y = 3[(x + 2)2 + 3 – 4] by factorising y = 3[(x + 2)2 – 1] y = 3(x + 2)2 – 3 We have now written y = 3x2 + 12x + 9 as y = ​3(x + 2)​2​– 3. Therefore, we have written y = ax2 + bx + c in the form y = a(x + p)2 + q with a = 3, p = 2 and q = –3. The quadratic equation y = 3x2 + 12x + 9 helps us to identify the y-intercept, while the form y = 3(x + 2)2 – 3 helps us to identify the turning point. Refer to graphs in unit 4 on Functions. 26 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 Activity 3 1. What term can be added to the following equations to make a complete square? b a) 0 = x2 – 8x + ? b) y = x2 + 9x + ? c) y = x2 – ​ __a ​x + ? 2. Solve for x by using the method of completing the square. a) –3x2 + 5x + 4 = 0 b) ax2 + bx + c = 0 [17] Solutions 1a) 0 = x2 – 8x + (–4)2 Use half of - 8 squared b) (2 ) y = x2 + 9x + ​​​  9_ ​  ​​ ​ 2 0 = x2 – 8x + 16 0 = (x – 4)2 3 (1) y = x2 + 9x + ​ 81 __ ​ 4 y = ​​ x + __ b b ​  2 ​  ​​ ​ 33 (2) ( 9 2 ) c) y = x2 – __ ​  a ​x + ? Use half of –​ __a ​squared y = x2 – __ b ​  a ​x + ​​ – ​ __ ( b) __ 2 __ bb 2a ​  ​​ ​ = x – ​  a ​ x + ​  a ​ 2 ( y = ​​ x – __ b ) ​  2a ​  ​​ ​ 33 2 (2) 2a) –3x2 + 5x + 4 = 0 Divide each term on both sides by –3 ​x2​− ​  5_ ​x − ​  4_ ​= 0 Get the constant value on its own on RHS 3 3 ​x2​− ​  5_3 ​x = ​  4_3 ​ 3 Add (½ coefficient of x term)2 to both sides ( ) x2 – ​ 5_3 ​x + ​​​  5_6 ​  ​​ ​= ​  4_3 ​ + ​​ __ 2 (5 )2 ​  6 ​  ​​ ​3 Complete the square by factorising the LHS __________ − b ± ​√​b​2​− 4ac ​ ∴ x = ​ _______________    2a ​ ( ) ​​ x − ​  5_6 ​  ​​ ​ 3 = ​  4_3 ​+ ​  25 2 __ ​ 36 Add the constant values on RHS is the formula that we ​​(x − ​  _5 ​  )​​ ​= ​  48 + 25 use to solve any quadratic 2 ______ ​ 6 36 equation, y=ax²+bx+c ​​(x − ​  5_6 ​  )​​ ​= ​  73 2 __ ​ 3 Take the square root of both sides where a = coefficient of 36___ x², x − ​  _5 ​=  ± ​ ​  73 6 √ __ ​ ​ ___ 36 Get x on its own b = coefficient of x and c = constant ​√___ x = ​  ​ ± ​  73 ​ ​ _5 Separate the two values of the square root value/term. 6 ___6 ___ √ ​ 73 ​ √ 73 ​ 5 x = ​  _6 ​+ ​  ___ ​ or x = ​  _5 ​− ​  ​___ ​33 (6) Use a calculator to get each value 6 6 6 x = 2,2573… or x = –0,5906… ∴ x = 2,26 or x = −0,59 Rounded off to two decimal places These are the roots of the equation. b) ax2 + bx +c = 0 Divide each term on both sides by a b c ​x2​ + ​ __a ​x + ​ __a ​= 0 Get the constant term on its own b c ​x2​+ ​ __a ​x = − ​ __a ​3 Add (½ coefficient of x term)2 to both sides b ( ) ( ) x2 + ​ __a ​x + ​​ __ b 2 ​  2a ​ ​ ​= ​​ __ b 2 c ​  2a ​ ​ ​− ​ __a ​3 This complete a perfect square on LHS ( ) b 2 ​b ​ ​ __ c 2 ​​ x + ​ __ 2a ​  ​ ​= ​  ___ 4​a2​ ​ a ​− ​  ​ 3   Add the constant values on RHS ( ​​ x + ​ __ ) b 2 ___ ​b​2​ ___ 4ac 2a ​  ​ ​= ​  4​a2​ ​​− ​  4​a2​ ​​ ( ​​ x + ​ __ ) b 2 ​b2​ ​− ______ 2a ​  ​ ​= ​  4​a​2​ ​ 3   4ac ________ _______ √( ) √ 4​a​​ b 2 ​b​ ​− ​​ x + ​ __ 4ac 2 ______ ​ 2a ​  ​ ​ ​ =  ± ​ ​  2 ​ ​ Take square root of both sides _______ ​= ± ​√​  ​b​ ​− __b ______ 4ac 2 ∴ x + ​  2a ​ ​ 2a _______ 2a ​± ​√ ​  2a ​ ​ Get x on its own b ​b​ ​− ∴ x =  − ​ __ 4ac 2 ______ _______ ∴ x =  − b ± ​√​  ​b​ ​− 4ac 2 ______ ​ ​  33 (6) Write the two fractions as one fraction [17] 2a © Department of Basic Education 2015 UNIT 2 ALGEBRA 27 Mind the Gap Mathematics

2 Unit PAY SPECIAL ATTENTION NB NB 2.6.3 Solving quadratic equations using the formula HINT! Some quadratic equations cannot be factorised, but there is another way hint to find the roots of the equation. e.g. 15 EG - worked examples Can you find factors for this quadratic equation: x2 – 5x + 3 = 0? exams There are no rational numbers that can be multiplied to get 3 and added to get 5, therefore use the quadratic formula to solve the equation. activity The standard form of theboyquadratic equation ax2 + bx + c = 0 is used ACTIVITIES and girl-left and from which the formula is derived: right of page _______ activity − b ± ​√​b2​ ​− 4ac ​ x = ​ ____________    2a ​ For x2 – 5x Step + by3step= 0 a = 1, b comment = –5 and c = 3 comment Substitute these values for a, b and c in the formula: _____________ − ( − 5) ± ​√ (− 5​)​2​−    ___________________ 4(1)(3) ​ x = ​        2(1) ​ _______ 5 ± ​√25 − 12 ​ x = ___________ ​  2 ​ ___ 5 ± ​√13 ​ x = _______ ​  2 ​ ___ ___ 5 + ​√13 ​ 5 − ​√13 ​ x = _______ ​  2 ​ OR x = _______ ​  2 ​ 28 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 Activity 4: Interpret a graph Solve for x (correct to two decimal places): 4x2 – 8x = 7 2x (3x + 5) – 11 = 0 [9] Solutions 1. 4x2 – 8x = 7 Write the equation in standard form PAY SPECIAL ATTENTION (ax2 + bx + c = 0) NB 4x2 – 8x – 7 = 0 List the values of a, b and c a = 4; b = _______ – 8; c = –7 Write down the formula − b ± ​√​b2​ ​− 4ac ​ x = ​  ____________    2a ​ Substitute the values for a, b and c into the formula. HINT! _____________ – ​(–8 )​± ​√(–8​)2​ ​–    ___________________ 4(4)(–7) ​ hint x = ​    2(4) ​33 Simplify the value under the square root sign ____ These roots are 8 ± ​√176 ​ x = _______ ​  8 ​ Separate the positive and negative value of irrational. Unless the ____ ____ the square root e.g. question asks forEG - worked examples decimal values, leave 8 + ​√176 ​ 8 – ​√176 ​ x = ​ _______ 8 ​ or x = _______ ​  8 ​ 3 Answers in surd form them in surd form (the x = 2,66 3 or x = –0,66 3 Answers correct to two decimal places (5) square root form). exams 2. 2x(3x + 5) – 11 = 0 Write the equation in standard form 6x2 + 10x – 11 = 0 (ax2 + bx + c = 0) _________ –10 ± ​√100 + 264 ​ x = ​ _____________    12 ​ 33 ____ PAY SPECIAL ATTENTION NB = –10 ± ​  √​____ 364 ​ NB ​ 3 12 activity ___ –5 ± ​√91 ​ = _______ ACTIVITIE ​  6 ​ 3 (4) and girl-le HINT! right of pa hint [9] activity e.g. 16 EG - worked examples 2 If __ ​  3 exams ​is a root of the equation 12x2 – kx – 8 = 0, determine the value of k. Step by step com Solution 2 2 2 If __ ​  3 ​is a root of the equation, then x = __ ​  3 ​. Therefore, we can substitute x = __ ​  3 ​into the equation: activity 12x2 – kx – 8 = 0 ACTIVITIES boy ∴ 12​​(​  3 ​  )​​ ​– k​(​  3 ​  )​– 8 = 0right of page and girl-left and 2 __ 2 2 __ activity 16 2 ​ __ __ 3 ​– ​  3 ​k – 8 = 0 8 ∴ – ​  2__3 ​k = __ ​  3 ​ ∴ k = –4comment Step by step comment © Department of Basic Education 2015 UNIT 2 ALGEBRA 29 Mind the Gap Mathematics

2 Unit 2.7 Quadratic inequalities Solving quadratic To solve quadratic inequalities PAY SPECIAL ATTENTION • NB NB Get the inequality into the standard form ax2 + bx + c > 0 or ax2 + bx + c < 0 or ax + bx + c ≤ 0 or ax + bx + c ≥ 0 2 2 • If the value of a < 0, multiply the equation by –1. If we multiply an inequality by a negative, the • Factorise HINT! the inequality when this is possible or hint inequality sign swaps: • use the quadratic formula to obtain the critical values. If -5 < 7, then after being multiplied by (-1), ∴ 5 > –7 e.g. 17 Solve for x if x2 < 25 EG - worked examples Method 1 exams x2 < 25 x2 – 25 < 0 (x – 5)(x + 5) < 0 The critical values are where the expression x²–25 is equal to zero. activity Therefore the critical values ACTIVITIES boy are –5 and 5. and girl-left and right of page We now indicate –5 and 5 on a number line. activity –5 5 Step by step comment comment We know that the expression x²–25 = 0 at –5 and 5. We can indicate this on the number line. 0 0 PAY SPECIAL ATTENTION NB NB –5 5 Our next step is to choose values less than –5, values between –5 and 5 HINT! and values greater than 5 and substitute it into the expression x²–25. If the hint answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line. If x = –10, then (–10)² – 25 = 75 > 0 ∴ + + 0 – 0 + e.g. If x = –6, EG - worked examples then (–6)² – 25 = 11 > 0 ∴ + –5 5 If x = –3, exams then (–3)² – 25= –16 < 0 ∴ – We have to solve for x where x2 – 25 < 0. The solution on the number line If x = 2, is the interval where we see a negative. This happens between –5 and 5. then (2)² – 25= –21 < 0 ∴ – + 0 – 0 + If x = 7, then (7)² – 25 = 24 > 0 ∴ + –5 5 If x = 9,activity 0 0 then (9)² – 25 = 56 > 0 ∴ + ACTIVITIES boy and girl-left and Therefore the solution is : –5 < x < 5 right of page activity 30 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 ORNB ALTERNATIVE PAY SPECIAL ATTENTION METHOD NB by using a rough sketch of the parabola: Above the x-axis y is positive y y = ax2 + bx + c ... y is equal to what is On the x-axis HINT!y is zero defined in terms hint of x Below the x-axis y is negative x e.g. 18 Solve for x if x2 < 25 EG - worked examples x2 < 25 x2exams – 25 < 0 Get 0 on RHS (x – 5)(x + 5) < 0 Factorise LHS Critical values of x: –5 and 5 Make a rough sketch of a parabola activity ACTIVITIES boy and girl-left and right of page If (***)(***) < 0 (it means where y activity x is negative) Read –5 5 off the x values of the graph under the x-axis Step by step comment comment –5 < x < 5 Activity 5 Solve for x if 1. (x+3)(x – 5) ≤–12 2. –x ≤ 2x2 – 3 [10] Solutions 1. (x+3)(x – 5) ≤ –12 ​ x2​ ​– 2x – 15 +12 ≤ 0 Get into the standard form (ax2 + bx + c ≤ 0) ​x​2​– 2x –3 ≤ 0 3 Factorise the trinomial: (x –3)(x + 1) ≤ 0 3 Critical values: x = 3 and x –1 We now indicate 3 and –1 on a number line. –1 3 We know that the expression x2 – 2x –3 = 0 at x = 3 and x = –1. We can indicate this on the number line. + 0 – 0 + –1 3 © Department of Basic Education 2015 UNIT 2 ALGEBRA 31 Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB 2 Unit HINT! Our next step is to choose values less than –1, values between –1 hint and 3 and values greater than 3 and substitute it into the expression ​ x​2​– 2x –3. If the answer is positive, then we indicate + on the number line. If the answer is negative, we indicate – on the number line If x = –10, then (–10)² e.g. 2(–10) – 3 =117 >EG0- ∴ + examples worked + 0 – 0 + If x = 1, then (1)² – 2(1) – 3 =–4 < 0 ∴ – –1 3 x = 5, then (5)² –2(5) – 3 If exams We have to solve for x where x2 – 2x –3 ≤ 0. The solution on the = 12 > 0 ∴ + number line is the interval where we see zero and a negative. This happen when the x values are less than or equal to 3 and are also more than or equal to –1. + 0 – 0 + activity –1 3 ACTIVITIES boy and girl-left and 3 right of page Whenever we multiply or divide an Therefore the solution is : − 1 ≤ × ≤ 3 33 (5) activity inequality by a negative, the inequality sign changes i.e. 2. –x ≤ 2x2 – 3 the less or equal to sign –2x2 – x + 3 ≤ 0 Get into the standard form (ax2 + bx + c ≤ 0) changes to a greater or –2x x 3 0 equal to sign. ​  –1 ​– __ ___ ​  –1 ​+ __ ​  –1 ​ ≥ ​ __ –1 ​ Divide both sides by –1 to change the Step by step comment comment coefficient of x2 to a positive 2x2 + x – 3 ≥ 0 3 Factorise the trinomial (2x + 3)(x – 1) ≥ 0 3 Critical values: –3 x = __ ​  2 ​and x = 1 –3 We now indicate __ ​  2 ​ and 1 on a number line. + 0 – 0 + 3 –​ __ 2 ​ 1 PAY SPECIAL ATTENTION –3 NB We knowNBthat the expression 2x2 + x – 3 = 0 at x = ​  __ ​ and x = 1. 2 We can indicate this on the number line. 0 0 HINT! 3 –​ __ 2 ​ 1 hint Our next step is to choose values less than ​ –3 __ ​, values between ​ –3 __ ​ 2 2 If x = –10, then 2(–10)² and 1 and values greater than 1 and substitute it into the expression 2x2 + x – 3. If the answer is positive, then we indicate + on the number e.g. +(–10) – 3 = 187 > ∴ +examples EG -0worked line. If the answer is negative, we indicate – on the number line If x = 0, then 2(0)² – 0 –3 = –3 < 0 ∴ – + 0 – 0 + x = 3, then 2(3)² + 3 – 3 If exams = 18 > 0 ∴ + 3 3 –​ __ 2 ​ 1 activity 32 UNIT 2 ALGEBRA © Department of Basic Education 2015  ACTIVITIES boy Mind the Gap Mathematics and girl-left and

Unit 2 We have to solve for x where 2x2 + x – 3 ≥ 0. The solution on the number line is the interval where we see zero and a positive. This happens for the x values less than or –3 equal to ​ ___ ​and for the x values greater than or equal to 1. 2 + 0 – 0 + 3 –​ __ 2 ​ 1 Therefore the solution is : x ≤ −​  3_ ​  or  x ≥ 1 33 (5) 2 OR ALTERNATIVE METHOD by using a rough sketch of the parabola: –x < 2x2 – 3 Get into the standard form ax2 + bx + c < 0 –2x2– x + 3 < 0 Divide both sides by –1. –2x ___ x ​  –1 ​– __ ​  –1 ​+ ​ 3 __ ​> 0 This is necessary to draw the rough sketch –1 of a “positive” parabola 2x2 + x – 3 > 0 3 With 0 on RHS (2x + 3)(x – 1) > 0 3 Factorise LHS Critical values of x: ​  –3 __ ​and = 1 Make a rough sketch of a parabola 2 If (***)(***) > 0 (it means where y is positive) x Read off the x values of the graph above the 3 x-axis –​ __ 2​ 1 3 ∴ x < – ​ _3 ​ or x > 1 33 (5) 2 [10] © Department of Basic Education 2015 UNIT 2 ALGEBRA 33 Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB 2 Unit 2.8 Simultaneous equations HINT! PAY SPECIAL ATTENTION hint NB NB e.g. 19 EG - worked examples HINT! Solve exams for x and y simultaneously: hint y + 2x – 2 = 0 and 2x² + y² = 3yx When you divide, you In this example, a quadratic equation and a linear equation must be solved sometimes need to round off simultaneously. Use the following steps: e.g. to the closest EG numbers that - worked are easier to divide. examples activity Step 1: Use the linear ACTIVITIES equation boy to make one of the unknowns the subject of the equation (i.e. get xpageor y alone on one side of the equation). and girl-left and right of Step 2: activity Substitute x or y (whichever is the subject of the equation) into the exams quadratic equation. The equation will contain only one unknown. PAY SPECIAL ATTENTION NB NBthe one unknown. Step 3: Solve Step by step comment comment Step 4: Substitute the solved unknown into the linear equation to solve for the other unknown. HINT! hint activity Solution ACTIVITIES boy Always label the equations: y + 2x – 2 = 0 ………….eqn (1) and girl-left and PAY SPECIAL ATTENTIONright of page NB equation 1 (eqn 1) and 2x² + y² =NB 3yx……………...eqn (2) e.g. activity equation 2 (eqn EG - worked examples 2) Step 1: y + 2x – 2 = 0..............from eqn (1) ∴ y = 2 – 2x................eqn (3) exams HINT! Step 2: Substitute eqn (3) in eqn (2) hint Step by step comment 2x + y2 = 3yx comment 2 ∴ 2x2 + ​(2 – 2x)​2​= 3x(2 – 2x) If the coefficient of y in the linear equation is one, get Step 3: 2x2 + (2 – 2x)(2 – 2x) = 3x(2 – 2x) e.g. y alone on one sideEG - worked examples of the equation. If the coefficient 2x2 + 4 – 8x + 4x2 = 6x – 6x2 activity of x in the linear equation is 12x2 – 14x + 4 = 0 one, get x alone on one side ACTIVITIES boy ÷ 2 ∴ 6x2 – 7x + 2 = 0 exams of the equation. This way, and girl-left and you will not have to deal with right of page ∴ (3x – 2)(2x – 1) = 0 2 1 activity fractions. ∴ x = ​ __3 ​or x = __ ​  2 ​ 2 1 Step 4: Substitute x = __ 3​  ​ in eqn (3) Substitute x = __ 2​  ​ in eqn( 3) (2 ) ∴ y = 2 – 2 ​​ __3 ​  ​ = __ ​  3 ​ 2 ∴ y = 2 – 2 ​__ (1 ) ​  2 ​  ​= 1 activity Step by step comment comment ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 34 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 2 e.g. 20 EG - worked examples In order to determine the points Given the functions y = ​ __ 6 of intersection of two x ​and y = x − 1, find the coordinates of the graphs, we solve the exams points of intersection of the two graphs algebraically. equations of the graphs 6 simultaneously. y = ​ __ x ​ … eqn (1) y=x–1 … eqn (2) activity Substitute eqn (2) into eqn (1): ACTIVITIES boy (Wherever there is a y,and replace girl-left and right of page it with (x – 1), using brackets) 6 x − 1 = __ ​  x ​ activity ∴ ​x​​− x = 6        LCD = x 2 ∴ ​x2​− x − 6 = 0 comment ∴ (x − 3)(x + 2) Step by step comment ∴ x = 3 and x = − 2 Subst. x = 3 in eqn(2):      Subst. x = − 2 in eqn(2): y = 3 − 1 = 2                          y =  − 2 − 1 =  − 3 ∴ (3;2) and (− 2; − 3) are the points of intersection Both these points satisfy the equations. The graphs of the two equations will intersect at two points. y 8 y=_ 6 7 ​  x ​ 6 y=x–1 5 4 3 2 1 x –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 –4 –5 –6 –7 –8 © Department of Basic Education 2015 UNIT 2 ALGEBRA 35 Mind the Gap Mathematics

2 Unit Activity 6 Solve the following equations simultaneously. 1. 2x + y = 3 and x2 + y + x = y2 −6 2. y = ​ ____ x + 1 ​  − 2 and y =  − 3x + 2 [14] Solutions 1. 2x + y = 3 ..................... eqn (1) x2 + y + x = y2 .............. eqn (2) y = –2x + 3 3 .............. eqn (3) Use the linear equation (1) to write y alone on one side of the equation. Substitute eqn (3) into eqn (2), to eliminate the y variable. x2 + (–2x + 3) + x = (–2x + 3)2 3 Simplify both sides. x – x + 3 = 4x – 12x + 9 2 2 Factorise the trinomial. 0 = 3x2 – 11x + 6 3 0 = (3x – 2)(x – 3) 3 ∴ 3x – 2 = 0 or x – 3 = 0 ∴ x = ​ _2 ​ or ∴ x=3 3 3 Substitute these two values of x into eqn (3) to find the values for y: 2 Subst x = __ ​  3 ​in eqn (3) Subst x = 3 in eqn (3) () ∴ y = –2 ​​  2__3 ​  ​+ 3 = ​  5__3 ​ 3 ∴ y = –2(3) + 3 = –3 3 (7) So there are two solutions: ​__ ( ) 2 5 ​  3 ​; __ ​  3 ​  ​and (3; –3) −6 2. y =  ​ ____ x + 1​ − 2 and y =  − 3x + 2 −6 y =  ​ ____ x + 1 ​ − 2.....................(eqn 1) y = –3x + 2.......................(eqn 2) y is alone on one side of both equations. −6 ∴ y =  ​ ____ x + 1 ​ − 2 = –3x + 2 3 ............LCD = x + 1 ∴ –6 –2(x + 1) = –3x(x + 1) + 2(x + 1) 3 ∴ –6 –2x –2 = –3x2 – 3x + 2x + 2 ∴ 3x2 – x – 10 = 0 3 ∴ (3x + 5)(x – 2) = 0 3 5 ∴ x = – __ ​  3 ​or x = 2 3 5 Subst x = – __ ​  3 ​in eqn (2) Subst x = 2 in eqn (2) ( ) y = –3 ​– __ 5 ​  3 ​  ​+ 2 = 7 3 y = –3(2) + 2 = –4 3 (7) [14] 36 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 2.9 The nature of the roots 2.9.1 Determine the nature of the roots ∆ = b2 – 4ac of any quadratic equation ax2 + bx + c = 0 can be found by The roots________ The value of the ∆ –b ± ​√b2 – 4ac ​ determines the x = ___________ ​    2a ​ nature of the roots • The roots of a quadratic equation are the x-values when the equation is zero. • The roots are the x-intercepts of the graph. • When you are asked to ‘determine the nature of the roots of an equation’, you are NOT asked to solve the equation. Summary: To find the nature of the roots of a quadratic equation ax2 + bx + c = 0, look at the value of D, the discriminant. • If Δ < 0 : The roots are non-real/no real roots. • If Δ = 0 : There are two equal, real and rational roots. • If Δ > 0 : There are two real roots which may be rational or irrational. • If D is a perfect square, the roots are rational. • If D is not a perfect square, then the roots are irrational The nature of the roots also tells us about the x-intercepts of the graph of the quadratic equation. Nature of roots Graphs Δ<0 Roots are non-real. There are no x-intercepts. Δ=0 Roots are real and equal. There is only one x-intercept and it is at the turning point of the graph. Δ>0 Roots are real and unequal (two roots): If Δ is a squared rational number, roots are rational. If Δ is not a squared number, the roots are irrational. © Department of Basic Education 2015 UNIT 2 ALGEBRA 37 Mind the Gap Mathematics

HINT! 2 Unit hint e.g. 21 EG - worked examples ___ − 6 ± ​√25 ​ x = ________ 1.exams ​  4 ​ Δ = 25 ∴ Δ > 0, so there are two real roots. ___ We can see that 25 is a perfect square (​√25 ​= 5) So the roots will be real, rational and unequal. ___ x =activity 4 ± ​√24 ​ _______ 2. ​  2 ​ ACTIVITIES boy Δ=___ 24 ∴ Δ > 0, soand there of page are two real roots. 24 is not a perfect square girl-left and right (​ √ 24 ​ = 4,898979486…) activityPAY SPECIAL ATTENTION So the roots will be real, NB NB irrational and unequal. ___ − 5 ± ​√−9 ​ 3. x = ​ ________ 8 by step ​ Step ___ comment comment ∴ Δ < 0, so ​√−9 ​is non-real. There are no real solutions for x, Δ = –9 HINT! hint so the roots are non-real. e.g. 22 EG - worked examples _______ The solutions to a quadratic equation are: x = 5 ± ​√10 + 2a ​. exams For which values of a will the equation have equal roots. Solution activitywill have equal roots if Δ = 0. The equation ACTIVITIES boy Δ = 10 + 2a and girl-left and right of page 0 = 10 + 2a activity 10 =  −2a ∴ a = −5 Step by step comment comment Activity 7 1. Show that the roots of x2 – 2x – 7 = 0, are irrational, without solving the equation. (3) 2. Show that x2 + x + 1 = 0 has no real roots. (3) 3. If x = 2 is a root of the equation 3x – 5x – 2k = 0, determine the 2 value of k. _______ (2) 4. The solutions to a quadratic equation are: x = 5 ± ​√12 − 3a ​. For which value(s) of a will the equation have equal roots. (3) 5. Determine the value(s) of k for which the equation ​3x​2​+ (k + 2) x + k = 0 has equal roots (4) [15] 38 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB Unit 2 Solutions HINT! 1. a = 1; b = –2 ; c = –7 hint Δ = b – 4ac = (–2) – 4(1)(–7) 3 2 2 32 is not a perfect square so = 4 + 28 the roots are irrational. = 32 3 e.g. PAY- worked EG SPECIAL examples ATTENTION ∴ The roots will be irrational 3 NB (Δ > 0 and not a perfect square) (3) exams 2. a = 1; b = 1; c = 1 HINT! Δ = b – 4ac = (1) – 4 (1)(1) 3 2 2 hint –3 < 0 =1–4=–3 3 ∴ is non-real ∴ There are no real roots (Δ < 0 ) 3 (3) e.g. activity EG - worked examples 3. If 2 is a root of the equation, then x = 2. Therefore, we can substitute ACTIVITIES boy and girl-left and x = 2 into the equation. right of page exams 3​x2 ​– 5x – 2k = 0 ∴ 3(2)2 – 5(2) –2k = 0    3 activity ∴ 12 − 10 − 2k = 0 ∴ 2k = 2 ∴ k = 1     3 (2) Step by step comment activity 4. The equation will have equal roots if Δ = 0 ACTIVITIES boy and girl-left and Δ = 12 – 3a right of page 0  3 = 12 − 3a   3 activity − 12 = − 3a ∴ a = 4​  3 (3) 5. 3x2 + (k + 2)x + k = 0 Step by step comment ∴ a = 3 ; b = (k + 2) ; c = k ∴ Δ = b2 – 4ac = (k + 2)2 – 4(3)(k) 3 = k2 + 4k + 4 – 12k = k2 – 8k + 4 3 For equal roots the Δ = 0 3 ∴ k2 – 8k + 4 = 0 ___________ 8±​√(–8​)​2​–    4(1)(4) ​ ∴ k = _____________ ​    2(1) ​ ___ √ 48 ​ ∴ k = ​ 8±​ _____ ​ 2 ∴ k = 7,46 or k = –0,54 3 (4) [15] © Department of Basic Education 2015 UNIT 2 ALGEBRA 39 Mind the Gap Mathematics

2 Unit PAY SPECIAL ATTENTION NB NB PAY SPECIAL ATTENTION 2.9.2 Problem solving NB NB using quadratic equations You can useHINT!an equation to represent a problem. Find what part of the HINT! hint is unknown and needs to be represented by a variable. problem hint PAY SPECIAL ATTENTION mNB2 here means suare metres. e.g. NB 23 EG - worked examples It is not a variable. e.g. EG - worked examples The area of a rectangle is 12 m2. exams The length is 4 metres longer than the breadth. Find the dimensions of HINT! the rectangle. hint exams We don’t know the length or the breadth of the rectangle. Dimensions: the measurements of the sides We do activity know that the length is 4 m longer than the breadth. ACTIVITIES boy e.g. and girl-left and EG - worked examples It makes sense to let the breadth be x metres. Then the length is right of page activity x + 4 metres. exams activity Draw a picture to help you. Let breadth be x metres comment Area of rectangle ACTIVITIES boy Step by step = length × breadth comment and girl-left and 12 = (x + 4)x right of page 12 = x2 + 4x activity x metres 0 = x2 + 4x – 12 0 = (x + 6)(x – 2) activity ∴x+6=0 or x – 2 = 0 ACTIVITIES boy comment x+4 Step by step comment and girl-left x = –6 and x=2 right of page Length and breadth must both be positive lengths. You can’t have a activity negative length! So x ≠ –6 ∴ x = 2 and so the breadth is 2 metres. The length is x + 4 and so the length is 6 metres. Step by step comment comment 40 UNIT 2 ALGEBRA  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 2 What you need to be able to do: • Solve quadratic equations by factorising when possible. • Rewrite a quadratic equation that is written in the general form y = a​x2​+ bx + c in the form y = a(x + p)2 + q by completing the square. • Use completing the square________ to solve quadratic equations. − b ± ​√​b2​ ​− 4ac ​ • Use the formula x = ____________ ​   2a ​to determine the roots of an equation. • Use the value of the discriminant (​b2​− 4ac) of a quadratic equation to determine the nature of the roots. • Solve linear and quadratic inequalities. • Solve simultaneous equations to find the points of intersection between two different functions. Feb/March 2014 Q1.1.1 & Q1.1.2 & Q1.2 & Q1.3 Nov 2013 Q1.1.1 & Q1.1.2a,b & Q1.1.3 & Q1.2 Feb/March 2013 Q1.1.1 & Q1.1.2 & Q1.1.4 & Q1.2.1 & Q1.2.2 & Q1.2.3 Nov 2012 Q1.1.1 & Q1.1.1 & Q1.1.3 & Q1.2.1 & Q1.3.1 & Q1.3.2 Feb/March 2012 1.1.1 & 1.1.2 & 1.1.3 & 1.2 Nov 2011 Q1.1.1 & Q1.1.2 7 Q1.1.3 & Q1.2 Feb/March 2011 Q1.1.1 & Q1.1.2 & Q1.1.3 & Q1.2 Keep going! © Department of Basic Education 2015 UNIT 2 ALGEBRA 41 Mind the Gap Mathematics

3 Unit 3 Unit Number patterns, sequences and series 3.1 Number patterns A list of numbers in order is called a number pattern or number sequence. We need at least three numbers in the list to work out if the numbers form a pattern. If we only have two numbers, we cannot be sure what the pattern is. For example, if we have the list 2; 4; … many different number patterns are possible: The pattern could be 2; 4; 6; … add 2 to each number to get the next number OR 2; 4; 8; … multiply each number by 2 to get the next number OR 2; 4; 2; 4; … repeat the pattern A single number in a pattern or sequence is called a term. PAY SPECIAL ATTENTION NB NB Term 1 is written as T1, term 2 is written as T2 and so on. The number of the term shows its position in the sequence. T10 is the HINT!10th term in the sequence. hint Tn is the nth term in a sequence. e.g. 1 EG - worked examples 1. Look at the number pattern 3; 8; 13; … exams T1 T2 T3 T4 T5 3 8 13 ... +5 +5 +5 +5 activity ACTIVITIES boy If we keep addingand5girl-left right ofto page each term we get the next term: and T4 activity = 13 + 5 = 18; T5 = 23; T6 = 28, etc. 2. Look at the number pattern 5; 15; 45; … T1 T2 T3 T4 T5 Step by step comment comment 5 15 45 ... ×3 ×3 ×3 ×3 42 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 In this pattern, each term is multiplied by 3 to get the next term. So T4 = 45 × 3 = 135; T5 = 405; T6 = 1 215, and so on. 3. Look at the sequence: 1; 4; 9; … T1 = 12; T2 = 22; T3 = 32 These are all perfect square numbers. Each number is the number of the term squared. So T4 = (4)2 = 16; T5 = (5)2 = 25; T6 = (6)2 = 36, and so on. It is important to learn to recognise square numbers. 3.2 NB Arithmetic NB sequences PAY SPECIAL ATTENTION Arithmetic sequence is a sequence where the common difference (d) between HINT! consecutive terms is constant. Thint 2 – T1 = T3 – T2 = Tn – Tn–1 = d (common difference) e.g. 2 Given the sequence: EG - worked examples 5; 9; 13; 17 ; . . . a) Determine the common difference exams b) Determine the next two terms Solution d = 9 – 5 = PAY13SPECIAL – 9ATTENTION =4 or by looking at NB T5 = 17 + 4 = 21 and T6 = 21 + 4 = 25 NB the structure, the activity numbers are 2 less ACTIVITIES boy than the multiples of 6 and girl-left and right of page If we use a HINT! for the first term T1, d for the common difference, then the i.e Tn = 6n – 2 hint activity general term Tn for an arithmetic sequence is: Tn = a + (n – 1)d e.g. 3 Given the sequencecomment EG - worked examples Step by step 4; 10; 16; . . . comment a) Determine a formula for the nth term of the sequence. exams b) Calculate the 50th term. c) Which term of the sequence is equal to 310 Solutions a) = 4 and d = 10 – 4 = 16 – 10 = 6 aactivity Tn = a + (n – 1) d ACTIVITIES boy and girl-left and = 4 + (n – 1) 6right of page activity = 4 + 6n – 6 = 6n – 2 b) T50 = 6Step×by step 50 – 2 comment comment = 300 – 2 = 298 c) 6n – 2 = 310 6n = 312 n = 52 © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 43 Mind the Gap Mathematics

3 Unit Activity 1 1. Given the sequence 6; 13; 20; … a) Determine a formula for the nth term of the sequence. b) Calculate the 21st term of this sequence. c) Determine which term of this sequence is 97. (5) 2. Consider this number pattern: 8; 5; 2; … a) Calculate the 15th term. b) Determine which term of this sequence is –289. (4) 3. a) Given the arithmetic sequence 1 − p; 2p − 3; p + 5; . . . determine the value of p. b) Determine the values of the first three terms of the sequence. (5) [14] Solutions 1. a) It is an arithmetic sequence because there is a common difference. a = 6; d = 7 Tn = a + (n – 1)d 3 Tn = 6 + (n – 1)(7) Tn = 7n – 1 3 b) T21 = 7(21) – 1 = 147 – 1 = 146 3 c) 97 = 7n – 1 3 ∴98 = 7n ∴14 = n 3 ∴97 is the 14th term of the sequence. (5) 2. a) It is an arithmetic sequence: a = 8; d = 5 – 8 = 2 – 5 = – 3 Tn = a + (n – 1)d ∴T15 = 8 + (15 – 1)(–3) 3 T15 = 8 + 14(–3) T15 = 8 – 42 = –34 3 b) Tn = a + (n – 1)d – 289 = 8 + (n – 1)(–3) 3 ∴– 289 = 8 – 3n +3 ∴– 300 = –3n ∴100 = n 3 ∴– 289 will be the 100th term (4) 3. a) Since this is an arithmetic sequence, you can assume that there is a common difference between the terms. d = T2 – T1 = T3 − T2​ ∴(2p – 3) – (1 – p) = ( p + 5) – (2p – 3) 3 3p – 4 = – p + 8 3 4p = 12 p=33 b) p = 3 T1 = 1 – p = 1 – 3 = –2 T2 = 2p – 3 = 2(3) – 3 = 3 3 T3 = p + 5 = 3 + 5 = 8 3 So the first three terms of the sequence are –2; 3; 8 (5) [14] 44 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 3.3 Quadratic sequences At least four numbers are needed to determine whether the sequence is For T1, n = 1; T2, quadratic or not. n = 2; T3, n = 3 ; . . . Consider this number pattern: There is no common difference between the numbers. The differences are 6; 10; 14; 18. T1 T2 T3 T4 T5 6 12 22 36 54 Now we can see if there (T2 – T1) (T3 – T2) (T4 – T3) (T5 – T4) First difference =6 = 10 = 14 = 18 is a second common Second difference 10 – 6 14 – 6 18 – 4 difference. =4 =4 =4 In this sequence, there is a second common difference of 4. The next term will be: T6 = 54 + (18 + 4) = 76 A pattern with a common second difference is called a quadratic number sequence. The general formula for any term of a quadratic sequence is: Tn = an2 + bn + c T1 = a + b + c T2 = 4a + 2b + c T3 = 9a + 3b + c T4 = 16a + 4b + c 1st difference 3a + b 5a + b 7a + b 2nd difference 2a 2a PAY SPECIAL ATTENTION If NB Tn = an2 + bn + c NB then 2a is the second difference 3a + b is T2 – T1 a + bHINT!+ c is the first term hint e.g. 4 Look at the number sequence 12; 20; 32; 48; . . . EG - worked examples 2nd common difference is 4 exams So 2a = 4 ∴a = 2 3. 12 20 32 48 T2 – T1 = 8 2. So 3a + b = 8 ∴3(2) + b = 8 T2 – T3 – T4 – activity ∴b = 2 ACTIVITIES boy =8 =12 =16 1st term is 12 3. and girl-left and right of page 2. So a + b + c = 12 ∴2 + 2 + c = 12 activity ∴c = 8 4 4 ∴Tn = 2n2 + 2n + 8 1. comment ∴T5 = Step 2(5) 2 by step + 2(5) + 8 = 68 comment ∴T6 = 2(6)2 + 2(6) + 8 = 92 © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 45 Mind the Gap Mathematics

3 Unit Activity 2 1. Consider the number pattern: 3; 13; 31; 57; 91; … a) Determine the general term for this pattern. b) Calculate the 7th term of this pattern. c) Which term is equal to 241? (9) 2. Find term 6 of this pattern and then find the rule in the form Tn = an2 + bn + c –1 ; 3; 9; 17; 27 ... (4) [13] Solutions 1. a) It helps to make a diagram: T1 T2 T3 T4 T5 3 13 31 57 91 (T2 – T1) (T3 – T2) (T4 – T3) (T5 – T4) First difference = 10 = 18 = 26 = 34 Second difference 18 – 10 26 – 18 34 – 26 =8 =8 =8 ∴ it is a quadratic sequence. 2a = 8 ∴ a = 4 3 ∴ 3a + b = 10 3(4) + b = 10 n = –7,5 not possible b = –2 3 because n is the position of the ∴ a + b + c = 3 4 + (–2) + c = 3 term so it must be a positive c=1 3 natural number. ✓ ∴Tn = 4n2 – 2n + 1 3 ∴241 is the 8th term of the sequence. b) T7 = 4(7)2 – 2(7) + 1 3 = 4(49) – 14 + 1 = 183 3 c) 241 = 4n2 – 2n + 1 0 = 4n2 – 2n + 1 – 241 3 make the equation = 0 to solve 0 = 4n2 – 2n – 240 0 = 2n2 – n – 120 divide through by 2 0 = (2n + 15)(n – 8) 3 factorise ∴ 2n + 15 = 0 OR n–8=0 ∴ n = –7,5 OR n=8 3 (9) 46 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 2. T1 T2 T3 T4 T5 –1 3 9 17 27 ... 4 6 8 10 2 2 32 ∴T6 = 27 + (10 + 2) = 39 3 use the pattern of the numbers 2a = 2 ∴a = 1 3a + b = 4 3(1) + b = 4 ∴b = 1 a + b + c = –1 1 + 1 + c = –1 ∴c = –3 Tn = n2 + n – 3 33 (4) [13] © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 47 Mind the Gap Mathematics

3 Unit 3.4 Geometric sequences When there is a common ratio (r) between consecutive terms, we can say this is a geometric sequence. If the first term (T1) is a, the common ratio is r, and the general term is Tn, then: T2 T3 Tn r = ​ __ ​ __ ​ ___ Given the sequence, T = ​ T = ​  T ​and 1 2 n–1 check whether it is Tn = ar n–1 arithmetic, geometric PAY SPECIAL ATTENTION NB NB or quadratic. Look at the sequence 5; 15; 45; 135; 405; … __15 45 135 ​  5 ​= 3 ​ __ ___ 15 ​= 3 and ​  45 ​= 3 and so the common ratio is 3. HINT! Therefore hint the sequence is geometric. To get the next term you multiply the preceding term by the common ratio. e.g. 5 EG - worked examples 2 4 Given the sequence 1; __ ​  3 ​; __ ​  9 ​; ... exams a) Determine the next two terms 32 b) Which term of the sequence is equal to ​ ___ 243 ​? Solutions activity 2 2 2 4 2 The common ratio is __ ​  3 ​ because __ ACTIVITIES boy ​  3 ​÷ 1 = ​ __3 ​= ​ __9 ​÷ ​ __3 ​ and girl-left and a) T4 = ar3 = 1​​(__​  ​  )​​ ​= ​ __ 27 ​and ​T​ = 1​​(​  3 ​  )​​ ​= ​  81 ​ 2 8 3 2 __ right of page ​ 16 __ 4 OR activity 3 5 T4 = T3 × r = 2 32 b) a = 1; r = __ ​  3 ​and Tn = arn–1 = ___ ​  243 ​ 4 2 8 ​ __ __ ____ 9 ​× ​  3 ​= ​  27 ​ (__2 ) ​= ​ __​2​3​​​​​= ​​(__​ 23​  )​​ ​ comment n−1 5 5 T5 = T4 × r = ∴ NBTn = (1) ​​ ​  3 ​  ​​ PAY SPECIAL ATTENTION 5 NB 8 2 16 Step by step comment ____ ​  27 ​  × ​ ​ __ ____ 3 ​ ​ =  ​ ​ 81 ​ ​ ∴​​(​  3 ​  )​​ ​= ​​(​  3 ​  )​​ ​ 2 2 n−1 5 __ __ ∴n – 1 =HINT!5 hint n=6 e.g. 6 EG - worked examples In a geometric sequence, the fifth term is 80 and the common ratio is –2. exams Determine the first three terms of the sequence. T5 = 80 and r = –2 T5 = ar4 = a(–2)4 = 80 16a = 80 a = 5 activity ∴T1 = 5; T2 = 5(–2)1 =ACTIVITIES and–10; boy T3 = 5(–2)2 = 20 girl-left and right of page activity Step by step comment comment 48 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 Activity 3 a) Determine the 10th term of the sequence: 3; 6; 12; . . . (2) b) Determine the number of terms in the sequence: 2; 4; 8; . . .; 1024 (2) c) If 5; x; 45 are the first three terms of a geometric sequence, determine the value of x. (2) d) Determine the geometric sequence whose 8th term is 9 and whose 10th term is 25. (3) [9] Solutions 6 12 a) a = 3; r = __ ​  3 ​= __ ​  6 ​= 2 Tn = arn−1 T10 = 3(2)10−1 = 3(2)9 = 3 × 512 = 1536 33 (2) 4 8 b) a = 2; r = __ ​  2 ​= __ ​  4 ​= 2 arn–1 = 1024 2(2)n–1 = 210 = 2n = 210 3 ∴ n = 10 3 (2) x 45 c) __ ​  5 ​= __ ​  x ​ 3 ____ x = ± ​√ 225 ​ = ± 15 3 (2) d) ar7 = 9 ar9 = 25 ar9 ___ 25 ​  ar7 ​= ​ __ ​ 9 ∴r = ​  ​ 2 25 __ 9 5 _ r = ​ 3 ​ 3 a = ___ 9 ​  __5 7 ​= 9 × ​​ __ () ​​ ​  3 ​  ​​ ​ 3 7 ​  5 ​  ​​ ​ 3() ( ) ( ) ( ) ( ) ( ) The sequence is: 9​​​  3_5 ​  ​​ ​; 9​​ ​  3_5 ​  ​​ ​ ; 9​​ ​  3_5 ​  ​​ ​; 9​​ ​  3_5 ​  ​​ ​ ; 9​​ ​  3_5 ​  ​​ ​ 3 7 6 5 4 3 (3) [9] © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 49 Mind the Gap Mathematics

3 Unit 3.5 Arithmetic and geometric series The proof must be learnt When we add the terms of a sequence together, we form a series. for exams We use the symbol Sn to show the sum of the first n terms of a series. So Sn = T1 + T2 + T3 + T4 + … + Tn 3.5.1 Arithmetic series n The formula is Sn = __ ​  2 ​[2a + (n − 1)d] where Sn is the sum of n terms, a is the first term, n is the number of terms and d is the common difference. Proof The general term of an arithmetic series is Tn = a + (n – 1)d So Sn = T1 + T2 + T3 + T4 + … + Tn Sn = a + [a + d] + a + 2d + … + [a + (n – 2)d] + [a + (n – 1)d] … equation 1 If we write the series in reverse we get: Add first terms: Sn = [a + (n – 1)d] + [a + (n – 2)d] + [a + (n – 3)d] + … + [a + d] + a … a + [a + (n – 1)d] equation 2 = 2a + (n – 1)d We can add equation 1 and equation 2. Add second terms: a + d + [a + (n – 2)d] So 2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + [2a + (n – 1)d] + … + = 2a + (n – 1)d [2a + (n – 1)d] + [2a + (n – 1)d] Add third terms: 2Sn = n [2a + (n – 1)d] a + 2d + [a + (n – 3)d] n Sn = __ ​  2​[2a + (n – 1)d] = 2a + (n – 1)d This formula is provided on the information sheet in the final exam. Add last terms: [a + (n – 1)d] + a = 2a + (n – 1)d Alternative Proof i.e (a + l), n times Or Sn = a + [a + d] + [a + 2d] + … + [l – d] + l … equation 1 In reverse Sn = [a + (n – 1)d] + [a + (n – 2)d] + [a + (n – 3)d] + … + [a + d] + a Sn = l + [l – d] + [l – 2d] + . . . + [a + d] + a … equation 2 PAY SPECIAL ATTENTION Adding equation 1 and equationNB NB 2 2Sn = [a + l] + [a + l] + . . . + [a + l] n times 2Sn = n[a + l] __n HINT! ∴Sn = ​  2 ​[a + l] hint e.g. 7 EG - worked examples 1. Determine the sum of the first 20 terms of the series: exams 3 + 7 + 11 + 15 + … 2. The sum of the series 5 + 3 + 1 + . . . is –216, determine the number of terms in the series activity ACTIVITIES boy and girl-left and right of page 50 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  activity Mind the Gap Mathematics

Unit 3 Solutions 1. a = 3, n = 20, d = 4 n Sn = __ ​  2 ​[2a + (n – 1)d] 20 S20 = __ ​  2 ​[2(3) + (19)4] S20 = 10(6 + 76) S20 = 820 The sum of the first 20 terms is 820 n 2. a = 5 d = –2 Sn = –216 Sn = __ ​  2 ​[2a + (n – 1)d] n=? Substitute into the formula: n –216 = ​ __2 ​[2(5) + (n – 1)(–2)] n –216 = __ ​  2 ​[10 + –2n + 2] n –216 = __ ​  2 ​[12 – 2n] –432 = 12n – 2n2 –432 = –2n2 + 12n …. Make equation = 0 2n – 12n – 432 = 0 …. Divide through by 2 (common factor) 2 n2 – 6n – 216 = 0 …. Factorise trinomial (n – 18)(n + 12) = 0 ∴ n – 18 = 0 or n + 12 = 0 n = 18 or n = –12 n>0 ∴n = 18 ∴18 terms of the series add up to –216. Activity 4 1. Determine the sum of the series: 19 + 22 + 25 + . . . + 121 (3) 2. The sum of the series 22 + 28 + 34 + . . . is 1870. Determine the number of terms. (2) 3. Given the arithmetic sequence -3; 1; 5; …,393 a) Determine a formula for the nth term of the sequence. b) Write down the 4th, 5th, 6th and 7th terms of the sequence. c) Write down the remainders when each of the first seven terms of the sequence is divided by 3. d) Calculate the sum of the terms in the arithmetic sequence that are divisible by 3. (10) n __ 4. The sum of n terms is given by ​Sn​  =​  ​  2 ​ ​​(1 + n)​. Determine T5. (3) 5. 3x + 1;  2x;  3x − 7 are the first three terms of an arithmetic sequence. Calculate the value of x. (3) 6. The first and second terms of an arithmetic sequence are 10 and 6 respectively. a) Calculate the 11th term of the sequence. b) The sum of the first n terms of this sequence is –560. Calculate n. (6) [27] © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 51 Mind the Gap Mathematics

3 Unit Solutions 1. a = 19 and d = 3 Tn = 3n + 16 = 121 3n = 105 n = 35 3 n ​Sn​= ​ __2 ​(a + l) 3 35 35 ​S35​= ​ __   2 ​ (19 + 121) = ​  __ 2 ​ (140) = 35 × 70 = 2450 3 (3) 2. a = 22 and d = 6 n ​Sn​ = ​ __2 ​[2a + (n − 1)d] __ n ​  2 ​[2 × 22 + (n − 1)6] = 1870 3 19n + 3​n2​= 1870 3​n2​+ 19n − 1870 = 0 (3n + 85)(n − 22) = 0 3  ∴ n = 22 n cannot be a negative because it is the number of terms (2) 3. a) Tn = –3 + (n − 1)4 3 4n – 7 = Tn 3 b) ​T4​= 5 + 4 = 9;      ​T5​= 9 + 4 = 13;    3      ​T6​= 13 + 4 = 17     and  ​ T7​= 17 + 4 = 21 3 c) 0; 1; 2; 0; 1; 2; 0 33 d) Tn = –3 + 12 (n − 1) 3 393 = 12n – 15 12n = 393 + 15 = 408 3 n = 34 34 ​S34​= ​ __ 2 ​× (–3 + 393)          = 17 × 390 33          = 6630 (10) 5 4. ​S5​= ​ __2 ​​( 1 + 5 )​= 15 3 4 ​S4​= ​ __2 ​​( 1+ 4 )​= 10 3 T5 = 15 – 10 = 5 3 (3) 5. T2 – T1 = T3 – T2 2x – (3x + 1) = (3x – 7) – 2x 3 2x – 3x – 1 = 3x – 7 – 2x –2x + 6 = 0 3 2x = 6 x = 3 3 (3) 52 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 6. a) Tn = a + (n – 1)d T11 = 10 + (11 – 1)(–4) 3 = –30 3 n b) Sn = __ ​  2 ​[2a + (n – 1)d] n –560 = ​ __2 ​[2(10) + (n – 1)(–4)] 3 –1120 = –4n2 + 24n 4n2 – 24n – 1120 = 0 n2 – 6n – 280 = 0 3 (n – 20)(n + 14) = 0 3 n = 20 or n = –14 n = 20 only 3 because number of terms cannot be a negative number (6) [27] 3.5.2 Geometric series The formula is a(​rn​− 1) a(1 − ​rn​) ​Sn​  = ​  ​ _______ _______ r − 1 ​ ​  for r > 1   or  ​Sn​  = ​  ​  1 − r ​ ​  for  r < 1 where a is the first term r is the common ratio n is the number of terms Sn is the sum of the terms Proof: The general term of a geometric series is Tn = arn – 1 So Sn = T1 + T2 + T3 + T4 + … + Tn Sn = a + ar + ar2 + … + arn–2 + arn–1 rSn = ar + ar2 + ar3 + … + arn–1 + arn multiply each term by r Sn = a + ar + ar + … + ar2 _____________________________ n–2 + arn–1 write down the series again with ​         rS – S = –a + 0 + 0 + … + 0 + 0 + arn ​ n n like terms under each other ∴rSn – Sn = arn – a subtract each bottom term from each top term Sn and a are common factors Sn (r – 1) = a(rn – 1) a(rn –1) So Sn = ______ ​  r – 1 ​ Divide through by (r – 1) The proof a(1 – rn) must be learnt We can also use for Sn = ______ ​  1 – r ​for r < 1 for exams © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 53 Mind the Gap Mathematics

HINT! 3 Unit hint e.g. 8 EG - worked examples Evaluate: 25 + 50 + 100 + … to 6 terms exams Solution We need to check if this is an arithmetic series or a geometric series first. 50 100 You should see that there is a common ratio of 2 because __ ​  2 ​= 2 and ___ ​  50 ​= 2 r=2 activity ACTIVITIES boy ∴ It is a geometric series and a = 25, n = 6, r = 2 and girl-left and right of page a(1 – rn) Sn = ______ 1–r ​ ​  activity 25(1 – 26) S6 = _______ ​  1 – 2 ​ 26 = 64 25(1 – Step 64)by step comment S6 = ________ comment ​  – 1 ​ 25(–63) S6 = ______ ​  –1 ​ = 1 575 So the sum of the first 6 terms of the series is 1 575. Activity 5 1. Determine 3 + 6 + 12 + 24 + . . . to 10 terms (2) 2. If 2 + 6 + 18 + . . . = 728, determine the value of n. (3) [5] Solutions 6 12 1. a = 3 and r = ​ __3 ​= ​ __ 6 ​= 2 a(​rn​ ​− 1) Sn = ​ _______ r−1 ​ ​ ​− 1) 3(​210 3 ​S​10​= ​ _______ 2−1 ​= 3(1024 − 1) = 3069 3 (2) 6 18 2. a = 2 and r = ​ __2 ​= ​ __ 6 ​= 3 2(​3n​ ​− 1) Sn = ​ _______ 3 − 1 ​= 728 3 2(​3n​ ​− 1) _______ ​  2 ​= 728 3n − 1 = 728 3n = 729 = ​36​ ​ 3 ∴n = 6 3 (3) [5] 54 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 3.5.3 Sigma notation Here is another useful way of representing a series. The sum of a series can be written in sigma notation. To determine the The symbol sigma is a Greek letter that stands for ‘the sum of’. number of terms: top PAY SPECIAL ATTENTION value minus bottom value NBthe symbol for ‘the sum of’  is NB plus 1 i.e the number of n terms in this case is ​∑​  Tk​ means ‘the sum of the terms Tk from k = 1 to k = n. (17 – 3 ) +1 = 15 k=1 HINT! n other words,​∑​  Tk​ = T1 + T2 + T3 + T4 + … + Tn Inhint k=1 e.g. 9 EG - worked examples 17 k​ ​∑​ ​p exams = ​p​3​+ ​p​4​+ ​p​5​+ . . . + ​p​17​ k=3 Activity 6 PAY SPECIAL ATTENTION activity70 NB 1. Evaluate:​∑​(2n –ACTIVITIES 4)​ boy (3) and girl-left and n=4 right of page m 2. What is the value of m for which ​∑​5(3)k–1 = 65​? activity (4) k=1 HINT! 1 1 1 3. Consider the sequence: __ ​  2 ​; 4; ​ __4 ​; 7; __ ​  8 ​; 10; . . . hint a) If the pattern Step by step continues comment comment same way, write down the next in the TWO terms in the sequence. Look for two different sequences in the pattern and b) Calculate the sum of the first 50 terms of the sequence. (5) [12] e.g. separate them EG - worked examples Solutions 1. The question asks you to find the sum of the terms from n = 4 to exams n = 70 if the nth term is 2n – 4. a = T1 = 2(4) – 4 = 4 Find the first term a T2 = 2(5) – 4 = 6 T3 = 2(6) – 4 = 8 So the sequence is 4; 6; 8; … and this is an arithmetic series. 3 activity To check d, calculate T2 – T1 ACTIVITIES boy and girl-left and d = T2 – T1 = 6 – 4 = 2 right of page n = (70 – 4) + 1 = 67 3 There are 67 terms activity Now we can substitute these values into the formula to find the sum of 67 terms. n Sn = __ ​  2 ​[2a + (n – 1)d] 67 S67 = __ ​  2 ​[2(4) + (67 – 1)2] Step by step comment S67 = 33.5 [8 + 132] = 4690 70 So ​∑​(2n – 4) = 4690 3​ (3) n=4 © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 55 Mind the Gap Mathematics

3 Unit 2. This is a geometric series because 5(3)k–1 has the form ark–1, T1 = 5(3)1–1 = 5 ; T2 = 5(3)2–1 = 15; T3 = 5(3)3–1 = 45 a = 5; r = 3; n = m and Sm = 65 a(​r​n​−1) Sn = ______ ​  r − 1 ​ 3 … substitute 5(​3m​ ​− 1) 65 = ​ _______ 3−1 ​ 3 5(​3m​ ​− 1) 65 = _______ ​  2 ​ … multiply through by 2 130 = 5.3m – 5 … add like terms 135 = 5.3 3 m … divide through by 5 27 = 3 m … write 27 as a power of 3 3 = 3 3 …m bases are the same, so the powers are equal ∴ m = 3 3 (4) 1 1 ​  2 ​, so T7 is __ 3. a) T1, T3 and T5 form a sequence with a common ratio of __ ​  16 ​. 3 T2, T4 and T6 form a sequence with a common difference of 3, so T8 is 13. b) S50 = 25 terms of 1st sequence + 25 terms of 2nd sequence 1 1 1 S50 = (​ __2 ​+​ __4 ​+ __ ​  8 ​+ … to 25 terms) + (4 + 7 + 10 + 13 + … to 25 terms) 3 [( ) ] 1 __ 25 ​ __​​ ​​ ​  1 2 2 S50 = ________ ​  __1 ​  ​​ ​−1 ​ 25 ​+ ​ __ ​  2 ​− 1 2 ​[2(4) + 24 (3)] 3 S50 = 0,99999997 + 1 000 3 S50 ≈ 1 001,00 3 (5) PAY SPECIAL ATTENTION [12] NB NB 3.5.4 Infinite geometric series An infinite series HINT! is one in which there is no last term, i.e. the series goes on hint without ending. e.g. 10 EG - worked examples 3 3 + 3 + ​ __2 ​+ ​ __4 ​+ ... 6 exams ∞ S∞ = ​∑​2(3)k–1 ​= 2 + 6 + 18 + 54 + . . . the sum from term1 to infinity of 2(3)k – 1 k=1 T1 = 2(3)0 = 2 T2 = 2(3)1 = 6 activity T3 = 2(3)2 = 18 ACTIVITIES boy and girl-left and T4 = 2(3)3 = 54 right of …page activity The terms of this series are all positive numbers and the sum will get bigger and bigger without any end. This is called a divergent series. Step by step comment comment 56 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 3 e.g. 11 EG - worked examples Look at this infinite series: exams 1 1 1 1 S∞ = 1 + ​ __2 ​+ __ ​  4 ​ + __ ​  8 ​ + __ ​  16 ​+ … 1 1 S2 = 1 + __ ​  2 ​= 1​ __2 ​= 1.5 1 1 3 S3 = 1​ __2 ​+ __ ​  4 ​= 1​ __4 ​= 1.75 activity 3 1 7 S4 = 1​ __4 ​+ __​  8 ​= 1​ __8 ​= 1.675 ACTIVITIES boy and girl-left and right of page 7 1 15 S5 = 1​ __8 ​+ __ ​  16 ​= 1​ __ 16 ​= ….. activity This series will converge to 2. It is therefore called a convergent series and we can write the sum to infinity equals 2: S∞ = 2 comment You can identify Step by step a convergent comment infinite series by looking at the value r An infinite series is convergent if –1 < r < 1, r ≠ 0 PAY SPECIAL ATTENTION NBformula for the sum of a convergent The NB infinite series: a S∞ = ​ ____ 1 − r​ where a is the HINT! first term, r is the common ratio hint formula is provided on the information sheet in the final exam. This e.g. 12 EG - worked examples 1 1 1 1 1. Look again at the example where ​S∞​= 1 + ​ __2 ​+ __ ​  4 ​ + __ ​  8 ​+ __ ​  16 ​+ … exams __ 1 a = 1 and r = ​  2 ​ 0 < r < 1 a S∞ = ​ ____ 1 − r​ 1 1 S∞ = ​ _____ __ 1 ​= 1 ÷ ​  2 ​ 1 − ​ __2 ​ activity S∞ = 1 × 2 = 2 ACTIVITIES boy and girl-left and right of page 2. Foractivity which value(s) of x will 8x2 + 4x3 + 2x4 + … be convergent? For convergent geometric series, –1 < r < 1 r = T2 ÷ T1 comment = 4x3 Step ÷by8x step2 comment x = __ ​  2 ​ x ∴–1 < __ ​  2 ​< 1 multiply through by 2 –2 < x < 2…………………..x ≠ 0 © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 57 Mind the Gap Mathematics

3 Unit Activity 7 ∞ 1. Calculate S∞ if ​∑​8(4)1–p​ (3) p=1 2. Given the series: 3(2x – 3)2 + 3(2x – 3)3 +3(2x – 3)4 + … for which values of x will the series converge? (4) m 3. Find the value of m if: ​∑​3(2)k–1 = 93​ (4) k=1 ∞ 4. For which values of x will ​∑​(4x – 1)k exists.​ (3) k=1 [14] Solutions 1. T1 = 8(4)1 – 1 = 8 = a 3 To find r, find the common ratio using T1 and T2, T2 and T3. 1 T2 = 8(4)1 – 2 = 8(4)–1 = 8 × __ ​  4 ​= 2 1 1 T3 = 8(4)1 – 3 = 8(4)–2 = 8 × __ ​  16 ​= __ ​  2 ​ 1 2 1 ​ __2 ​ 1 1 1 T2 ÷ T1 = __ ​  8 ​= ​ __4 ​and T3 ÷ T2 =​ __ __ __ __ 2 ​= ​  2 ​× ​  2 ​= ​  4 ​ 1 so r = __ ​  4 ​and a = 8 3 a 8 8 ∴S∞ = ​ ____ ____ __ 1 − r ​= ​  1 − ​ __1 ​​= ​  ​  3__​​ 3  When dividing by a fraction, you can 4 4 multiply by the inverse 4 32 = 8 × __ ​  3 ​= __ ​  3 ​ 32 2 ∴S∞ = __ ​  3 ​or 10 __ ​  3 ​ (3) 2. This is a geometric series with r = 2x – 3 3 To converge –1 < r < 1 3 –1 < 2x – 3 < 1 Add 3 to both sides 2 < 2x < 4 Divide by 2 on both sides 3 1<x<2 3 x ≠ ​ __2 ​ 3 (4) The series will converge for 1 < x < 2 3. a = 3; r = 2; Sm = 93 4. r = 4x – 1 3 ​S​n​= ​ _______ a(1 − ​r​n​) 1−r ​ 3 –1 < r < 1 3​(​1 − 2​m​  )​ 1 93 = _______ ​  1 − 2 ​ 3 –1 < 4x – 1 < 1; x ≠ ​ __4 ​ 3 3​(​1 − 2​m​  )​ 93 = _______ ​  − 1 ​ 0 < 4x < 2 1 0 < x < __ –93 = 3(1 – 2m) ​  2 ​ 3 (3) –31 = 1 – 2 m 2m = 32 3 2m = ​25​ ​ ∴ m = 5 3 (4) [14] 58 UNIT 3 number patterns, sequences and series  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 3 What you need to be able to do: • Find the next few terms in a given sequence. • Identify arithmetic sequences, quadratic sequences and geometric sequences • Apply knowledge of sequences and series to solve real life problems • Find the first difference and the second common difference in a quadratic sequence. • Find the general terms of sequences. • Know how to derive the formulae for the sum of Arithmetic or Geometric Series. • Solve problems using these sum formulae. • Work with the sum of infinite geometric sequences that are convergent. February/March 2014 Questions 2; 3 and 4 November 2013 Questions 2 and 3 February/March 2013 Questions 2 and 3 February/March 2012 Questions 2, 3 and 4 November 2012 Questions 2, 3 and 4 November 2010 Questions 2 and 3 Keep going! © Department of Basic Education 2015 UNIT 3 number patterns , sequences and series 59 Mind the Gap Mathematics

4 Unit 4 Unit Functions 4.1 What is a function? If you are given a set of x-values, you can work out the set of y-values or answers that came from using a given rule on each x-value. So there is a relationship between the x-values RULE and the y-values that is described by the rule. The x-values are the input values and the y-values are the output values. In this flow diagram, the x–values y - values rule is y = 2x – 1 So for every x-value, we multiply it by 2 and subtract 1 to find the corresponding y-value. –1 –3 0 –1 1 2x – 1 1 2 3 3 5 x–values rule y–values domain range The input values or x-values are the elements of the domain of this set and the output values or y-values are the elements of the range of this set. 60 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 We can plot these values on the Cartesian plane. y = 2x – 1 y If we extend the domain so that x∈핉, we get the (3; 5) graph for y = 2x – 1. (2; 3) Look at the graph. For every x-value on this graph, (1; 1) there is only one y-value. If a rule or a formula 0 produces only one y-value for each x-value, then (0; –1) we have a function. (–1; –3) A function is a relationship between x and y, where for every x-value there is only one y-value. PAY SPECIAL ATTENTION OneNBway to decide whether or not NB a graph represents a function is to use the vertical line test. If any line drawn HINT! parallel to the y-axis cuts the graph only once, then the graph hint represents a function. e.g. 1 EG - worked examples exams Graph A Graph B Graph C y y y x x activity x ACTIVITIES boy and girl-left and right of page activity Graph A and Graph B are functions. Graph C is not a function because the vertical cuts the graph twice. So for comment an x-value on the graph, there are two y-values. Step by step comment © Department of Basic Education 2015 UNIT 4 functions 61 Mind the Gap Mathematics

4 Unit 4.2 Function notation We use function notation f(x) to show that each y-value is a function of an x-value. We can also use other letters too, such as g(x), h(x), etc. So y = 2x – 1 can be written as f(x) = 2x – 1. The value of f(x) for any x-value can be worked out by substitution: For example, at x = –3 we can find f(–3) = 2(–3) – 1 = –7 So the point (–3; –7) lies on the graph of f(x) = 2x – 1 Activity 1 If h(x) = ​​(__ ​  2 ​  )​ ​determine the value of h(–4). 1 x 1. (3) 2. If the function g(x) = –​x​ ​–3x, find g(x + h) 2 (2) 3. If f(x) = 4x + 1, determine the value of: 3.1 f(x + a) 3.2 f(x) + a 3.3 af(x) (3) 4. If g(x) = 2x2, determine the value of: 4.1 g(–x) 4.2 –g(x)(2) [10] Solutions 1. h(x) = ​​(__ ​  2 ​  )​ ​ 1 x ∴h(–4) = ​​(​  1__2 ​  )​​ ​3 (​​2–1 )–4 –4 ​​ ​= 24 = 16 3 So when x = –4, y = 16 and the point (–4; 16) lies on the graph PAY SPECIAL ATTENTION of the function 3h. (3) NB NB 2. g(x) = –x2 – 3x ∴g(x + h) = –(x + h)2 –3(x + h) 3 wherever there is an x, replace it with (x + h) HINT! = –(x2 + 2xh + h2) – 3x – 3h hint = –x2 – 2xh – h2 – 3x – 3h 3 In each example, there is only This means that when x = x + h, y = –x2 – 2xh – h2 – 3x – 3h (2) one possible y-value for each e.g. EG - worked examples x-value, so f(x); h(x) and g(x) 3.1 f(x) = 4x + 1 3.2 f(x) = 4x + 1 3.3 f(x) = 4x + 1 are functions. f(x + a) = 4(x + a)+1 f(x) + a = 4x + 1 + a af(x) = a(4x + 1) = 4x + 4a + 1 3 3 = 4ax + a 3 (3) exams 4.1 g(x) = 2x2 4.2 g(x) = 2x2 g(–x) = 2(–x)2 –g(x) = –2x2 3 = 2x2 3 (2) [10] activity ACTIVITIES boy 62 UNIT 4 functions  and girl-left and © Department of Basic Education 2015  right of page Mind the Gap Mathematics

Unit 4 4.3 The basic functions, formulas and graphs Important terms to remember: Domain: the set of possible x-values Range: the set of possible y-values Axis of symmetry: an imaginary line that divides a graph into two mirror images of each other. Maximum: the highest possible y-value of a function. Minimum: the lowest possible y-value of a function. Asymptote: an imaginary line that a graph approaches but never touches. Turning point: The point at which a graph reaches its maximum or minimum value and changes direction. 4.3.1 The linear function (straight line) Linear functions have the form y y=x f(x) = ax + q where a represents the gradient of a straight-line graph and q represents the y-intercept when x = 0. The graph of y is a straight line with a = 1 and q = 0 Domain: x ∈ 핉 x Range: y ∈ 핉 Also note the shape of the following linear functions y y y y x x x x a<0 a=0 a>0 a is undefined q<0 y=q q<0 there is no q-value SKETCHING THE LINEAR FUNCTION To sketch the linear function using the dual intercept method. • Determine the x-intercept (let y = 0) • Determine the y-intercept (let x = 0) • Plot these two points and draw a straight line through them. © Department of Basic Education 2015 UNIT 4 functions 63 Mind the Gap Mathematics

4 Unit DETERMINING THE EQUATION OF A LINEAR FUNCTION To determine the equation of the linear function follow the following steps: PAY SPECIAL ATTENTION • NB Determine the gradient of NB the function. • Substitute the value of the gradient into the general formula for the linear function. • Solve for HINT!q. • hint Write the equation in the form f(x) = ax + q e.g. 2 EG - worked examples 1.exams 2. y y (–1; 2) 0 2 x activity x (1; –1) ACTIVITIES boy and girl-left and right of page activity Step by step comment comment Solutions 1. 2. y2 – y1 y2 – y1 a = _____ ​  x2– x1 ​ a = ​ _____ x2– x1 ​ –1 _____ –0 2–0 = ​  1 – 2 ​ 3 = ​ _____ –1– 0 ​ 3 3 a = 1 a = –2 ∴y = –2x + c ∴y = 1x + c 0 = 1(2) + c 0 = –2(0) + c c = –2 3 c=0 3 ∴f(x) = x – 2 ∴f(x) = x – 2x [5] 4.3.2 The quadratic functions (parabola) A quadratic function is a parabola and can be represented with a general formula y = ax2 + bx + c or y = a(x + p)2 + q [PROPERTIES OF A PARABOLA] 1. Shape a<0 a>0 64 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 2. ​  –b ​or The graph has an axis of symmetry at x = __ 2a 3. The function has one turning point given by ​– ​  b ( 2a ( 2a ) ) ​  b ​  ​  ​. __​; f ​– __ 4. The function may have either a maximum or a minimum value but never both. 5. Domain: x ∈ 핉 ( 2a ) ​  b ​  ​or y < f ​– ​  b Range: y > f ​– __ __​  ​ ( 2a ) SKETCHING THE QUADRATIC FUNCTION To sketch any quadratic function, follow the following steps: • Write down the y-intercept (let x = 0) • To calculate the x-intercepts, • Write the equation in the form ax2 + bx + c = 0 • Factorise the left hand side of the equation. • Use the fact that if (x – p)(x – q) = 0, then x = p or x = q, to calculate the x-intercepts. PAY SPECIAL ATTENTION • NB NB Determine the axis of symmetry. • Substitute the x-value of the axis of symmetry into the original equation of the function to calculate the co-ordinates of the turning point. HINT! hint • Plot the points and then draw the function using free hand. e.g. 3 EG - worked examples Sketch the graph of f(x) = x2 – 5x – 6 exams 1. y-intercept f(0) = –6 Therefore the co-ordinates of the y-intercept are (0; –6) 3 2. activity x-intercept ACTIVITIES boy x2 – 5x – 6 = 0and girl-left and right of page 3 (x – 6)(x + 1) = 0 3 activity x = 6 or x = –1 3 (6; 0) and (–1; 0) 3. Axis ofStepsymmetry by step comment comment –b x = ___ ​  2a ​ 3 – (–5) = _____ ​  2(1) ​ 3 5 = __​  2 ​ 3 4. Turning point (5 ) (5 ) (5 ) 2 ​  2 ​  ​= ​​ __ f ​__ ​  2 ​  ​​–5​__ ​  2 ​  ​– 6 3 1 = –12 __ ​  4 ​ 3 ∴TP ​ __5 (2 1 ​  ​;–12​ __​  ​ 4 ) 3 © Department of Basic Education 2015 UNIT 4 functions 65 Mind the Gap Mathematics

4 Unit 5. Sketch graph y (–1; 0) 0 (6; 0) x (0; –6) ​(__ 4 ​  )​ 5 49 ​  2 ​; –​ ___ 3x-intercepts 3y-intercept 3shape 3turning point Determining the equation of a quadratic function Given the x-intercept and one point Given the turning point and one point • Use the formula: y = a(x – x1)(x – x2). • Use the formula: y = a(x + p)2 + q. • Substitute the values of the x-intercepts. • Substitute the co-ordinates of the turning • Substitute the given point which is not the point (p; q). x-intercept. • Substitute the given point. • Solve for a. • Solve for a. • Write the equation in the form • Write the equation in the form f(x) = ax2 + bx + c. y = a(x +p)2 + q or f(x) = ax2 + bx + c depending on the instruction in the question. Given the co-ordinates of three points on the parabola • Use the formula: y = ax2 + bx + c. • One of the given point is the y-intercept, therefore c is given, so substitute its value. • Substitute the co-ordinates of the other two points into y = ax2 + bx + c. • Solve the two equations simultaneously for a and b. Nature of the roots and the quadratic function Nature of roots Quadratic function Real roots y y Δ>0 x or x NOTE: there are two x-intercepts. 66 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 Equal roots y y Δ=0 x or x NOTE: there is only one x-intercept. Non-real roots y y Δ>0 x or x NOTE: there are no x-intercepts. Activity 2 The sketch represents the graph of the parabola given by f(x) = 2 – x – x2. Points A, B and C are the intercepts on the axes and D is the turning point of the graph. y D B A C x f 1.1 Determine the co-ordinates of A, B and C. (4) 1.2 Determine the co-ordinates of the turning point D. (3) 1.3 Write down the equation of the axes of symmetry of f(x– 5). (1) 1.4 Determine the values of x for which – f(x) > 0. (2)  [10] © Department of Basic Education 2015 UNIT 4 functions 67 Mind the Gap Mathematics

4 Unit Solutions 1.1 B(0; 2) 2 – x – ​x2​​= 0 3 ​x​​+ x – 2 = 0 2 (x – 1)(x + 2) = 0 3 x = 1 or x = –2 3 A(–2; 0) and C(1; 0) 3 (4) –b 1.2 x = __ ​  2a ​ – (–1) = _____ ​  2(–1) ​ 3 1 = – __ ​  2 ​ 3 ( ) 1 f ​– ​ __ ( ) ( ) 1 __ 1 __ 2 2 ​  ​= 2 – ​– ​  2 ​  ​– ​​ – ​  2 ​  ​​ ​ 9 1 = __ ​  4 ​= 2​ __4 ​ ( D​– __ )1 9 ​  2 ​; __ ​  4 ​  ​ 3 (3) 9 1 1.3 x = __ ​  2 ​or x = 4​ __2 ​3 (1) 1.4 x < –2 3 or x > 1 3 (2) [10] Activity 3 The sketch represents the graph of the parabola given by f(x) = ax2 + bx + c and the straight line defined by g(x) = mx + c Points A, B ,C and D are the intercepts on the axes. E is the point of intersection of the two graphs. y g C(0; 3) E D x A(–1; 0) B(3; 0) f 2.1 Write down the co-ordinates of point D if D is the image of B after B has been translated two units to the right. (1) 2.2 Determine the equation of g.(3) 2.3 Determine the equation of the function f in the form f(x) = ax2 + bx + c.  (4) 68 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 2.4 Determine the coordinates of E. (4) 2.5 Write down the values of x for which f(x) ≥ g(x).(2) [14] Solutions 2.1 D(5; 0) 3 (1) 2.2 g(x) = mx + 3 3–0 3 0 = m(5) + 3 or mg = ____ ​  0– 5 ​ = – __ ​  5 ​ 3 3 m = –​ __5 ​ 3 3 g(x) = – __ ​  5 ​x + 3 3 (3) 2.3 f(x) = a(x + 1)(x – 3) 3 3 = a(0 + 1)(0 – 3) 3 a=13 f(x) = –(x + 1)(x – 3) f(x) = –x2 + 2x + 3 3 (4) 3 2.4 –​ __5 ​x + 3 = –x2 + 2x + 3 3 13 x2 – __ ​  5 ​x = 0 x ​(x – __ 13 ) ​  5 ​  ​= 0 3 13 x=0 or x = __ ​  5 ​= 2,60 3 (13 ) g ​__ ​  5 ​  ​= – __ 3 13 ​  5 ​​__( ) ​  5 ​  ​+ 3 36 = __ ​  25 ​ = 1,44 3 ( ∴E ​​  13 __ 36__ ) 3 ( __ 11 __ ) 5 ​;​  25 ​  ​ or E ​2​  5 ​; 1​  25 ​  ​ or E (2,60; 1,44) (4) 13 2.5 0 < x < ​ ___ 5 ​33 (2) [14] 4.3.3 The hyperbolic function a Hyperbola of the form y = __ ​  x ​or xy = a where a ≠ 0; x ≠ 0; y ≠ 0. Properties Shape 1. a>0 a<0 y y y = –x y=x 0 x x 0 © Department of Basic Education 2015 UNIT 4 functions 69 Mind the Gap Mathematics

4 Unit 2. (i) Domain : x ∈ 핉; x ≠ 0 (i) Range: y ∈ 핉; y ≠ 0 3. The horizontal asymptote is the x-axis 4. The vertical asymptote is the y-axis 5. If a < 0, the graph lies in the 2nd and 4th quadrant 6. If a > 0 , the graph lies in the 1st and 3rd quadrant 7. The lines of symmetry are: y = x and y = –x. SKETCHING THE HYPERBOLA OF THE FORM: a y = __ ​  x ​or xy = a • The graph does not cut the x-axis and the y-axis (asymptotes) • Use the table and consider both the negative and positive x-values • a determine two quadrants where the graph will be drawn Activity 4 1 –4 1. Sketch the graph of y = __ ​  x ​by plotting points. 2. Sketch the graph of y = __ ​  x ​by plotting the Describe the main features of the graph. (4) points. Describe the main features of the graphs. (4) Solution a=1 Solution a > 0, the graph lies in the 1 and 3 quadrant st rd a = –4 a < 0, the graph lies in the 2nd and 4th quadrant – 3 – 2 – 1 – _​  1 ​ 0 ​ _1 ​ 1 2 3 2 2 –4 –2 –1 0 1 2 4 –​ _13 ​ –​ _1 ​ – 1 2 –2 undefined 2 1 ​  1 ​ _ 2 ​ _1 ​ 3 1 2 4 undefined –4 –2 –1 y 3 y 4 – ​  x ​ y=_ 1 y = __ x ​ ​  –4 2 3 – 1 2 – 1 – – – – – – – -3 -2 -1 0 1 2 3 x – –4 –3 –2 –1 0 1 2 3 4 x -1 –1 – -2 –2 – -3 –3 – –4 • Domain: x ∈ 핉; x ≠ 0 3 • Domain: x ∈ 핉; x ≠ 0 3 • Range: y ∈ 핉; y ≠ 0 3 • Range: y ∈ 핉; y ≠ 0 3 • Asymptotes: x = 0 and y = 0 3 • Asymptotes: x = 0 and y = 0 3 • Lines of symmetry y = x and y = –x 3 (4) • Lines of symmetry y = x and y = –x 3 (4) [8] 70 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 4.3.4 The hyperbola a a Hyperbola of the form y = __ ​  x ​+ q is the translation of the graph of y = __ ​  x ​ vertically by q units. The Horizontal asymptote (x–axis) will also shift q units vertically (up or down). Activity 5 1 –4 1. Consider the function y = __ ​  x ​– 2 2. Consider the function f(x) = __ ​  x ​+ 1 1.1 Determine : 2.1 Determine: a) the equations of the a) the equations of the asymptotes asymptotes b) the coordinates of the b) the coordinates of the x–intercepts x–intercepts 1.2 Sketch the graph 2.2 Sketch the graph 1.3 Write down: 2.3 Write down the domain and a) the domain and range range b) the lines of symmetry 2.4 If the graph of f is reflected by y = x + c and y = –x + c the line having the equation y = –x + c, the new graph (10) coincides with the graph of f(x). Solutions Determine the value of c. 1.1 (9) a)  The horizontal asymptote is y = –2 since the graph Solutions moved 2 units down and the 2.1 vertical asymptote is x = 0 3 a) The horizontal asymptote is denominator cannot equal to y = 1 3 since the graph moved zero. 1 units up and the vertical b) For x – intercepts let y = 0 asymptote is x = 0 denominator 1 cannot equal to zero. 0 = __ ​  x ​– 2 3 3 b) For x–intercepts let y = 0 0 = 1 – 2x (multiplying by LCD –4 which is x) 0 = ​ __ x ​+ 1 3 3 2x = 1 3 0 = –4 + x (multiplying by LCD 1 which is x x = __ ​  2 ​ 3 x=4 3 ​(__ 1 ​  2 ​; 0 )​ (4; 0) © Department of Basic Education 2015 UNIT 4 functions 71 Mind the Gap Mathematics

4 Unit 1.2 2.2 x –4 –2 –1 0 1 2 4 x –4 –2 –1 0 1 2 4 y – 2​ _14 ​ – 2​ _12 ​ –3 undefined –1 – 1​ _1 ​ 2 – 1​ _3 ​ 4 y 2 2 5 undefined –3 –1 0 y y 2 3 shape ​  x ​– 2 y=_ 1 4 ✓ shape 3 shape 3 asymptotes 1 ✓ asymptotes 3 asymptotes ​  x ​+ 1 y = __ –4 3 –4 –3 –2 –1 0 1 2 3 4 x 2 –1 1 –2 –4 –3 –2 –1 0 1 2 3 4 x –3 –1 –4 –2 1.3 –3 a) Domain: x ∈ 핉; x ≠ 0 3 –4 Range: y ∈ 핉; y ≠ 2 3 b) y = x and y = –x 2.3 Domain: x ∈ 핉; y ≠ 0 3 translation 2 units down therefore Range: y ∈ 핉; y ≠ 1 3 y = x – 2 and y = –x – 2 3 2.4 The asymptotes are ∴c = –2 x = 0 and y = 1 Or substitute (0; 2) point of intersection of y = –x + c the two asymptotes in y = x + c or y = –x + c 1 = –(0) + c And PAY SPECIAL ATTENTION calculate the value of c 1=c PAY SPECIAL ATTENTION NB NB NB 3 NB lines are y = –x + 1 and y = x + 1 [10] [9] HINT! HINT! hint Compare this graph with the one in activity 4 (a) hint Compare this graph with the one in activity 4 (b) e.g. EG - worked examples e.g. EG - worked examples exams exams activity activity ACTIVITIES boy ACTIVITIES boy and girl-left and and girl-left and 72 UNIT 4 functions  right of page © Department right of Basic Education 2015 of page  Mind the Gap Mathematics activity activity

Unit 4 4.3.5 Hyperbola of the form a y = ____ ​  x+ p ​+ q where a ≠ 0, x ≠ 0, y ≠ 0 1. Shape a>0 a<0 The dotted lines are the asymptotes PAY SPECIAL ATTENTION 2. NB NB y ∈ 핉; y ≠ q Domain: x ∈ 핉; x ≠ –p. Range: 3. The horizontal asymptote is y = q 4. The vertical HINT! asymptote is x + p = 0 ∴x = –p 5.hint The lines of symmetry are y = x + c and y = x + c e.g. 4 EG - worked examples 8 Consider exams g(x) = ____ ​  x – 2 ​–3 has the horizontal asymptote at y = –3 and 8 x – 2 ≠ 0 ∴x ≠ 2 because if x = 2 the denominator of the expression ____ ​  x – 2 ​ 8 8 would be ____ ​  2 – 2 ​ = __ ​  0 ​ which is undefined because the denominator is zero. activity Thus the graph is undefined ACTIVITIES boy for x – 2 = 0 ∴x = 2 is the vertical asymptote and girl-left and 8 The graph y = __ right of page ​  x ​shift 2 units to the right and 3 units down to form the activity 8 graph g(x) = ____ ​  x – 2 ​– 3 SKETCHING THE HYPERBOLA comment OF THE FORM Step by step comment a y = ____ ​  x + p ​+ q • Write down the asymptotes • Draw the asymptotes on the set of axes as dotted lines • Use a to determine the two quadrants where the graph will be drawn • Determine the x – intercept(s) let y = 0 • Determine the y – intercept(s) let x = 0 • Plot the points and then draw the graph using free hand © Department of Basic Education 2015 UNIT 4 functions 73 Mind the Gap Mathematics

4 Unit Activity 6 2 3 1. Consider the function f(x) = ____ ​  x – 3 ​+ 1 2. Consider the function f(x) = ____ ​  x – 1 ​– 2 a) Write down the equations of the a) Write down the equation of the asymptotes. (2) asymptotes of f (2) b) Calculate the coordinates of the intercepts b) Calculate the coordinates of of the graph of f with the axes. (3) the x and y-intercepts of f(4) c) Sketch the graph of f clearly showing c) Write the domain and range (2) the intercepts with the axes and the d) Sketch the graph of f clearly asymptotes.(3) showing ALL asymptotes and d) Write down the range of y = –f(x).(1) intercepts with the axes. (3) e) Describe, in words, the transformation –3 of f to g if g(x) = ____​  x + 1 ​– 2 (2) [22] Solution Solution 1. a) x = 3 and y = 1 3 3 (2) 2. a) 3 x = –1 y = –2 3 (2) 2 b) f(x) = ____ ​  x– 3 ​+ 1 b) y – intercept 3 y – intercept y = ____ 2 ​  0 – 3 ​+ 1 = __ 1 ​  3 ​ 3 y = ___ ​  0–1 ​– 2 = –5 ( 1) ​0; __ ​  3 ​  ​ (0; –5) 3 ____ 3 x – intercept 3 0 = ​  x– 1 ​–2 2 x – intercept 0 = ____ ​  x – 3 ​+ 1 3 2 = ____ 3 ​  x – 1 ​ 3 2(x – 1) = 3 0 = 2 + 1(x – 3) 2x – 2 = 3 0=2+x–3 2x = 5 3 x = 1 ∴(1; 0) (4) 3 x = ​  5__2 ​ ​(__ ​  2 ​; 0 )​ c) Domain: x ∈ 핉; x ≠ 3 3 5 (3) Range: y ∈ 핉; y ≠ 1 3 (2) c) a > 0 d) a > 0 y y (2; 1) (4; 3) 0 ​  5__2 ​ x 1 __ (3; 1) (1; –2) ​  3 ​ 0 1 x (–2; –3) (2; –1) –5 PAY SPECIAL ATTENTION NB NB 3 intercepts 3 asymptotes 3 shape (3) 3 intercepts 3 asymptotes 3 shape (3) [11] HINT! hint Inx =the3 graph 1 (d) the points (4; 3), x = 4 was chosen because it has x-coordinate greater than the vertical asymptote. The point (2; –1), was chosen because has x-coordinate x = 2 is less than x = 3 the vertical asymptote. These points can also be used to help determining in which quadrants the graph must be drawn. The points (2; 1) and (–2; –3) on graph 2 (iii) were chosen similarly. e.g. EG - worked examples 74 UNIT 4 functions  © Department of Basic Education 2015  exams Mind the Gap Mathematics

Unit 4 3 d) f(x) = ____ ​  x – 1 ​– 2 ( 3 – f(x) = – ​____ ) ​  x – 1 ​– 2 ​ –3 – f(x) = ____ ​  x – 1 ​+ 2 Range: y ∈ 핉; y ≠ 2 3 (1) –3 e) g(x) = ____ ​  x + 1 ​– 2 3 g(x) = _____ ​  –x – 1 ​– 2 Since x is negative this is the reflection 3 of f about the y-axis 3 (2) [11] Activity 7 a The diagram below represents the graph of f(x) = ____ ​  x + p ​ + q. T(5; 3) is a point on f. y– 4– T(5; 3) 2– 10 x – – – – – – – –4 –2 0 2 4 6 8 –2 – –4– 4.1 Determine the values of a, p and q(4) 4.2 If the graph of f is reflected across the line having the equation y = –x + c, the new graph coincides with the graph of y = f(x). Determine the value of c.(3) [7] Solutions 4.1 3p = 4 and q = 2 3using the asymptotes a Substitute T(5; 3) into y = ____ ​  x – 4 ​+ 2 a 3 = ____ ​  5 – 4 ​+ 2 3 3 = a + 2 a=13 (4) 4.2 Substitute (4; 2) 3into y = –x + c 32 = –(4) + c ∴c = 6 3 (3) [7] © Department of Basic Education 2015 UNIT 4 functions 75 Mind the Gap Mathematics

4 Unit Activity 8 a Sketched below are the graphs of f(x) = (x + p)2 + q and g(x) = ____ ​  x + b ​+ c ( 1 ) A ​2 __ ​  2 ​; 0 ​ is a point on the graph of f. P is the turning point of f. The asymptotes of g are represented by the dotted lines. The graph of g passes through the origin y 4– 3– g 2 g f 1– x – – – – – – – – – –3 –2 0 –1 1 2 3 4 5 6 A ​(__ ​  2 ​;0 )​ 5 –1– –2– P –3– 5.1 Determine the equation of g.(4) 5.2 Determine the coordinates of P, the turning point of f.(4) 5.3 Write down the equation of the asymptotes of g(x – 1). (2) 5.4 Write down the equation of h, if h is the image of f reflected about the x–axis.(1) [11] Solutions 5.1 Using the asymptotes 3b = 1 and c = 2 3 a Substitute (0; 0) into y = ____ ​  x – 1 ​+ 2 30 = ​  0 a ____ – 1 ​+ 2 ⇒ 0 = –a + 2 ∴a = 2 3 2 y = ____ ​  x – 1 ​+ 2 (4) 5.2 Axis of symmetry p = 1 3 f(x) = (x – 1)2 + q ​(__​  2 ​;0 )​ 3 5 30 = (​​​  5_2 ​– 1 )​​ ​ + q 2 0 = ​ 9_ ​ + q 4 q = – ​ 9_4 ​ ( ) ∴P ​1;–​ __94 ​  ​3 (4) 76 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 2 5.3 g(x) = ____ ​  x –1 ​+ 2 2 g(x – 1) = _______ ​  (x– 1) – 1 ​+ 2 substitute x with (x – 1) 2 g(x – 1) = ____ ​  x – 2 ​+ 2 3 x = 2 and y = 2 3 (2) 5.4 f(x) = (x – 1)2 – ​ _94 ​ Reflection about the x – axis y changes the sign – y = (x – 1)2 – ​ 9_ ​ 4 [ y = – ​ (x – 1) – ​ 9_ ​  ​ 4 2 ] 9 _ y = –(x – 1) + ​ 4 ​ 3 2 (1) [11] 4.3.6 The exponential function An exponential function can be represented with a general formula y = abx+p + q; b > 0 Shape and properties of an exponential function x x y=b;b>1 y=b;0<b>1 5_ y 5_ y 4_ 4_ 3_ 3_ 2_ 2_ 1 1 x _ _ _ _ _ _ _ _ _ _ –3 –2 –1 0 1 x –3 –2 –1 0 1 2 • The graph passes through the • The graph passes through the point (0; 1). point (0; 1). • Domain: x ∈ 핉 • Domain: x ∈ 핉 • Range: y > 0 but for y + bx + q, • Range: y > 0 but for y = bx + q, the range will be at y > q. the range will be at y > q. • The graph is smooth, continuous • The graph is smooth, continuous and an increasing function. and a decreasing function. • Asymptote is at y = 0 but for • Asymptote is at y = 0 but for y = bx + q, the horizontal y = bx + q, the horizontal asymptote will be at y = q. asymptote will be at y = q. NOTE: The two functions are a reflection of each other about the y-axis. © Department of Basic Education 2015 UNIT 4 functions 77 Mind the Gap Mathematics

HINT! 4 Unit hint e.g. 5 EG - worked examples Given: f(x) = 2x exams 1.1 Draw the graph of f(x) = 2x, show at least three points on the sketch. 1.2 Draw, on the same system of axes the graph of f –1, the inverse of f. 1.3 Write down the equation of f –1 in the form y = ... activity Solutions ACTIVITIES boy and girl-left and right of page 1.1 Start by drawing the table: activity x –1 0 1 f(x) 0,5 1 2 Step by step comment comment Then plot the graph using the points y 4 f 3 f​ –1 (1,2) 2 ​​ (0,1) (–1,0.5) 1 (2,1) x (1,0) –3 –2 –1 1 2 3 4 (0,5,–1) 1.2 The sketch of f –1 is obtained by interchanging the x and y co-ordinates of f. 1.3 y = 2x x = 2y 3 y = log2 x 3 [2] 78 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 4 e.g. 6 EG - worked examples The sketch represents the graph given by f(x) = ax. exams y f activity ACTIVITIES boy and girl-left and right of page activity A B x Step by step comment comment 2.1 Write down the coordinates of point A. (1) 2.2 How can we tell that 0 < a < 1? (1) __1 2.3 Determine a if B is the point ​3; ​  27 ​  ​.(2) ( ) 2.4 Determine the equation of the graph obtained if f is reflected about the y–axis.(2) 2.5 What are the coordinates of the point of intersection of the two graphs?(1)  [7] Solutions 2.1 A(0; 1) 3 2.2 Because the graph is a decreasing function. 3 2.3 f(x) = ax 1 __ ​  27 ​= a3 3 (​​ ​3–1 ​ ​  )3​​ ​= a3 1 a = __ ​  3 ​ 3 (1 ) x ​  3 ​  ​​ ​ 2.4 f(x) = ​​ __ y = ​​(__ ​  3 ​  )​​ ​becomes y = ​​(__ ​  3 ​  )​​ ​3 x –x 1 1 ​ ​  )–​​ x​ ∴y = (​​ ​3–1 y = 3x 3 2.5 (0; 1) 3 [7] © Department of Basic Education 2015 UNIT 4 functions 79 Mind the Gap Mathematics

4 Unit Activity 9 The curve of an exponential function is given by f(x) = kx and cuts the 9 ( ) y–axis at A (0; 1) while B ​2:​ __4 ​  ​lies on the curve. y f B A x Determine 1.1 the equation of the function f.(3) 1.2 the equation of the asymptote of h if h(x) = –f(x).(2) 1.3 the range of h.(1) 1.4 The equation of the function g of which the curve is the reflection of the curve of f in the line y = x.(2) Solutions 1.1 f(x) = kx __ 9 ​  4 ​= a2 3 ()​​ __ 3 2 ​  2 ​  ​​ ​= a2 3 () x a = __ 3 ∴f(x) = ​​ ​  3__2 ​  ​​ ​ ​  2 ​ 3 (3) 1.2 y = 0 33 (2) 1.3 y ≤ 0 3 (1) 1.4 g(x) = log ​ __3​x 33 (2) 2 [8] 80 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 4.4 Inverse functions • The inverse of a PAY SPECIAL function takes the y-values (range) of the function to ATTENTION NB NB the corresponding x-values (domain) and vice versa. Therefore the x and y values are interchanged. • The function is reflected along the line y = x to form the inverse. The notation for the inverse of a function is f –1. HINT! •hint e.g. 7 EG - worked examples Given f(x) = 2x + 6. exams 1. Determine f –1(x) 2. Sketch the graphs of f(x), f –1 (x) and y = x on the same set of axis Solutions activity 1. In order to find theACTIVITIES inverse boy of a function, there are two steps: and girl-left and right of page STEP 1: Swap the x and y activity y = 2x + 6 3 becomes x = 2y + 6 3 We then rewrite the equation to make y the subject of the formula. Step by step comment comment Therefore, STEP 2: make y the subject of the formula x = 2y + 6 x – 6 = 2y 3 So y = ​ _12 ​x – 3 3 We can say that the inverse function f –1(x) = ​ 1_ ​x –3 2 2. y 6 y=x y = 2x +6 3 axis of symmetry –3 0 3 6 x 3 –3 y = ½x – 3 3 • Every point on the function has the same coordinates as the corresponding point on the inverse function, except that they are swapped around. • Example: (–3; 0) on the function is reflected to become (0; –3) on the inverse function. • Any point (a; b) on the function becomes the point (b; a) on the inverse. • To find the equation of an inverse function algebraically, we interchange x and y and then solve for y. • To draw the graph of the inverse function, we reflect the original graph about the line y = x, the axis of symmetry of the two graphs. © Department of Basic Education 2015 UNIT 4 functions 81 Mind the Gap Mathematics

HINT! 4 Unit hint e.g. 8 EG - worked examples 1. a) Sketch f(x) = 2x2 exams b) Determine the inverse of f(x) c) Sketch f –1(x) and y = x on the same axes as f(x) Solution y activity f(x) = 2x2 1. b) y = 2x2 ACTIVITIES boy and girl-left and x = 2y2 3 right of page __ 4 y=x ​  x2 ​ ​ 3 y = ± ​√__ activity • This is not a function. 2 __ _x • Check it with a vertical line Step test.by step comment comment 3 √ y = ± ​ ​  2 ​ ​ There are two y-values for one x x-value. -4 -2 2 4 • Not all inverses of functions are also –2 3 functions. Some inverses of functions 3 are relations. • If an inverse is not a function, then –4 we can restrict the domain of the function in order for the inverse to be a function. • To make the inverse a function, we need to choose a set of x-values in the function and work only with those. We call this ‘restricting the domain’. • A one to one function has an inverse that is a function Example: y = 3x + 4 is a one to one function. For every x value there is one and only one y value The inverse of is a function. • A many to one function has an inverse that is not a function. However, we can restrict the domain of the function to make its inverse a function. Example: y = 2x2 is a many to one function. For two or many x values there is one y value. (if x = 2, then y = 8. __ ​  x2 ​ ​, is not a function. If x = –2, then y = 8). Therefore, its inversey = ± ​√__ • To check for a function, draw a vertical line. If any vertical line cuts the graph in only one place, the graph is a function. If any vertical line cuts the graph in more than one place, then the graph is not a function. • To check for a one-to-one function, draw a horizontal line. If any horizontal line cuts the graph in only one place, the graph is a one-to-one function. If any horizontal line cuts the graph in more than one place, then the graph is a many-to-one function. [5] Activity 10 1. a) If f(x) = –3x2 , write down the equation for the inverse function in the form y = ….................  (2) b) Determine the domain and range of f(x) and f –1 (x)(4) c) Determine the points of intersection of f(x) and f –1 (x)(4) 2. a) If g(x) = 3x + 2, find g –1(x)  (2) b) Sketch g, g and the line y = x on the same set of axes.  –1 (3)  [15] 82 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 Solutions 1. a) For f(x) = – 3x2. f –1 (x): x = –3y2 3 x – __ ​  3 ​= y2 ____ x √ y = ± ​ − __ ​  3 ​ ​ 3 (2) b) f(x) f–1(x) Domain x∈ 핉 3 x≥0 3 Range y≥0 3 y∈ 핉 3 (4) c) To determine the points of intersection, we equate the two equations.  The line y = x, the axis of symmetry of f(x) and f –1(x), can also be used to determine the points of intersection of f(x) and f –1(x). y = x and f(x) = − 3x2 ∴ x = −3x2 ∴ 3x2 + x = 0 3 ∴ x(3x + 1) = 0 3 1 ∴ x = 0 or x = – __ ​  3 ​ 3 Substitute x = 0 in y = x ∴ y = 0 ∴ (0; 0) 3 Substitute x = – __ 1 ​  3 ​in y = x ∴ y = – __ 1 ​  3 ​∴ ​– __ 1 1 ( ​  3 ​;– __ ) ​  3 ​  ​ (4) 2. a) g(x) = 3x + 2 3 b) y 4 For g (x), x = 3y + 2 −1 g(x) = 3x + 2 3 3 x – 2 = 3y y=x x−2 2 y = ____ ​  3 ​ 3 1 x 2 y = __ ​  3 ​− __ ​  3 ​ 3 3 –3 –2 –1 0 1 2 3 4x –1 ​  x ​ – ​  2__​ g–1(x) = _3 3 –2 –3 (4) [15] Given: g (x) = – x2 where x < 0 and y < 0 (a) Write down the inverse of g, g–1 in the form h(x) = ........... (3) (b) Sketch the graphs of g, h and y = x on the same set of axis. (4) Solutions (a) y = –x2 x = –y2 – x = y2 3 ___ ± ​√–x ​= y 3 ___ – ​√–x ​= y where x < 0 and y < 0 ___ ∴ h(x) = –​√–x ​ 3 (3) © Department of Basic Education 2015 UNIT 4 functions 83 Mind the Gap Mathematics

4 Unit (b) Y y=x X 0 h g For g correct shape 3and the intercept 3 For h correct shape 3and the intercept 3 (4) [7] 4.5 The logarithmic function PAY SPECIAL ATTENTION NB NB • y = logxa is a logarithmic function with a = log number, x=log base • y = logxa Reads “y is equal to log a base x” • The logarithmic HINT! function is only defined if a > 0, a ≠ 1 and x > 0 • hint An exponential equation can be written as a logarithmic equation and vice versa e.g. 9 EG - worked examples Write each of the following exponential equations as logarithmic exams equations: • ​26​ ​= 64 • 5³ = 125 activity Solutions 1. 26 = 64 ACTIVITIES boy and girl-left and ∴ 6 = log2 64 right of page activity 2. 53 = 125 ∴ 3 = log5 125 log Step PAY by step ATTENTION SPECIAL comment comment NB 64 = 2 6 NB log2 64 = 6 PAY SPECIAL ATTENTION base NB NB number HINT! The inverse of the exponential hint function y = ax is x = ay In order to make y the subject HINT! of thehint 10 formula, x = ay, we use e.g. EG - worked examples the log function. Given: f(x) = 2x y = log ax is the inverse of y exams a) Determine f –1 in the form y = …… = ax. e.g. EG - worked examples b) Sketch the graphs of f(x), f –1 (x) and y = x on the same set of axes. c) Write the domain and range of f (x) and f –1 (x) activity exams ACTIVITIES boy and girl-left and right of page activity 84 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 Solutions a) The inverse of the exponential function y = 2x is x = 2y which can be written as y = log2 x. 3 b) To plot the graph, use a table of values: First make a table for y = x –2 –1 0 1 2 3 y = ​2​x​ ¼ ½ 1 2 4 8 Make a table for y = log2 x x ¼ ½ 1 2 4 8 y = log2x –2 –1 0 1 2 3 y 5 4 3 3 y = log2x 2 3 1 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 x –1 –2 [3] Let’s compare the two graphs on the Cartesian plane. 9 y = 2x y=x 8 7 6 5 3 3 4 3 y = log2x 2 3 1 3 –3 –2 –1 0 1 ✓2 3 4 5 6 7 8 9 –1 –2 The graph of y = log2 x is a reflection about the y = x axis of the exponential graph of y = 2x. © Department of Basic Education 2015 UNIT 4 functions 85 Mind the Gap Mathematics

4 Unit Activity 11 The graph of h(x) = ax is sketched below. A​(–1; ½ )​is a point on the graph of h. 1. Explain why the coordinates y of Q are (0; 1). (2) h 2. Calculate the value of a. (2) 3. Write down the equation for the inverse function, h–1 in the form y=… (1) 4. Draw a sketch graph of h–1. Indicate the coordinates A(–1; ½) of two points that lie Q on this graph. x (2) 5. Read off from your graph the values of x for which log2 x > –1. (1) [8] Solutions 1. h(0) = a0 = 1. 3 Any base raised to the power of 0 is 1. 3 (2) 2. h(x) = ax and A(–1; ½) so a –1 = ½3 a –1 = 2 –1 so a = 23 and y = 2x (2) 3. Interchange x and y, so x = 2y and y = log2 x 3 (1) 4. y 4 3 h–1 2 3 1 (2; 1) 3 0 1 2 3 4 5 6 7 8 x (1; 0) –1 –2 –3 –4 (2) 5. x > 0.5 3 (1) [8] 86 UNIT 4 functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 4 What you need to be able to do: • Understand the concept of the inverse of a function and find the equations of the inverses. • The line y = x is the line of symmetry of the function and the inverse of the function • The logarithmic function and the exponential function are inverse functions of each other. • If the inverse is not a function, restrict the domain of a function in order to make the inverse a function • Identify axes of symmetry for parabolas and hyperbolas • Sketch the graphs of different functions using their characteristics e.g. asymptotes, x- and y –intercepts and turning points • Determine the functions equations from a graph • Solve problems involving two or more graphs • Understand the concept of the inverse of a function and the equation of the inverses • The line y = x is the line of symmetry of the function and the inverse of the function • The logarithmic functions and the exponential function are inverse function of each other • If the inverse is not a function, restrict the domain of a function in order to make the inverse a function. Keep going! © Department of Basic Education 2015 UNIT 4 functions 87 Mind the Gap Mathematics

5 Unit 5 Unit Trig functions To ensure that all the critical values are 5.1 NB Graphs ofNB trigonometric PAY SPECIAL ATTENTION indicated on the graph, we have to use the correct x-values. functions If y = a sinbx, then HINT! ​  b ​ will give us the intervals ____0 90 hint we have to use from 0. In Graph 1. The sine function: y = a sinb (x + p) + q our example b = 1, therefore ​  90 _____0 ​= 900. 1 e.g. 1 EG - worked examples Therefore we will use x-values of (00,900, 1800, Sketch the graph of y = sin x for x 270o, 360o etc) exams If we use a calculator, • We can make use of a table or a calculator to determine the critical points we will use 90 as the “step”. on the graph. • The endpoints of the domain must be included i.e. x = −360° and x = 360° • Allactivity intercepts with the x and y axis must be indicated as well as all minimum ACTIVITIES boy and maximum pointsand girl-left(turning right of page and points) activity Solution x –360° –270° –180° –90° 0° 90° 180° 270° 360° comment y Step by step comment 0 1 0 –1 0 1 0 –1 0 (–270°; 1) Y (90°; 1) y = sinx X –360° –270° –180° –90° 0 90° 180° 270° 360° (–90°; –1) (270°; –1) Domain: all the possible x values on the graph Range: all the possible y-values on the graph Amplitude: the maximum distance from the equilibrium position Period: number of degrees to complete a wave or a cycle. 88 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 Activity 1 Use the graph y = sin x above to answer these questions: 1. What are the maximum and minimum values of y = sin x? (2) 2. Write down the domain and the range of y = sin x. (4) 3. Write down the x-intercepts of y = sin x. (2) 4. What is the amplitude of the graph of y = sin x? (1) 5. What is the period of the graph of y = sin x? (1) [10] Solutions y = sinx 1 Maximum Values 13 , at x = –270° and 90° Minimum Values –1 3, at x = –90° and 270° (2) 2 Domain x ∈ [− 360°;360°], x ∈ ℝ33 Range [–1; 1] y ∈ ℝ33 (4) 3 x-intercepts –360°, –180°, 0°, 180° and 360°.33 (2) 4 Amplitude 13 (1) 5 Period 360°3 (1) PAY SPECIAL ATTENTION NB NB [10] Graph 2. The cosine function: HINT! To ensure that yhint = a cos b(x + p) + q all the critical values are indicated on the graph, we have to use the correct x-values. e.g. 2 EG - worked examples If y = acosbx, then ​ ____ 900 b ​ will give us the intervals we have to use from 0°. In our example b = 1, therefore Sketch the graph of y = cos x for x ∈ [–360°;360°] ____ ​  90 ​ = 900. 0 exams 1 • We can make use of a table or a calculator to determine Therefore we will use x-values of the critical points on the graph. (00,900, 1800, 270o, 360o etc) • The endpoints of the domain must be included i.e. If we use a calculator, we will use 900 as the “step”. x = –360° and x = 360° • Allactivity intercepts with the x and y axis must be indicated as well ACTIVITIES boy as all minimum and maximum points (turning points) and girl-left and right of page x activity –360° –270° –180° –90° 0° 90° 180° 270° 360° y 1 0 –1 0 1 0 –1 0 1 comment Step by step comment y (360°; 1) (–360°; 1) 1 y = cosx x –360° –270° –180° –90° 0 90° 180° 270° 360° –1 (–180°; –1) (180°; –1) © Department of Basic Education 2015 UNIT 5 trig functions 89 Mind the Gap Mathematics

5 Unit y = cos x To ensure that all the 1 Maximum Values 1, at x = 0° and 360° critical values are indicated 2 Minimum Values –1, at x = –180° and 180° on the graph, we have to use the correct x-values. 3 x-intercepts –270°, –90°, 90° and 270°. 450 ​  b ​ will If y = atanbx, then _____ 4 Amplitude 1 give us the intervals we have to use from 0. In our example 5 Period 360° 450 b = 1, therefore ​ _____ 1 ​= 45 . 0 6 Domain x ∈ [− 360°;360°], x ∈ 핉 Therefore we will use x-values NB PAY SPECIAL ATTENTION NB of (00,450, 900, 135o etc) 7 Range [–1; 1] y ∈ 핉 If we use a calculator, we will use 450 as the “step” Graph 3. The tangent function: HINT! yhint = a tan b(x + p) + q e.g. 3 EG - worked examples Sketch the graph of y = tan x for x ∈ [− 180°;180°] exams • All intercepts with the x and y axis must be indicated. • The endpoints of the domain must be included i.e. x = − 180°and x = 360° • The equations of the asymptotes must be written on the graph. activity ACTIVITIES boy Solution and girl-left and right of page activity x –180° –135° –90° –45° 0° 45° 90° 135° 180° 225° 270° 315° 360° unde-Step by step comment comment unde- unde- y 0 1 –1 0 1 –1 0 1 –1 0 fined fined fined y _ 2_ y = tanx _ 1_ (225°; 1) (–135°; 1) (45°; 1) _ _ x –180° –135° –90° –45° 0 45° 90° 135° 180° 225° 270° 315° 360° _ (–45°; –1) –1_ (135°; –1) (315°; –1) _ –2_ _ _ y =  tan x 1 Asymptotes x = –90°, x = 90° and x = 270° 2 x-intercepts –180°, 0°, 180° and 360°. 3 Period 180° 4 Domain x ∈ [− 180°;360°], x ∈ 핉 5 Range (–∞; ∞). y ∈ 핉 90 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 5.2 The effect of a on the y = -1sinx …. Amplitude = 1 y = 2sinx….Amplitude = 2 NB shape of NBthe graph: PAY SPECIAL ATTENTION y = sinx…Amplitude = 1 y = asinbx…..Amplitude = a change in amplitude HINT! (The amplitude value is always positive, irrespective if a is hint negative.. Consider the graphs of y = a sin x, y = a cos x and y = a tan x Example: if a = -2, then the amplitude is 2. e.g. The parameter a changes 4 EG - worked examples the amplitude of the graph. 1. Sketch the following graphs: exams a) on the same set of axes y = –sin x and y = 2sin x for x ∈ [–360°; 360°], b) on the same set of axes y = –2 cos x and y = ½ cos x for x ∈ [–360°; 360°], activity c) y = 2 tan x for x ∈boy [–180°; 180°], ACTIVITIES and girl-left and right of page activity Solutions a) y Step by step comment 2 comment y = 2sinx 1 x –360° –270° –180° –90° 0 90° 180° 270° 360° –1 y = – sinx –2 y = –2 cosx ... Amplitude = 2 y = ​ 1 __​cos ... Amplitude = ​ 1 x __​ b) y 2 2 y = –2cosx y = cosx ... Amplitude = 1 2 y = acosbx ... Amplitude = a The parameter a changes 1 the amplitude of 1 the graph. ​ __ 2​ x –360° –270° –180° –90° 0 90° 180° 270° 360° 1 __ ​  2 ​ –1 1 y = __ ​  2 ​cosx –2 © Department of Basic Education 2015 UNIT 5 trig functions 91 Mind the Gap Mathematics

5 Unit c) y 3 (–135°; 2) 2 (45°; 2) y = 2tanx 1 x –180o –135o –90o –45o 0 45o 90o 135o 180o –1 –2 (–45°; –2) (135°; –2) –3 Conclusion The parameter a changes the amplitude of the graph in y = asinbx and y = acosbx. The graph y = atanbx has no maximum or maximum value. The value does not change the amplitude of y = atanbx as there is no amplitude. The value of a affects the y-value of each point. Each y-value is multiplied by a. 92 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 5.3 The effect of q on the NB shape of NBthe graph: PAY SPECIAL ATTENTION vertical shift HINT! hint Consider the graphs of y = sin x + q, y = cos x + q and y = tan x + q. e.g. 5 EG - worked examples 1. Sketch the following graphs on the same set of axes for the domain exams [–360°; 360°]: a) y = –sin x and y = – sin x + 1 b) y = cos x, y = cos x + 1, y = cos x – 2 c) y = tan x and y = tan x + 1 activity ACTIVITIES boy and girl-left and Solutions right of page activity a) b) y 2 2 y comment y = cosx + 1 y = 1 – sinx Step by step comment 1 1 x x –360° –270° –180° –90° 0 90° 180° 270° 360° –360° –270° –180° –90° 0 90° 180° 270° 360° –1 –1 y = cosx y = –sinx y = cosx – 2 –2 –3 c) y y = tanx + 1 y = tanx (–315o; 2) (–135o; 2) 2 (45o; 2) (225o; 2) (–180o;1) (–135o;1) 1 (45o;1) (225o;1) (–315o;1) x –360o –315o –270o –225o –180o –135o –90o –45o 45o 90o 135o 180o 225o 270o 315o 360o 0 –1 (–225o; –1) (135o;–1) (–45o;–1) (315o;–1) –2 Conclusion The parameter q shifts the whole graph up or down by q units. © Department of Basic Education 2015 UNIT 5 trig functions 93 Mind the Gap Mathematics

5 Unit 5.4 The effect of b on the NB shape of NBthe graph: PAY SPECIAL ATTENTION change in period HINT! hint Consider the graphs of y = sin bx, y = cos bx and y = tan bx. e.g. 6 EG - worked examples 1. Draw the graphs on separate set of axes: exams a) y = sin 2x for x ∈ [–90°, 180°] b) y = cos 3x for x ∈ [0°; 360°] 1 c) y = tan __ 2​  ​ x for x ∈ [–360°; 360°] activity Solutions ACTIVITIES boy and girl-left and 1. a) For y = sin 2x, the period is 360° ÷ 2 = 180°. right of page y activity y = sin 2 x 1 0,5 Step by step comment comment –90° –75° –60° –45° –30° –15° 0 15° 30° 45° 60° 75° 90° 105° 120° 135° 150° 165° 180° x –0,5 –1 Period = 180° b) For y = cos 3x, the period is c) For y = tan ​ 1_ ​ x, the period 2 360° ÷ 3 = 120°. = 180° × 2 = 360° y 360° ​  3 ​= 120° period = ____ y = tan ½x y 4 3 1 2 y = cos (3x) 1 x –360° –270° –180° –90° 0 90° 180° 270° 360° x 0 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° –1 –2 –3 –1 –4 Period = 360° Conclusion • The period of the graph is the number of degrees it takes to complete one wavelength. • The value of b, affects the period of the graph ​360​0​ • For y = sin bx and y = cosbx, the period = ____ ​  b ​ 18​0​0​ • For y = tan bx, the period = ​ ____ b ​ 94 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 5.5 The effect of p on the NB shape of NBthe graph: PAY SPECIAL ATTENTION horizontal shift HINT! Consider hint the graphs of the form y = sin (x + p), y = cos (x + p) and y = tan (x + p). e.g. 7 EG - worked examples 1. Draw the following graphs on the same set of axes and for x ∈ exams [–180°, 180°]: a) y = sin x and y = sin (x + 60°) b) y = cos x and y = cos (x – 45°) c) y = tan x and y = tan (x + 45°) activity ACTIVITIES boy Solutions and girl-left and right of page 1. a) y = sin x; y = sin (x + 60°) activity y shifted 60° to the left Step by step comment comment (30°; 1) 1 (90°; 1) y = sinx x –180° –150°–120° –90° –60° –30° 0 30 °60° 90° 120° 150° 180° y = sin(x + 60°) –1 (–150°; –1) (–90°; –1) SPECIAL ATTENTION NB The graph of y = sin x has shifted 60 to the left to form y = sin (x + 60°) ! b) y = cos x, y = cos (x – 45) y shift of 45° to the right (0°;1) (45°;1) 1 worked examples y = cos(x – 45°) x –180° –135° –90° –45° 0 45° 90° 135° 180° y = cosx –1 PAY SPECIAL ATTENTION NB NB The graph of y = cos x has shifted 45 to the right to form y = cos (x – 45). y ACTIVITIES boy and girl-left and HINT! right of page hint © Department of Basic Education 2015 UNIT 5 trig functions 95 y Mind the Gap Mathematics

5 Unit c) y = tan x and y = tan (x + 45°) y 3 2 y = tanx 1 (–135o; 1) (45o; 1) x 0 –180o –135o –90o –45o 45o 90o 135o 180o (–45o; –1) (135o; –1) –1 –2 –3 The graph of y = tan x has shifted 45 to the left to form y = tan (x + 45°). The asymptotes have also shifted 45 to the left. y 2 y = tan(x + 45°) 1 (0o; 1) (–180o; 1) x 0 –180o –135o –90o –45o 45o 90o 135o 180o (–90o; –1) –1 (90o; –1) –2 96 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 Conclusion For graphs of the form y = sin (x + p), y = cos (x + p) and y = tan (x + p), p affects the horizontal shift of the graph. • If p > 0 (positive), the graph moves p degrees to the left. • If p < 0 (negative), the graph moves p degrees to the right. • In all graphs the x-intercepts, y-intercepts, maximum and minimum points must be indicated on the graph. If the value of b changes, the x-intercepts, y-intercepts, maximum and minimum points also change. In order to ensure that these points are always indicated, use the following x-values for plotting the graph: Equation b=1 b=2 b=3 b=½ y = sinbx From 0°, use From 0°, use From 0°, use From 0°, use or y = cosbx intervals of 90° intervals of 45° intervals of 30° intervals of 180° ​90​0​ ​90​0​ ​90​0​ ​90​0​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ ​360​0​ ​360​0​ ​360​0​ ​360​0​ Period = ​ ____ b ​ Period = ​ ____ b ​ Period = ​ ____ b ​ Period = ​ ____ b ​ y = tanbx From 0°, use From 0°, use From 0°, use From 0°, use intervals of 45° intervals of 22,5° intervals of 15° intervals of 90° ​45​0​ ​45​0​ ​45​0​ ​45​0​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ Interval = ___ ​  b ​ ​180​0​ ​180​0​ ​180​0​ ​180​0​ Period = ​ ____ b ​ Period = ​ ____ b ​ Period = ​ ____ b ​ Period = ​ ____ b ​ y = sin(x+p) From 0°, use or y = cos(x+p) intervals of (90°– p°). with p > 0 The intervals for y = sin(x – 30) and y = sin(x + 30) will be the same. The intervals will be 90 – 30=60. y = tan(x+p) From 0°, use intervals of (45°– p°). with p>0 The intervals for y = tan(x – 30) and y = tan(x + 30) will be the same. The intervals will be 45 – 30 = 15. © Department of Basic Education 2015 UNIT 5 trig functions 97 Mind the Gap Mathematics

5 Unit Activity 2 1. Given f(x) = 2cos x and g(x) = sin (x + 30°) a) Sketch the graphs of f and g on the same set of axes for x  ∈ [− 150°; 180°] Clearly show all intercepts with the axes and the coordinates of turning points. (7) Use your graph to answer the following questions: b) Write down the period of f. (1) c) For which values of x is f(x) = g(x)? (2) d) For which values of x is f(x) > 0? (2) e) For which values of x is g(x) increasing? (2) f) Determine one value of x for which f(x) − g(x) = 1,5. (1) g) If the curve of f is moved down one unit, write down the new equation of f. (2) h) If the curve of g is moved 45° to the left, write down the new equation of g.(2) 2. Sketch below are the graphs of g(x) = asin(x+p) and f(x) = bcosqx for x ∈ [0°; 180°] y g(x) = b cosqx f(x) = asin(x + p) x 0 30o 60o 90o 120o 150o 180o 210o 240o 270o 300o 330o 360o a) Determine the numerical values of a, p, b and q. (5) b) If the graph of g(x) is shifted two units down: 1) Write down the amplitude of the new graph (1) 2) Write down the equation of the new graph (2) c) If the graph of f(x) is shifted 60 to the left, write down two possible equations of the new graph. (2) [29] 98 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 5 Solutions 1. a) 333 for g(x) = 2 cos x and 3333 for f(x) = sin(x + 30°) y 2 (0o; 2) y = 2 cosx 1,5 1 (60o; 1) y = sin (x + 30°) (0o; 0,5) –150o –120o –90o –60o –30o 0 30o 60o 90o 120o 150o 180o x –0,5 –1 (–120o; –1) –1,5 –2 (7) b) period = 360°3 (1) c) x = –120° or 60° 33 (2) d) for f(x) >0; x ∈ (–90°; 90°) 33 (2) e) g(x) increasing when x ∈ (–120°; 60°) 33 (2) f) x = 0° 3 (1) g) New f(x) = 2cos x –133 (2) h) Original equation: g(x) = sin (x + 30°), with 45° shift to the left: g(x) = sin (x + 30° + 45°) so g(x) = sin (x + 75°) 33 (2) 2. a) a = 2 (amplitude of the f(x)) 3 f(x) = 2sin(x+p)… Substitute 60° ∴ 2 = 2sin(60° + p)3 ÷ 2 ∴ 1 = sin(60° + p) Pressing shift ​sin​−1​(1) = 90° ∴ 60° + p = 90° ∴ p = 30°3∴ f(x) = 2sin(x + 30°) 3 b=3 (amplitude of the g(x)) period = 720 360° 1 1 720° = ​ ____ __ __ q ​∴ q = ​  2 ​∴ g(x) = 3 cos ​  2 ​x 3 (5) b) (1) Amplitude = 3 (shift up or down has no effect on the amplitude)3 (1) 1 (2) g(x) = 3 cos __ ​  2 ​x − 233 (2) c) f(x) = 2sin(x + 90°) = 2 cos x 33 (2) [29] © Department of Basic Education 2015 UNIT 5 trig functions 99 Mind the Gap Mathematics

5 Unit What you need to be able to do • Recognise the basic shapes of the graphs associated with their equations. • Sketch functions and show the effect of different parameters a, p and q. • Draw each graph using the critical points: intercepts with the axes and turning points, where applicable • Show any asymptotes and include any other points you might need. • Determine the features of graphs including • domain and range of functions • turning points • asymptotes • intercepts with axes • Find the equation from the graph. • Sketch trig functions, any shifts and changes in amplitude and period. In Unit 10, we will discuss the solutions to trigonometric equations. You will be shown how to determine the solution of 2cosx=sin(x+30) algebraically. In this question the solutions can be read off the graphs. Keep going! 100 UNIT 5 trig functions  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 6 Finance, growth and decay PAY SPECIAL ATTENTION NB 6.1 Revision: Simple and HINT! hint compound interest In all calculations, round off your final answer only. Financial terms NB PAY SPECIAL ATTENTION NB e.g. EG - worked examples • Interest is a fee paid for the use of borrowed money, or money earned on money saved. It is calculated as a percentage of the money borrowed or lent. interest is the interest on an initial (principal) sum of money. Each exams • Simple HINT! hint year you receive or you are charged the same amount of interest. e.g. 1 EG - worked examples Simple interest of 6% p.a. (per year) on R100 means that if you borrow R100 for exams a year, you owe that R100 and another R6 back. So you owe R106. activity ACTIVITIES boy If you borrow R100 for 2 years, you will owe R100 + R6 + R6 = R112. and girl-left and right of page Micro-lenders and Hire Purchase agreements often work on simple interest at a monthly or at a yearly interest rate. activity activity ACTIVITIES boy Percentage increase or decrease in populations, number of learners etc., can also and girl-left and right of page be calculated using simple interest formula. activity • Compound interest is also interest on a principal amount P. For each year, Step by step comment the previous year’s final amount becomes the new principal amount. So the interestStepisbycalculated step on the comment principal and the interest from the previous comment year. Compound interest of 6% p.a. (per year) on R100 means that if you borrow R100 for 2 years, you owe R100 + R6 = R106 in the first year. In the second year, you owe R106 + 6% of R106. R106 + (6% × R106) = R106 + R6,36 = R112,36 Here are the formulae for simple and compound interest. Simple interest: A = P(1 + ni) where P is the principal (original sum of money invested or borrowed) i is the interest rate n is the number of years A is the final amount © Department of Basic Education 2015 UNIT 6 finance, growth and decay 101 Mind the Gap Mathematics

6 Unit Compound PAY interest: A = P(1 +i)n SPECIAL ATTENTION NB NB where P is the principal (original sum of money invested or borrowed) i is the interest rate is the number of years nHINT! hint A is the final amount e.g. 2 EG - worked examples If you borrow R300 at 9% p.a simple interest, how much will you owe after exams 7 years? Solution 9 A=? P = R300 i = 9% = ​  ___ ​ = 0,09 n = 7 years 100 activity A = P(1 + ni) ACTIVITIES boy • Write down what is given. A = 300(1 + 7 × 0,09)and=girl-left 489 and right of page • Decide what you need to find. After 9 years you will owe R489. • Solve for that variable. activity If you borrow R300 at 9% p.a compound interest, how much will you owe after 7 years? Step by step comment comment 9 A=? P = 300 i = 9% = ​ ___ 100 ​= 0,09 n = 7 years A = ​P(1 + i)​n​ A = ​300(1 + 0,09)​7​ A = ​300(1,09)​7​ A = 548,411736… A ≈NB R548,41PAYto the nearest cent. SPECIAL ATTENTION NB Which is the better option? R548,41 – HINT! R489 = R59,41 hint So compound interest is R59,41 more than simple interest after 7 years. e.g. 3 EG - worked examples 1. You invest R1 570 at 11% p.a. compounded monthly. exams a) How much will you receive after 7 years? b) How much interest have you earned after 7 years? Solutions Compounded monthly means activity the interest is calculated at 1. a) A = P(1 + i) n ACTIVITIES boy and girl-left and the end of every month. So A=? P = R1 570 n = 7 years × 12 months right= 84 time periods of page 0,11 convert years to months. i = 11% ÷ 12 months = ____ ​  12 ​ activity (1 + 0,11 ) 7×12 A = 1570​​ ______ ​  12 ​  ​​ ​ 11% per annum compounded monthly, so Step by step comment we divide the interest rate by 12 months. comment A = 3 378,959672… You will receive R3 378,96 (to the nearest cent) after 7 years. b) You have earned R3 378,96 – R1 570 = R1 808,96 interest 102 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 PAY SPECIAL ATTENTION NB NB Interest per annum compounded: i ​ monthly ➞ ​ __ n years × 12 months 12 HINT! hint quarterly ➞ ​ _i ​ 4 n years × 4 quarters in the year semi-annually or half-yearly, (every 6 months) ➞ ​ _i ​ n years × 2 2 e.g. EG - worked examples Activity 1 6.1.1 exams You invest R1 700 at an interest rate of 10% compounded quarterly. Calculate how much your investment is worth after 6 years.  (3) 6.1.2 R25 000 is invested into a savings account. Calculate the value of the investment of the savings after 5 years if interest rates are: a) 11% compounded monthly b) 11% activity compounded semi-annually (5) ACTIVITIES boy [8] and girl-left and right of page Solutions activity 6.1.1 A = ? P = R1 700 n = 6 years × 4 = 24 i = 10% compounded quarterly 0,10 so divide by 4 i = ​ ____ 4 ​3 A = P(1 +i) Step by step n comment comment ( 0,10 24 A = 1 700 ​​ 1 + ​ ____ ) 4 ​  ​​ ​3 = R3 074,83 (to nearest cent) 3  (3) 0,11 6.1.2 a) A = R25 000 i = ​  ____ 12 ​ 3 n = 5 × 12 ( 0,11 ) 5 × 12 A = 25 000 ​​ 1 + ​  ____ 12 ​  ​​ ​3 = R43 222,89 3 0,11 b) A = R25 000 i = ​  ____ 2 ​n = 5 × 2 ( A = 25 000 ​​ 1 + ​  ____ ) 0,11 5 × 2 2 ​  ​​ ​3 = R42 703,61 3  (5) [8] © Department of Basic Education 2015 UNIT 6 finance, growth and decay 103 Mind the Gap Grade 12 Mathematics

6 Unit 6.2 Calculating the value of P, i and n PAY SPECIAL ATTENTION NB NB We can also HINT! use the formulae for compound and simple interest to hint calculate the principal P, the rate of interest i, or the time period n. e.g. 4 EG - worked examples 1. How much must John invest now so that after 5 years at 8% simple exams interest, he will have R4 200? Solution 1. A = R4 200 n=5 i = 8% P=? activity A = P(1PAY+SPECIAL n.i)ATTENTIONACTIVITIES boy NB and girl-left and NB 4 200 = P(1 +5(0,08)) right of page activity 4 200 = P(1,4) 4 200 P = ​ _____ ​= 3 000 1,4HINT! hint ∴ John must invest R3 000. Step by step comment comment e.g. 5 (finding i) EG - worked examples A population increases from 12 000 to 214 000 in 10 years. At what annual exams (compound) rate does the population grow? (Give your answer correct to one decimal place.) Solution A = 214 000 activity P = 12 000 n = 10 i=? A = P(1 + i) n ACTIVITIES boy and girl-left and right of page 214000 = 12000(1 + i)10 activity ​  214000 ______ ​= (1 + i)10 12000 ______ 10 √ 12000 ​ ​  214000 ______ ​ ​ = 1 + i Step by step comment comment 1,333899939… –1 = i 0,333899939... = i ∴i = 33,389..% The population grew at an annual (compound) rate of 33,4% (correct to one decimal place). 104 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

HINT! hint Unit 6 e.g. 6 (finding n) EG - worked examples Ms Gumede puts R3 500 into a savings account which pays 7,5% p.a. exams compound interest. After some years, her account is worth R4 044,69. For how long did she invest the money? Solution activity A = R4 044,69 Pand= R3boy500 ACTIVITIES girl-left and n=? i = 7,5% p.a. = 0,075 right of page A = P(1 +i) n activity 4 044,69 = 3 500(1 +0,075)n 4 044,69 = 3 500(1,075)n 4044,69 ______ ​  3500 ​= Step (1,075) by step n comment comment 1,155625714 = (1,075)n n = log1,075 1,155625714 n = 2,000008543 n = 2 years So Ms Gumede invested the money for 2 years. • Substitute for A, P and i • Simplify • Divide by 3500 • Write in logarithmic form • Use the log keys on the calculator • Round off the answer to the nearest year © Department of Basic Education 2015 UNIT 6 finance, growth and decay 105 Mind the Gap Grade 12 Mathematics

6 Unit Activity 2 1. Mary borrowed a certain sum of money from a bank at a compound interest rate of 15% calculated quarterly. After 3 years she now owes R7 000. How much did she borrow? (3) 2. R1 570 is invested at 12% p.a. compound interest. After how many years will the investment be worth R23 000? (4) 3. R2 000 was invested in a fund paying interest compounded monthly. After 18 months the value of the fund was R2 860, 00. Calculate the interest rate. (4) [11] Solutions 1. A = R7000 i = ​ 0.15 ____ ​ 3 4 n =3 × 4 P =? ( 7000 = P​​ 1 + ​ ____ 4 ​  ​​ ​3 ) 0,15 3×4 7000 = P(1,555454331) ………… Divide both sides by 1,555454331 P = R4500,29 3 (3) 2. A = P (1 + i)n substitute for A, P and i 23 000 = 1 570(1 + 0,12) 33 n simplify and divide 23000 ​ _____ 1570 ​= (1,12) n keep the number on your calculator without rounding 14,6496… = (1,12)n use log laws n = log1,12 14,6496… 3 Use the log keys on your calculator. n = 23,69 years n ≈ 24 years to the nearest year 3 (4) 3. A = 2860 P = 2000 i=? n = 18 A = P (1 + i) n 2000​​(1 +​ __ 12 ​  )​​ ​= 2860 3 i 18 ​​(1 + __ ​  12 ​  )​​ ​= ​ ____ i 18 2860 2000 ​3 ____ i 1 + __ ​  12 ​= 18​√1,43 ​ i ​ __ 12 ​= 0,020069541 i = 0,020069541... 3 i =0,2408344924 × 100 i = 24,08% 3 (4) [11] 106 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 6.3 Simple and compound decay formulae Decay or depreciation is when a quantity decreases by a percentage of the amount present. For example, your assets (house, car) and machinery lose value through age and use. Ways NBof calculating depreciation. PAY SPECIAL ATTENTION NB Simple decay or depreciation: A = P(1 –ni) HINT! This hintis also called straight line depreciation because it can be represented with a straight line graph. e.g. 7 EG - worked examples A car worth R120 000 depreciates at a rate of 12% (simple interest) p.a. exams How much will the car be worth after 5 years? Solution: A = P(1 – ni) A=? P = 120 000 i = 12% = 0,12 n = 5 years activity A = 120 000 (1 – 5 × 0,12) A = 48 000 ACTIVITIES boy and girl-left and right of page The car will be worth R48 000 after 5 years. activity PAY SPECIAL ATTENTION NB NB Compound decay or depreciation: A = P(1 – i)n comment This is also called Step by step depreciation comment on a reducing balance because the interest HINT! ishint calculated on the amount left over as it decreases. The amount left over is ‘the reducing balance’. e.g. 8 EG - worked examples A car worth R120 000 depreciates at a rate of 12% p.a. (on a reducing exams balance). How much will the car be worth after 5 years? Solution activity A = P (1 – i)n A = ? P = 120 000 i = 12% = 0,12 n = 5 years A = 120 000 (1 – 0,12)ACTIVITIES 5 boy and girl-left and right of page A = 63 327,83002… activity A = R63 327,83 (to the nearest cent) ***Compare this with simple depreciation: The car’s Step value is R63 327,83 by step comment – R48 000 = R15 327,83 less on simple decay comment than on compound decay. © Department of Basic Education 2015 UNIT 6 finance, growth and decay 107 Mind the Gap Grade 12 Mathematics

6 Unit Activity 3 The value of a piece of machinery depreciates from R10 000 to R 5 000 in 4 years. What is the rate of depreciation, correct to two decimal places, if calculated on the: a) Straight line method (i.e. simple depreciation)  (3) b) Reducing balance (i.e. compound depreciation)  (3) [6] Solutions a) A = 5 000 P = 10 000 n=4 i =? Note: A is less than P Straight line method: A = P(1 – ni) 5 000 = 10 000 (1 – 4i) 3 5000 _____ ​  10000 ​= (1 − 4i) 3 0,5 – 1 = – 4i − 0,5 ____ ​  − 4 ​= i 0,125 = i i =12,5% 3(3) b) Reducing balance: A = P (1 – i)n 5 000 = 10 000(1 – i)4 3 5000 _____ ​  10000 ​= (1 − i)4 3 0,5 = (1 – i)4 ___ 4 ​√0,5 ​= 1 – i i = 1 – 0,8408… i = 0,1591035… i = 15,9%3(3) [6] 108 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 6.4 Nominal and effective interest rates 1. A nominal interest rate is the quoted interest rate. 2. An effective interest rate is the actual interest rate received. If you are quoted a nominal interest rate of 8% p.a., the resulting effective rate will be different depending on if it is worked out annually, monthly or semi-annually 3. We use PAYthe following formula to calculate the effective interest rate SPECIAL ATTENTION NB from the nominal interest NB rate or vice versa: ( 1 + ieffective = ​​ 1 + _____ HINT! ​  k ​  ​ ​ inominal k ) hintIf k is the number of times per year the interest is calculated. e.g. 9 EG - worked examples 1. You borrow R500 at 8% p.a. compounded for one year. exams At the end of the year you owe 500(1+ 0,08)1 = R 540 2. You borrow R500 at 8% p.a. compounded monthly for one year At the end of the year you owe 500 ​​ 1 + ____ ​  12 ​  ​​ ( ) 0,08 1×12 ​= R541,50 Soactivity effectively, you are charged R41,50 interest on R500. R41,50 100 actually ​ ______ ___ R500 ​× ​  1 ​= 8,3% . ACTIVITIES boy Your interest rateand isgirl-left and right of page So the effective interest rate is 8,3% p.a. but the nominal interest rate activity is 8% p.a. 3. What is the effective interest rate if 7,5% p.a. is calculated monthly? Step by step comment comment Solution ( 1 + ieffective = ​​ 1 + _____ ​  12 ​  ​​ ​ 0,075 12 ) 1+i effective =1,07763 i = 0,07763 eff ∴ ieff = 7,76% © Department of Basic Education 2015 UNIT 6 finance, growth and decay 109 Mind the Gap Grade 12 Mathematics

6 Unit Activity 4 1. Khosi wants to invest R5 000 for 3 years. Which is the better investment for her if the interest is 10,5% p.a compounded quarterly or 10,5% p.a. compounded monthly?(7) 2. Convert a nominal interest rate of 9% per annum compounded semi-annually to the effective annual interest rate. (2) [9] Solutions 1. 0,105 First option: A=? P = R5 000 i =​  ____ 4 ​3 n=3×4 A = P (1 + i) n ( 0,105 3×4 A = 5 000 ​​1 + ​ ____ ​  ​​ ​3 4 ) Use a calculator to work out the whole answer. A = R6823,513 0,105 Second option: A=? P = R5 000 i = ​ ____ 12 ​3 n = 3 × 12 A = P (1 + i) n ( 0,105 3×12 A = 5000​​ 1 + ​  ____ ​  ​​ ​3 12 ) Use a calculator to work out the whole answer. A = R6 841,923 ∴10,5% compounded monthly gives Khosi a better interest on her investment. 3(7) ( inominal k ) ​  k ​  ​ ​if k is the number of times per year the interest is calculated. 2. 1 + ieffective = ​​ 1 + _____ ( 0,09 2 1 + i =​​ 1 + ​  ____ 2 ) ​  ​​ ​3 1 + i =1,092025 … i = 0,092025 ∴9,20% is the effective annual interest rate. 3 (2) [9] 110 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 6.5 Investments with time PAY SPECIAL ATTENTION and interest rate changes NB NB CalculationsHINT!of more than one interest, deposits and withdrawals are best hintusing timeline. done e.g. 10 EG - worked examples Thabo invested R 1 000 in a bank for 10 years. The interest rate was 6,5% exams compounded quarterly for the first 3 years. For the next 5 years, the interest was calculated at 7,2% compounded monthly and for the remainder of the investment, the interest was at 7,8% compounded semi-annually. How much money would Thabo get at the end of the investment? activity Solution ACTIVITIES boy and girl-left and right of page First draw a timeline so that you understand the question. activity Over 10 years, the interest rates are different over different time periods. R1000 n =Step3 byyears step × 4 = 5 years × 12 comment ncomment n = 2 years × 2 6.5% 7.2% 7.8% T0 i = ____ ​  4 ​ T1 i = ____ ​  12 ​ T2 i = ____ ​  2 ​ T3 This can all be worked out in one calculation or you can do separate calculations. Remember not to round off answers until the end so that you can have accurate answers. METHOD 1 ( ) ( ) ​​​(1+​ 0,078 2 ​  )​​ ​333 2×2 0.065 3×4 0.072 5×12 A = 1000​​ 1+​ _____ _____ 4 ​  ​​ ​​​ 1+​  12 ​  ​​ _____ = R2024,643 METHOD 2 ( 0,065 3×4 Time1: A = 1000 ​​ 1 + ​ _____ 4 ​  ​​ ​= R1 213,4075793) This amount of R1 213,407579 becomes P for the next calculation Time2: A = 1213,407579 ​​ 1 + ​ _____ 12 ​  ​​ ( ) 0,072 5×12 ​= R1 737,3429113 This amount of R1 735,911122 becomes P for the next calculation Time3:A = 1 737,342911 ​​ 1 + _____ ( ​  2 ​  ​​ ​= R2 024,643 ) 0,078 2×2 After 10 years Thabo will get R 2 024,64 (to the nearest cent) 3 NOTE: Rounding off was done on the final answer only. [8] © Department of Basic Education 2015 UNIT 6 finance, growth and decay 111 Mind the Gap Grade 12 Mathematics

6 Unit Activity 5 Mr. Sithole invests R50 000 in an account which offers 8% p.a. interest compounded quarterly for the first 18 months. The interest then changes to 6% p.a. compounded monthly. Two years after the money is invested, R10 000 is withdrawn. How much will be in the account after 4 years? [5] Solution Draw a timeline. The total length of time is 4 years. deposit R50 000 withdraw R10 000 year T0 1,5 years × 4 0,5 years × 12 2 years × 12 T4 8% ÷ 4 (quarterly) 6% ÷ 12 (monthly) 6% ÷ 12 (monthly) METHOD 1 DEPOSIT for the period of 4 years ( ) 0.08 1.5×4 A = 50000 ​​ 1 + ​ ____ 4 ​  ​​ ( ) 0.06 2.5×12 ​​​ 1 + ​ ____ 12 ​  ​​ ( ) 0.06 2×12 ​− 10000 ​​ 1 + ​ ____ 12 ​  ​​ ​3333 =R54 124,66 3 OR METHOD 2 18 First 18 months (​ __ 12 ​=1,5 years): 0,08 P = R50 000 i = 8% compounded quarterly = ____ ​  4 ​ n = 1,5 years × 4 ( ) 0,08 1,5×4 A = 50000 ​​ 1 + ​ ____ 4 ​  ​​ ​= R56 308,12096 33 Next 6 months (0,5 years): 0,06 P = R56 308,12096 i = 6% compounded monthly = ____ ​  12 ​ n = 0,5 × 12 ( 0,06 ) 0,5×12 A = 56308,12096 ​​ 1 + ​ ____ 12 ​  ​​ ​= R58 018,62143 3 R10 000 withdrawn, so R48 018,62143 remains as the new P value. 3 Next 2 years 0,06 P = 48 018,62134 i = 6% compounded monthly = ____ ​  12 ​ n = 2 × 12 ( 0,06 2×12 A = 48018,62134 ​​ 1 + ​ ____ ) 12 ​  ​​ ​= R54 124,66 3 [5] Deposit of 50 000 with two different interest for the whole period minus the withdrawal with interest for the remaining period 112 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 6.6 Annuities Annuities are number of equal payments made at regular intervals and subject to a rate of interest. Types of annuities are: Future value annuities and Present value annuities. 6.6.1 Using the Future Value formula • You can save money by putting away the same amount of money every month for use in the future. This can be done through an annuity fund, a retirement fund, a savings account, or a sinking fund. Compound interest is earned on your savings, so you will receive, at some time in the future, the total of all your monthly installments, as well as interest calculated every month on an increasing monthly balance. Future value formula When you pay equal monthly installments in order to save money for the future, you can calculate using the future value formula. ​x[(1 + i)​n​− 1] F = ​ __________ i ​ where F is the total accumulated at the end of the time period x is the monthly installment i is the interest rate per annum PAY SPECIAL ATTENTION NB NB n is the number of installments/payments This formula is provided on the information sheet in the final exam. NOTE: The formula assumes that payments start at the end of the first HINT! hint month. e.g. 11 EG - worked examples Sipho plans to save a fixed amount from his salary each month. He starts exams at the end of the month of his first salary. The bank offers an interest rate of 4,7% p.a. compounded monthly. a) Determine the amount he has to save every month if he wants to have R30 000 in his savings account at the end of 4 years. activity b) What is the total amount ACTIVITIES boy of interest he will receive after 4 years? and girl-left and right of page Solutions activity a) Sipho is saving for the future so use the Future Value formula. F = R30 000 x is the monthly installment comment 0,047 i = 4,7% compounded monthly = _____ Step by step comment ​  12 ​ n = 4 × 12 = 48 months ​[(1 + i)​ ​− 1]n F = x ​ _________ i ​ [( x ​ ​​ 1 + ​ _____ 30000 = ______________ ​    ) 0,047 48 12 ​  ​​ ​− 1 ​ ​ ] 0,047 _____ ​  12 ​ 0,047 30000 × ​ _____ _____ [( 12 ​= x ​ ​​ 1 + ​  12 ​  ​​ ​− 1 ​ 0,047 48 ) ] © Department of Basic Education 2015 UNIT 6 finance, growth and decay 113 Mind the Gap Grade 12 Mathematics

6 Unit 0,047 30000 × ​ _____ 12 ​ x = ​ _____________       ​ [​​​(1 + ​ _____ 12 ​  )​​ ​− 1 ]​ 0,047 48 x = 569,30932... b) He must save R569,31 every month for 4 years. PAY SPECIAL ATTENTION He will have paid R569,31 ×NB NB 48 months = R27 326,88 So the total interest he will receive after 4 years is R30 000 – R27 326,88 = R2 673,12 HINT! hint 6.6.2 Sinking fund e.g. 12 EG - worked examples A printing company buys new printers at a cost of R3,2 million. exams [Give all answers to the closest rand.] a) Calculate the book value of their printers after 5 years, if the depreciation is calculated at 16% p.a. on a reducing balance. activitythe cost of replacing the printers at the end of 5 years, if b) Calculate the price of new ACTIVITIES printers boy and girl-left and increases by 8,5% p.a. right of page c) How much more money would the company need if they sell the old activity printers at their book value and the money received is used towards the purchase of the new equipment? d) The company sets up a fund to make provision for replacing the old equipment Step by step at the end of comment comment 5 years. They deposit R240 000 at the end of the 1st year, R370 000 at the end of the 2nd year, R420 000 at the end of the 3rd year and R500 000 at the end of the 4th year. Determine the total amount of money accumulated in the fund at the end of 5 years if the interest paid on money in the fund is 11,5% p.a. compounded annually. e) How much additional money will they need to buy the replacement printers at the end of the replacement period? Solutions: 1. a) P = R3 200 000 i = 16% = 0,16 n=5 A = P (1 – i)n A = 3 200 000 (1 – 0,16)5 A = R1 338 278 Book value of R1 338 278 b) P = R3 200 000 i = 8,5% = 0,085 n=5 A = P (1 + i)n A = 3 200 000 (1 + 0,085)5 A = R4 811 701 Cost of replacing printers c) 4 811 701 – 1 338 278 = R3 473 423 They need R3 473 423 114 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 d) timeline year 0 1 2 3 4 5 deposit R240 000 R370 000 R420 000 R500 000 End of year 2: P = R240 000 i = 11,5% p.a. = 0,115 n=1 A = P (1 + i)n A = 240 000 (1 + 0,115)1 = R267600 R267600+ R370000 = R637600 End of year 3: A = 637600 (1 + 0,115)1 = R710924 R710 924 + R420 000 = R1130 924 End of year 4: A = R1130 924 (1 + 0,115)1 = R1260980,26 R1260980,26 + 500 000 = R1760980,26 End of year 5: A = 1760980,26 (1 + 0,115)1 = R1963492,99 After 5 years, they have R1963492,99 in the fund. e) R4 811 701 – R1963492,99 – R1338278 = R1509930,01 still needed for new printers. Activity 6: Interpret a graph 1. Ntsako invests R50 000 at 14% p.a. compounded annually. Liz saves R50 000 at 13,7% p.a. compounded monthly. a) Who has the most money at the end of 20 years? b) Calculate the difference in their investments after 20 years. [6] Solutions 1. a) Ntsako: A = ​50 000(1 + 0,14)​20​ 3 = R687 174,49 3 ( Liz: A = 50000 ​​ 1 + ​  ____ 12 ​  ​​ ) 0,137 20×12 ​= R762421,9984 = R762 422,00 33 Liz has the most money. 3 b) The difference is R762 422,00 – R687 174,49 = R75 247,51. 3 [6] 6.6.3 Using the Present Value formula • You can borrow a large amount of money from the bank. This is called a loan. For example, there are student loans to pay for further studies, vehicle loans to buy a car and loans to buy a house. • A bond or mortgage or a home loan is a loan used to buy a house or other property. • The amount you must pay back is the total of the loan and the interest charged on it. You must pay back an equal amount each month called a monthly installment. © Department of Basic Education 2015 UNIT 6 finance, growth and decay 115 Mind the Gap Grade 12 Mathematics

6 Unit Each month, the interest is calculated on the amount you still owe. Because you are paying back the same amount monthly, the amount you owe is decreasing. Here is a formula to work out your monthly installments. It is called the Present Value Formula. It is present because you receive the money now in the present. You start paying it back at the end of the first month of the loan. x ​[ 1 − ​(1 + i)​−n​) ]​ P = ​ ____________    i ​ where P is the Present PAY SPECIAL ATTENTION value NB NB x is the monthly installment i is the interest rate p.a. n is the number HINT! of time periods to repay the loan hint formula is provided on the information sheet in the final exam. This e.g. 13 EG - worked examples A loan of R240 000 is repaid over 5 years with equal monthly payments exams (installments), starting one month after the loan was granted. Notice: it is normal to start paying back a loan one month after it was granted. a) Calculate activitythe monthly repayments if the interest on the loan is 9% p.a. compoundedACTIVITIES monthly. boy and girl-left and b) The client has financial difficulties and makes only 17 payments. right of page activitythe balance of the loan at the end of the 17th month. Calculate Solutions Step by step comment comment 1. a) P= R 240 000; x is the monthly installment 0,09 The balance of a loan i = 9% p.a. monthly = ____ ​  12 ​ n = 5 × 12 = 60 that must be paid at any x [ ​ 1 − ​ ( 1 + −n i)​ ​  ]​ particular time during the P = ​ ___________ i ​ agreed loan time, can be calculated by using the [ ( x ​ 1 − ​​ 1 + ​ ____ ______________ ) ] 0.09 −60 12 ​  ​​ ​  ​ present value formula 240000 = ​    ​33 ​  0.09 ____​ for the remaining number 12 of installments. 240000 ​____ _____________ (0.09 ) ​  12 ​  ​ x = ​      ​ [ ( 0.09 −60 ​ 1 − ​​ 1 + ​ ____ ) ] 12 ​  ​​ ​  ​ 3 x = R4 982,0052… ≈ R4 982,01 (to the nearest cent) So the monthly installment is R4 982, 013 0,09 b) P= Balance of Loan x = R4 982,01 i = ____ ​  12 ​3 n = 60 – 17 = 43 monthly installments still to pay 3 x [​ 1 − ​(1 + i)​−n​  ]​ P = ___________ ​  i ​ [ ( 4982,01​ 1 − ​​ 1 + ​ ____ __________________ ) ] 0,09 −43 12 ​  ​​ ​  ​ P = ​    0,09 ​33 ____ ​  12 ​ P= 182 535,4693… After 17 months he will still owe R 182 535,47 3 [9] 116 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 Activity 7 1. In order to buy a car, Zack takes out a loan of R25 000 from the bank. The bank charges an annual interest rate of 11% compounded monthly. The installments start a month after he has received the money from the bank. a) Calculate his monthly installments if he has to pay back the loan over a period of 5 years. b) Calculate the outstanding balance of his loan after two years (immediately after the 24thinstalment).(8) 2. Jill negotiates a loan of R300 000 with a bank which has to be repaid by means of monthly payments of R5 000 and a final payment which is less than R5 000.The repayments start one month after the granting of the loan. Interest is fixed at 18% per annum, compounded monthly a) Determine the number of payments required to settle the loan. b) Calculate the balance outstanding after Jill has paid the last R5 000. c) Calculate the value of the final payment made by Jill to settle the loan. d) Calculate the total amount Jill repaid to the bank. (13) [21] Solutions 0,11 1. a) P = R25 000; i = 11% monthly = ____ ​  12 ​3 x is the monthly installment n = 5 × 12 = 60 x [​ 1 – ​(1 + i)​–n​  ]​ P = ___________ ​  i ​ [ ( x ​ 1 – ​​ 1 + ____ ) ] 0,11 –60 ​  12 ​  ​​ ​  ​ 25000 = ______________ ​    0,11 ​33 ​ ____ 12 ​ 25000 ​____ (0,11 ) ​  12 ​  ​ x = _____________ ​   –60 ​ [ ( 0,11 ____ ) ] ​ 1 – ​​ 1 + ​  12 ​  ​​ ​  ​ x = 543,5605768 His monthly installment will be R543,56 (to the nearest cent) 3 b) 5 years × 12 months = 60 months. He still has to pay for 60 – 24 = 36 months. 0,11 P=? i = 11% monthly = ____ ​  12 ​ x = R543,56 n = 36 3 x [ ​ 1 – ​ ( 1 + –n i)​ ​  ]​ P = ___________ ​  i ​ [ ( 543,56 ​ 1 – ​​ 1 + ____ ) ] 0,11 –36 ​  12 ​  ​​ ​  ​ P = __________________ ​    0,11 ​ 33 ​ ____ 12 ​ P = 16602,8718... The outstanding balance after two years will be R16 602,97 (to the nearest cent).3 (8) © Department of Basic Education 2015 UNIT 6 finance, growth and decay 117 Mind the Gap Grade 12 Mathematics

6 Unit 0,18 2. a) P = 300 000 x = 5 000 i = ____ ​  12 ​= 0,015 3 n=? x [​ 1 – ​(1 + i)​–n​  ]​ P = ___________ ​  i ​3 5000 [​ 1 – ​(1 + 0,015)​–n​  ]​ 300 000 = _________________ ​   ​3 Since 0,015 n = 154,6541086, the outstanding period to 300 000 × 0,015 = 5 000 [1 – (1,015)–n] cover the whole loan is 4500 0,6541086 ​ ____ 5000 ​– 1 = – (1,015) –n – ​(1,015)​–n​= –0,1 log 0,1 – n = _______ ​  log 1,015 ​3 n =154,65 ∴Number of payments = 155 3 [ 5 000 ​ 1 – (1 + ____ ____________________ 0,18 ] ​  12 ​​)​–0,6541086​  ​ b) Balance outstanding = ​     ​3333 0,18 ____ ​  12 ​ = R3230, 503 c) Amount paid in last month = 3230,50 ​1 + ____ ( ​  12 ​  ​3 0,18 ) = R3278, 96 3 d) Total repaid = (154 × 5000) + 3278,96 = R773278,96 3 (13) [21] There are 154 equal monthly payments of R5 000 plus the last instalment which is less than R5 000 118 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 Activity 8 1. A farmer buys a tractor for R450 000. a) How much will the tractor be worth in 5 years’ time if its value depreciates at 9% per annum on a reducing balance? (3) b) After 5 years, the tractor needs to be replaced. During this time, inflation remains constant at 7% per annum. Determine the cost of a new tractor after 5 years. (3) c) He plans to sell this tractor at its book value and use the money towards a new tractor. Calculate how much money he will need to put into a Sinking Fund to buy a new tractor in 5 years’ time. (1) d) Calculate the value of the monthly payment into the sinking fund if the interest is 8,5% p.a. compounded monthly over the next 5 years.  (4) 2. Timothy buys furniture to the value of R10 000. He borrows the money on 1 February 2010 from a financial institution that charges interest at a rate of 9,5% p.a. compounded monthly. Timothy agrees to pay monthly installments of R450. The agreement of the loan allows Timothy to start paying these equal monthly installments from 1 August 2010. a) Calculate the total amount owing to the financial institution on 1 July 2010. (2) b)  How many months will it take to pay back the loan? (6) c) What is the balance of the loan immediately after Timothy has made the 25th payment?(4) 3. Calculate how many years it will take for an investment to treble (becomes three times as big) if it is invested at 12% per annum compounded half-yearly.(5) [28] Solutions 1. a) Use compound depreciation with P = R450 000, i = 0,09, n = 5 years. A = P (1 – i)n A = 450 000 ​(1 – 0,09)​5​ 33 A = 280 814,4653 The tractor will be worth R 280 814,47 in 5 years’ time. 3 (This is what its ‘book value’ or ‘scrap value’ will be in 5 years’ time) (3) b) Use compound interest for inflation with P = R450 000, i = 0,07, n = 5 years. A = P(1 + i)n A = 450 000 (1 + 0,07)5 33 A = 631 148.2788 A new tractor will cost R631 148,29 in 5 years’ time. 3 (3) © Department of Basic Education 2015 UNIT 6 finance, growth and decay 119 Mind the Gap Grade 12 Mathematics

6 Unit c) Cost of new tractor – book value of old tractor = R631 148,29 – R280 814,47 = R350 333,82 in a Sinking Fund. 3 (1) d) Use the Future value formula and solve for x. F = R350 333,82x is the monthly installment 0,085 i = 8,5% compounded monthly = _____ ​  12 ​ n = 5 × 12 = 60 months x [​ ​(1 + i)​ ​– 1 ]​ n F = __________ ​  i ​ x ​ ​​ 1 + _____ 350333,82 = ______________ ​    [( ) 0,085 60 ​  12 ​  ​​ ​– 1 ​ ​333 ] 0,085 _____ ​  12 ​ 350333,82 ​_____ ​  12 ​  ​ (0,085 ) x = _____________ ​    ​ [​​​(1 + _____ 0,085 60 ​  12 ​  )​​ ​– 1 ]​ x = 4 706,103568… The monthly payment into the sinking fund over the next 5 years needs to be R4 706,10 (rounded off to the nearest cent) 3 (4) 2. a) A = 10000 ​​ 1 + _____ ( ​  12 ​  ​​ ​3 0,095 5 ) = R10 402,15 3 (2) Multiply 10 402,15 (0,095 ) by ​_______ ​  12 ​  ​and then 450 ​ 1 – ​​ 1 + _____ ________________ b) 10 402,15 = ​    [ ( ​  12 ​  ​​ ​  ​ ​333 )] 0,095 –n 0,095 divide by 450. _____ ​  12 ​ Write in log form to calculate the value of n 0,183000787 = 1– ​_______ ​  12 ​  ​ (1 + 0,095 ) –n (the number of months to pay ( ​1 + _____ 0,095 –n ) ​  12 ​  ​ = 0, 816999213 3 back the loan). ( log ​1 + _____ 0,095 –n ) ​  12 ​  ​ = log 0,816999213 3 –nlog ​1 + _____ ( 0,095 ) ​  12 ​  ​= log 0,816999213… n = 25,63151282… n = 25,63 months n = 26 3 (6) c) Balance outstanding after 25 months = 25, 6315128204… – 25 = 0, 6315128204 3 450 ​ 1 – ​​ 1 + _____ Balance outstanding = _____________________ ​       ​  12 ​  ​​ ​  ​ [ ( ​33 ) 0,095 –0,6315128204 ] 0,095 ​ _____ 12 ​ = R282, 36 3 (4) 120 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 6 3. Let x be the P, the investment in rand. So the final amount A will be three times as much: 3x Rand. 0,12 i = 12% compounded half-yearly (twice a year) = ​  ____ 2 ​3 A = P (1 + i)n ( ) 0,12 n×2 3x = x​​ 1 + ____ ​  2 ​  ​​ ​33 3x ​ __ x ​= (1,06) 2n 3 = (1,06)2n Use logs to find n 2n = log1,06 3 3 2n = 18,85 n = 9,42708834… It will take more than 9 years, so we say that the answer is 10 years. 3 It will take 10 years for an investment to treble if interest of 12% is compounded half-yearly. (5) [28] © Department of Basic Education 2015 UNIT 6 finance, growth and decay 121 Mind the Gap Grade 12 Mathematics

6 Unit What you need to be able to do: • Use the simple and compound growth formulae to solve problems. • Use the simple and compound decay formulae to solve problems. • Calculate the effect of different compounding periods on the effective interest rate when given the nominal interest rate and calculate the nominal interest rate when given the effective interest rate. • Use the Present Value Formula for loans, etc. • Use the Future Value Formula for annuities, savings, etc. • To calculate the outstanding balance at any given time. • To calculate the sinking fund. Keep going! 122 UNIT 6 finance, growth and decay  © Department of Basic Education 2015  Mind the Gap Grade 12 Mathematics

Unit 7 Calculus 7.1 Average gradient ​y​2​ − ​y​1​ The gradient of a straight line can be calculated using m = ______ ​ ​x2​ ​ − ​x​1 ​​ Activity 1 NB PAY SPECIAL ATTENTION 1. a) Determine the average gradient of the graph of y = 5x2 – 4 between x = – 4 and x = –1 b) Is the function increasing or decreasing between x = – 4 and x = –1?(3) HINT! 2. Determine the average gradient of the graph of y = 5x2 – 4 between: a) x = 1 and x = 3 hint b) x = 2 and x = 3 Use the equation of the c) x = 2,5 and x = 3 curve y = 5x2 –4 to 3. d) x = 2,99 and x = 3(8) a) Calculate the average gradient of the curve f(x) = x(x + 3) e.g. ​y​​ − ​y​​ calculate the y-values. EG - worked examples _____1 ​formula to Use the ​ ​x2 between x = 5 and x = 3. ​ ​ − ​x​ ​ 2 1 b) What can you deduce about the function f between x = 5 calculate the gradient. and x = 3? (3) exams [14] Solutions 1. a) At x = –4 b) The function is decreasing between x = –4 and x = –1 because the gradient is y = ​5(–4)​2​– 4 = 80 – 4 = 76 3 activity At x = –1 negative. 3 (1) ACTIVITIE and girl-l y = ​5(–1)​2​– 4 = 5 – 4 = 1 right of p ​y​2​ − ​y​1​ _____ ________ 76 − 1 75 ​  ​x​ ​− ​x​ ​​= ​  −4 − ( −1) ​ = ___ ​  −3 ​= − 25 3 (2) activity 2 1 2. a) The points at x = 1 and x = 3 are (1; 1) b) The points at x = 2 and x = 3 are (2; 16) and (3; 41) 3 and (3; 41) 3 y2 − y1 ​y​ ​ − ​y​ ​ 41 − 16 25 m = ______ 41 − 1 __ ​  x2 − x1 ​= _____ ​  3 − 1 ​= ​  40 ​ = 20 3 (2) m = ______ ​  ​x​ ​ − ​x​ ​​= ______ 2 1 ​  3 − 2 ​= __ ​  1 ​= 25 3 (2) Step by step com 2 2 1 c) The points at x = 2,5 and x = 3 are d) The points at x = 2,99 and x = 3 are (2,5; 27,25) and (3; 41) 3 (2,99; 40,7) and (3; 41) 3 ​y​ ​ − ​y​1​ 41 − ______ 27.25 13.75 ________ _____ ​y​ ​ − ​y​ ​ 41 − 40.7 0.3 ______1 _______ ____ m = ​  ​x2​2​ − ​x​1​​= ​  3 − 2.5 ​= ​  0.5 ​= 27.5 3 (2) ​2​ − ​x​1​​= ​  3 − 2.99 ​= ​  0.01 ​= 30 3 m = ​  ​x2 (2) 3. a) The points are (5; 40) and (3; 18). 3 b) The function is increasing between x = 5 ​y​2​ − ​y​1​ 18 − 40 − 22 and x = 3. 3 (1) m = ______ ​  ​x​ ​ − ​x​ ​​= ______ ​  3 − 5 ​= ____ ​  − 2 ​= 11 3 (2) 2 1 [14] © Department of Basic Education 2015 UNIT 7 calculus 123 Mind the Gap Mathematics

7 Unit Can we calculate the gradient of a curve? 33 30 27 B 24 21 18 tangent to the curve at B 15 12 9 D 6 3 A C 0 0 0.5 1 1.5 2 2.5 3 3.5 4 • The average gradient between two points on a curve is equal to the gradient of the straight line through the points. So the average gradient of curve AB is 12. • As the two points are moved closer together, the average gradient approaches the gradient of the curve which is also the gradient of the tangent to the curve at that point. So the gradient of the curve AB at point B is 30. • Remember that the tangent is a line that touches a curve at one point only. • The average gradient tells us whether the graph is increasing or decreasing between those points. • If the function is decreasing between two points, the average gradient will be negative. • If the function is increasing between two points, the average gradient will be positive. 124 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 7.2 Average rate of change The average rate of change between two points on a graph is the average gradient of PAY the line SPECIAL joining the two points. ATTENTION NB NB If the graph shows distance as a function of time, the average gradient change of distance is ​ ______________      time ​ change inHINT! hintis the average speed = ​ ________​ This Δ distance Δ time e.g. 1 EG - worked examples 1. The average rate of change between A and B shown in the graph exams 27 − 3 24 is _____ ​  3 − 1 ​= __ ​  2 ​= 12. 33 30 activity 27 B 24 ACTIVITIES boy and girl-left and right of page 21 activity 18 15 12 9 Step by step comment comment 6 3 A 0 0 0.5 1 1.5 2 2.5 3 3.5 4 2. If the distance travelled (in metres) is given by the equation s(t) = t2, where t is the time in seconds, then the average speed between ​52​ ​− ​32​ ​ 25 − 9 16 t = 3 seconds and t = 5 seconds is _____ ​  5 − 3 ​= _____ ​  2 ​= __ ​  2 ​= 8 m/s. © Department of Basic Education 2015 UNIT 7 calculus 125 Mind the Gap Mathematics

7 Unit 7.3 The derivative of a function at a point The rate of change of a function at a point is called a derivative. The derivative of a function at a point gives f ´(x) is the rate of change of f at x • the rate of change of the function at the point • the slope (gradient) of the tangent to the function at the point Definition of a derivative f(x + h) − f(x) The derivative of a function y = f (x) is defined as f ´(x) = ​lim​_____________ ​    ​ h➞0 h This formula is provided on the information sheet in the final exam. NOTE: The notation we use for the derivative of y = f(x) is dy f ´(x) or y´ or ​ ___​ or Dx[f(x)]. dx When we find the derivative of a function, we say we differentiate the function. 7.3.1 The derivative from first principles PAY SPECIAL ATTENTION NB (Definition) NB To differentiate from first principles (definition) use the formula below f(x + h) − f(x) f ´hint (x) = l​im​ __________ HINT! ​  ​ h➞0 h e.g. 2 EG - worked examples Determine f ´(x) from first principles if f(x) = –3x2 exams Solution Method 1 f(x + h) = −​3(x + h)​2​ = −3(​x2​ ​ + 2activity xh + ​h2​ ​) ACTIVITIES boy = −​3x​2​ − 6xh − ​3h​2​to get fright (andxgirl-left + hand) we replace x with x + h and get of page ​–3(x + h)​activity 2 ​ Expand the brackets and Make sure you multiply the –3 with each term in the brackets f(x + h) − f(x) Substituting into f ´(x) = ​lim​ ___________ ​  ​ the definition of the derivative gives h➞0 Step by step h comment comment − ​3x​ ​− 6xh − ​3h​ ​− ( − ​3x​ ​) 2 2 2 f ´(x) = ​lim ​ ​ ​ ____________________       ​ f (x) = –3x2 so h f(x) = –3x 2 h➞0 Take out a common factor of − ​3x​ ​ − 6xh 2 − ​3h​ ​ + ​3x​ ​ __________________ = ​lim ​​​   2 ​ 2 h so you can cancel it with the h➞0 h h in the denominator. As h goes to 0, –6x – 3h goes to –6x. 126 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 h(− 6x − 3h) = ​lim ​ ​​ __________ h ​ h➞0 = ​lim ​(− 6x − 3h) h➞0 = –6x Activity 2 1. Determine f ´(x) from first principles if f(x) = 5x2 – 4x + 2 (6) 2 2. Determine f ´(x) from first principles if f(x) = __ ​  x ​ (6) [12] Solutions 1. f(x + h) = ​5(x + h)​2​ − 4(x + h) + 2 = 5(​x2​ ​ + 2xh + ​h2​ ​) − 4x − 4h + 2 = ​5x​2​ + 10xh + ​5h2​ ​ − 4x − 4h + 2 3 f(x + h) − f(x) ​5x2​ ​+ 10xh + ​5h2​ ​− 4x − 4h + 2 − (​5x2​ ​− 4x + 2) f ´(x) = ​lim ​​ ___________ h ​= ​lim ​__________________________________ ​       ​3 h h➞0 h➞0 10xh + ​5h​ ​− 4h 2 = ​lim ​____________ ​    h ​3 h➞0 h(10x + 5h − 4) = ​lim​​ ____________    ​3 h h➞0 = ​lim​(10x + 5h − 4) 3 h➞0 = 10x – 4 3 (6) 2 2. f(x + h) = ​ ____ x+h ​ 2 f(x + h) − f(x) _____ ​  ​− ​  x2__​ f ´(x) = ​lim​​ ___________ h _______ ​= ​lim​​  x + h h ​3 3 h➞0 h➞0 2x 2(x + h) ​ _______ ​− _______ ​  x(x + h) x(x + h) ​ _____________ = ​lim​​    h ​3 h➞0 2x − 2x − 2h __________ ​  x(x + h) ​ _________ = ​lim​​  h ​ h➞0 –2h _______ ​  x(x + h) ​ ______ = ​lim​​  h ​ h➞0 –2h 1 = ​lim​​ _______ ​× __ x(x + h) h ​  ​3 h➞0 –2 = ​lim​_______ ​  x(x + h) ​3 h➞0 –2 −2 ≃ ​ ____ ___ x(x) ​= ​  x2 ​3 (6) [12] © Department of Basic Education 2015 UNIT 7 calculus 127 Mind the Gap Mathematics

7 Unit 7.3.2 The rules of differentiation PAY SPECIAL ATTENTION You could find any derivative fromNB first principles, but there are NBsome quick rules to find the derivative. Unless a question asks you to use the definition or to ‘differentiate from first principles’, it is easier to use the rules. HINT! You need to know and be able to use the following rules for differentiating: hint Rules 1. If f(x) = b then f ´(x) = 0 e.g. 3 EG - worked examples where b is a constant 2. If f(x) = xn then f ´(x) = nxn–1 h(x) = 12, then h ´(x) = 0 If exams d d d The derivative of a constant is always = 0. 3. ___ ​  dx ​[ f(x) ± g(x)] = ___ ​  dx ​[ f(x)] ± ___ ​  dx ​[g(x)] If k(x) = x5, then k ´(x) = 5x4 d d d If f(x) = x5 + x4, then ___ ​  dx ​f(x) = 5x4 + 4x3 4. ​ ___ dx ​[kf(x)] = k​ ___ dx ​[ f(x)] If f(x) = 3x5 then activity d ___ d ​  dx ​ f(x) = 3 × ___ ​  dx ​f(x) (xand 5 )= girl-left3 ACTIVITIES boy and× 5x = 15x 4 4 right of page activity Before NB you use differentiation NB you might need to simplify or change the PAY SPECIAL ATTENTION format of the expressions: Step by step comment comment 1. Expand brackets e.g. expand (3x + 2)(x – 5) to 3x2 – 13x – 10 because you have no rule for differentiating a product. HINT! hint So you need separate terms before you can differentiate. e.g. 4 EG - worked examples Determine f ´(x) if f(x) = (3x + 2)(x – 5) exams Solution f(x) = 3x2 – 13x – 10 ´(xactivity ∴fNB ) = 6x – 13 PAY SPECIAL ATTENTION NB ACTIVITIES boy and girl-left and right of page 2. Rewrite terms which are square roots, cube roots or other roots activity as exponentials HINT! so that you can use the rule: if f(x) = xn then hint f ´(x) = nx n–1 Step by step comment comment e.g. 5 EG - worked examples __ d __ 1 1 ​√x ​ = ​x2​ ​​​ so __ ​  dx ​ ​√x ​ = __ 1 __ __ 2​  ​x​ ​ – ​   ​ 2 exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 128 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 Activity 3 3 __ a) Evaluate Dx[​(x3 – 3)​2​] b) Find f ´(x) if f(x)= ​√x ​ __ d c) Find __ ​  dx ​3​√​x5​ ​  ​ __ ____ d) Differentiate f(x) if f(x) = ​√​x4​ ​  ​ e) Find f ´(x) if f(x) = ​√16​x3​ ​  ​ [11] Solutions a) Dx[(x3 – 3)2] First multiply out = Dx[x – 6x + 9] 3 6 3 = 6x5 – 18x2 3 3 (3) Apply the rules of differentiation __ __ 3 __ d b) ​√x ​ = ​x​ ​so f ´(x) = ​  1_3 ​ ​x​ ​3 3 c) 3​√​x5​ ​  ​= ​x​ ​3 ​​ so ___ ​  dx ​(3​√​x5​ ​ ​) = ​  5_ ​ ​x​ ​3 ​​ 3 3 5 2 ​  13 ​ __ 2 −​ __3 ​ _ _ (2) (2) 3 __ 4 _ ____ 1 _ 3 _ d) ​√​x4​ ​ ​ = ​x​ ​2 ​​= ​x2​ ​ 3 e) f(x) = ​√16​x3​ ​  ​= ​4(​x3​ ​)​ ​2 ​​= 4​x​ ​2 ​​ 3 so f ´(x) = 2​x1​ ​= 2x 3 (2) So f ´(x) = ​  3_ ​.​4.x​ ​2 ​ −1​= ​6x​ ​2 ​​ 3 3 _ 1 _ (2) 2 __ ​  _1 ​ You can write the answer as 6​√x ​ or as ​6x​ ​ 2 PAY SPECIAL ATTENTION [11] NB NB Rewrite terms which are ‘fractions’ where x is part of the __ 1 denominator, ​  xn​ ​​as ​x−n HINT! ​ ​ ​so that you can use the rule: if f(x) = xn hintf ´(x) = nxn–1 then e.g. 6 EG - worked examples 3​x2​ ​ Find f ´(x) if f(x) = ___ ​  4​x​3​​ exams Solution 3​x2​ ​ 3 f(x) = ___ ​  4​x3​ ​​= __ ​  4 ​​x​−1​ activity 3 3 boy So f ´(x) = − __ ​ ​= −and___ ​  4 ​​x−2 ​ ACTIVITIES 4​x​2 ​ ​ and girl-left right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 7 calculus 129 Mind the Gap Mathematics

7 Unit Activity 4 __ dy ​√ x ​ 1 1. Determine, using the rules of differentiation: __ ​  dx ​if y = ___ ​  2 ​− ​ ___ ​ ​6x​3​ (3) dy 4 ​x​ ​ 3 2. Evaluate __ ​  dx ​if y = ___ ​  √​ __ __ x ​​− ​  9 ​  (3) 3. Determine Dx ​ _____ [ 6x + 5 ] ​  3​x​2​ ​  ​ (4) [10] Solutions __ ​√x ​ 1 1. y = ___ ​  2 ​− ​ ___ ​ 6​x​3​ y = ​  1_2 ​​x ​ ​2 ​​− ​  1_6 ​ ​x−3 1 _ ​ ​ 3 First rewrite the terms in the form kxn ___ dy 3 ​  dx ​= ​  1_ ​​x −​ ​ 2 ​​+ ​ __6 ​​x−4 1 _ ​ ​ Use the rules of differentiation 4 ___ dy ​  dx ​= ​  1_ ​​x−​ ​ 2 ​​+ ​  1_ ​​x−4 1 _ ​ ​ Simplify 4 2 dy ____ ​ ___ ​= ​  1__ ​+ ​ ___ dx 4​√x ​ ​2x4​ ​ 1 ​3 3 Change back to surds and positive (3) exponents 4 ​x3​ ​ 2. y = ___ __ x ​​− ​  9 ​ ​  ​√__ y = 4​x−​ ​ 2 ​​− ​  1_9 ​​x3​ ​ 3 1 _ First rewrite the terms in the form kxn dy ​  dx ​= −​  1_ ​∙ 4 ​(​x−​ ​ 2 ​−1​  )​− 3 ∙ ​  1_ ​​x2​ ​ Use the rules of differentiation ___ 1 _ 2 9 ___ dy 1 ​  dx ​= −2​x−​ ​ 2 ​​− ​ __ 3 _ x ​x​ ​ 3 3 Simplify (3) 2 Often, the question will ask you to leave the answer with positive exponents. 2 = − __ ​  ​ 3_​​− ​  _1 ​​x​2​ ​x​2 ​ 3 [ 6x + 5 ] 3. Dx ​ _____ [ 6x 5 ] ​  3​x2​ ​ ​  ​= Dx ​ ___ ​  3​x2​ ​​+ ___ ​  3​x2​ ​​  ​3 = D [​ 2​x​ ​+ ​  5_ ​​x​ ​  ]​ 3 x −1 −2 3 ​ ​ − ​  10 = − 2​x−2 __ ​​x−3 ​ ​33 (4) 3 [10] 130 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 7.4 Uses of the derivative The derivative has many uses. It can be used to: • find the gradient of the equation of a tangent line • identify stationary points on a graph • find a maximum or minimum value • describe rates of change • draw graphs of cubic functions. • (A cubic function has the form f(x) = ax3 + bx2 + cx + d) 7.4.1 Finding the equation of a tangent line The slope of the tangent line to the graph at a point is equal to the derivative of the function at that point. So, to find the equation of the tangent line to f(x) at x = a, we must: 1. Find the derivative f ´(x) 2. Work out the derivative at x = a → i.e calculate f ´(a) to get the gradientPAYof the tangent line. SPECIAL ATTENTION 3. Calculate the y value at x =NB NB a →i.e calculate f(a). 4. The tangent line is a straight line. We can find the equation of a straight line using y – y1 = m(x – x1) if HINT! we know the gradient m and a point (x1; y1) on the line. hint e.g. 7 EG - worked examples Find the equation of the tangent to the function f(x) = x3 + 2x + 4 at the exams point where x = 1. Solution f ´(x) = 3x2 + 2 1. Take the derivative activity f ´(1) = 3(1) + 2 = 5 ACTIVITIES boy 2 2. Find the gradient of the tangent at x = 1 by and girl-left and so m = 5 evaluating the derivative at x = 1. right of page f(1) = activity 1+2+4=7 3. Calculate the y-value at x = 1 Tangent line: y – 7 = 5(x – 1) 4. Use y – y1 = m(x – x1) to give the equation of the tangent line comment y – 7 = 5x – 5 Step by step comment y = 5x + 2 So the equation of the tangent at x = 1 is y = 5x + 2 © Department of Basic Education 2015 UNIT 7 calculus 131 Mind the Gap Mathematics

7 Unit 7.5 Drawing the graph of a cubic polynomial A cubic polynomial is a function of the form f(x) = ax3 + bx2 + cx + d and we can represent it with a graph. In order to draw the graph, we need to work out the characteristics of the graph. • We can use the derivative to identify the slope of the graph at certain points. PAY SPECIAL ATTENTION • We also need to know howNB NB to solve equations in the third degree, so that we can work out the x- and y-intercepts of the graph. 7.5.1 hint Solving equations in the third degree: HINT! ax3 + bx2 + cx + d = 0 e.g. 8 EG - worked examples Work through this example: exams Factorise and solve for x: x3 – x2 – 5x = 3 Solutions 1. Get ax3 + bx2 + cx + d = 0 i.e. x3 – x2 – 5x – 3 = 0 (Standard form) activity 2. Use the Factor and Remainder theorem to find oneACTIVITIES factor. boy and girl-left and Use trial and error. right of page This step can also be calculated on aactivity calculator – see below. The factor theorem states: If f(k) = 0, then x – k is a factor of f(k) Step by step comment comment So if f(x) = x – x – 5x – 3, we want to find an x-value that makes f(x) = 0. 3 2 f(x) has a constant value of –3. If this expression can be factorised, then at least one of its factors will use a factor of –3 in it. The factors of –3 are –3; –1; 1; 3 By trial and error, test these factors to find which value of x gives f(x) = 0. f(x) = x3 – x2 – 5x – 3 If x = –3, then f(–3) = –27 – 9 + 15 – 3 = –24 ≠ 0 If x = –1, then f(–1) = –1 –1 + 5 – 3 = 0 ∴ x –(–1) is a factor of f(x) ∴ x + 1 is a factor of x3 – x2 – 5x – 3. We will use x + 1 to find the other factors. 3. Divide ​x3​​– ​x2​​– 5x – 3 by x + 1 to find the other factors. You can use the algebraic method, long division or synthetic division at this point. Method I: Using algebra x3 – x2 – 5x – 3 = (x + 1)(x2 + px – 3) Check this: First terms give x3, last terms give –3. We don’t know the middle terms, so we have used px in the second bracket. 132 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 To calculate the value of p: The x2 term in the expression has a coefficient of –1 ​x​3​ – x2 – 5x – 3 So the x2 part of the factorised expression must make – x2 (x + 1)(x2 + px – 3) x (px) + 1(x2) = px2 + x2 ∴ px2 + x2 = – x2 px2 = – 2x2 ∴ p = –2 ∴ x3 – x2 – 5x – 3 = (x + 1)(x2 –2x – 3) Method II: Long division [Divide, Multiply, Subtract, Bring down] x2 – 2x – 3 x+1 x3 – x2 – 5x – 3 divide: x3 ÷ x answer x2 on top x3 + x2 multiply: x2(x + 1) –2x2 – 5x subtract & bring down divide: –2x2 ÷ x answer –2x on top –2x2 – 2x multiply: –2x(x + 1) – 3x – 3 subtract & bring down – 3x – 3 divide: –3x ÷ x answer –3 on top multiply: –3(x + 1) 0 subtract ∴ x3 – x2 – 5x – 3 = (x + 1)(x2 –2x – 3) = (x +1)(x – 3)(x +1) Method III: Synthetic division Write down the coefficients of the terms in the original equation, x3 – x2 – 5x – 3. a) On the left place x = –1 d) Multiply –1 by –3 = +3 –1 1 –1 5 –3 Put this under the 4th coefficient (–3) Add the 4th column: –3 + 3 = 0 b) Write down the first coefficient (1) and 1st 2nd 3rd 4th multiply: coeff coeff coeff coeff –1 × 1 = –1 –1 1 –1 –5 –3 Put this under the 2nd coefficient (–1) 1st 2nd 3rd 4th 1 –1 +2 +3 coeff coeff coeff coeff –2 –3 0 –1 1 –1 –5 –3 1 –1 answer © Department of Basic Education 2015 UNIT 7 calculus 133 Mind the Gap Mathematics

7 Unit Add the second column: –1 + –1 = –2 1. You know you are correct when the final sum is 0. These numbers form the 1st 2nd 3rd 4th coeff coeff coeff coeff coefficients of the answer of the division: –1 1 –1 –5 –3 1st 2nd 3rd 4th coeff coeff coeff coeff 1 –1 –1 1 –1 –5 –3 –2 1 –1 +2 +3 c) Multiply –1 by –2 = +2 –2 –3 0 Put this under the third coefficient (–5) 1x2 – 2x – 3 Add the third column: –5 + 2 = –3 ∴ x3 – x2 – 5x – 3 = (x + 1)(x2 –2x – 3) 1st 2nd 3rd 4th = (x +1)(x – 3)(x +1) coeff coeff coeff coeff –1 1 –1 –5 –3 1 –1 +2 –2 –3  Now you have found the first factor (x + 1) using one of the three methods. 4. Factorise the answer further by factorising the trinomial. x3 – x2 – 5x – 3 = (x + 1)(x2 –2x – 3) = (x + 1)(x –3)(x + 1) 5. Determine the three solutions. If (x + 1)(x –3)(x + 1) = 0, Then (x + 1) = 0 or (x –3) = 0 or (x + 1) = 0 x = –1 or x = 3 or x = –1 These are the x-intercepts of a cubic graph with the equation: f(x) = ​x3​​– ​x2​​– 5x – 3 134 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 7.5.2 Stationary points of a cubic function • Stationary points on a graph are points where the gradient of the graph is 0. This is at points where the direction of the curve of the graph changes. PAY SPECIAL ATTENTION NBOn a cubic function, the NB stationary points are at local maximum or minimum turning point. There are also situations where a point of inflection can also be a stationery point as indicated on figure 2 of the example HINT! below. hint NOTE: A point of inflection is not always a stationary point. e.g. 9 EG - worked examples The turning points are only local because the end points of the graphs are exams often greater than the local maximum or less than the local minimum. figure 1 Stationary point is activity a turning point and y y endpoint a local maximum. ACTIVITIES boy and girl-left and right of page point of local maximum activity inflection. turning point x x Step by step comment comment Stationary point is local minimum a turning point and turning point a local minimum. endpoint figure 2 Stationary point is point of inflection. y x The derivative f ´(x) gives us the slope of a graph. So to find the coordinates of the turning points of a function f(x), we solve f ´(x) = 0. To find the coordinates of the point of inflection, find the derivative of the derivative, f ´´(x). This is called the second derivative. Solve for f ´´(x) = 0. © Department of Basic Education 2015 UNIT 7 calculus 135 Mind the Gap Mathematics

7 Unit 7.5.3 Drawing the graph of a cubic function To draw a graph of a cubic function, follow these steps: 1. Find the y-intercept by finding f(0). When x = 0, what is the value of y? You need to learn the steps of drawing a cubic 2. Find the x-intercepts by finding x value(s) polynomial! where f(x) = 0. Factorise f(x) to be able to work out these values. Identify one factor using the factor theorem. The factor theorem: If f(k) = 0, then x – k is a factor of f(x). 3. Find the stationary points or turning point by solving f ´(x) = 0. NB: The three steps indicated above are very important. Sketch graph should have all the above points with correct identification of the shape explained below. 4. Identify the end behavior i.e. identify what happens to the graph for very large positive and negative values of x. • If a > 0, then f(x) is Turning point positive for very big values of x and negative for very big PAY SPECIAL negative values of x. ATTENTION NB • If a < 0, then f(x) is NB negative for very big values of x and positive for very big negative HINT! hint A positive cubic A negative cubic values of x. a>0 a<0 e.g. 10 EG - worked examples Sketch the graph of f(x) = x3 – 4x2 – 11x + 30. exams 1. y-intercept: When x = 0, f(0) = 30 so y-intercept is at (0; 30). 2. x-intercepts: Test some values of x (choose factors of 30) f(1) = 16 so (x – 1) is not a factor. f(–1) = 36 so (x + 1) is not activity a factor. f(2) = 0 so (x – 2) is a boy ACTIVITIES factor. and girl-left and right of page Choose Method I, II or III above to continue. Here is the synthetic method. activity This method is very quick once you can use it accurately. 1st 2nd 3rd 4th Step by step coeffcomment comment coeff coeff coeff 2 1 –4 –11 30 1 2 –4 –30 –2 –15 0 ∴x3 – 4x2 – 11x + 30 = (x – 2)(x2 – 2x – 15) ∴x3 – 4x2 – 11x + 30 = (x – 2)(x – 5)(x + 3) Factorise the trinomial So when y = 0, (x – 2) = 0 or (x – 5) = 0 or (x + 3) = 0 ∴x = 2, x = 5 or x = –3. x-intercepts are at x = 2, x = 5 or x = –3 i.e (2; 0) ; (5; 0) or (-3; 0) 136 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 3. Stationary points or turning points: f ´(x) = 3x2 – 8x – 11 When f ´(x) = 0, then 3x2 – 8x – 11 = 0 (x + 1)(3x – 11) = 0 11 x = –1 or x = __ ​  3 ​ y-values at stationary points: f(–1) = –1 – 4 + 11 + 30 = 36 and (11 ) (11 ) (11 ) f ​__ 3 ​  3 ​  ​= ​​ __ 2 ​  3 ​  ​​ ​= 4​​ __ (11 ) ​  3 ​  ​​ ​− 11 ​__ ​  3 ​  ​+ 30 ≈ − 14.81 ∴ (− 1;36) and ​(__ ​  3 ​; − 14.81 )​ 11 4. Point of inflection: (–1, 36) 40 – f ´(x) = 3x2 – 8x – 11 (0, 30) f ´´(x) = 6x – 8. 30 – 8 4 6x – 8 = 0 where x = __ ​  6 ​= __ ​  3 ​, so point of 20 – 4 inflection is at x = __ ​  3 ​ 10 – (1.33; 10.59) y-value at point of inflection: (4 ) f ​__ 4 ​  3 ​  ​= 10,59. ∴ (​ __3 ​; 10,59) –4 (–3, 0) –3 –2 –1 0 0 1 2 (2, 0) 3 4 5 (5, 0) 6 7 5. End behaviour: a > 0 is positive for very –10 – big values of x and negative for very big negative values of x. (3.67, –14.81) – 20 – 6. Plot the points and the end behaviour. Join the points in a smooth curve. Activity 5 1. f(x) = − ​x3​ ​ − ​x2​ ​ + x + 10 a) Write down the coordinates of the y-intercept of f b) Show that (2; 0) is the only x-intercept. c) Calculate the coordinates of the turning points of f d) Sketch the graph of f. Show all intercepts with axes and all turning points. e) Determine the point of inflection. (17) 2. Sketched below is the graph of g(x) = –2x – 3x + 12x + 20 = 3 2 –(2x – 5)(x + 2)2. A and T are turning points of g. A and B are the x-intercepts of g. P(–3; 11) is a point on the graph. y T P(–3; 11) A B x © Department of Basic Education 2015 UNIT 7 calculus 137 Mind the Gap Mathematics

7 Unit a) Determine the length of AB. b) Determine the x-coordinate of T. c) Determine the equation of the tangent to g at P(–3; 11) in the form y = … d) Determine the value(s) of k for which –2x3 – 3x2 + 12x + 20 = k has three distinct roots. e) Determine the x-coordinate of the point of inflection. (14) [31] Solutions 1. a) When x = 0, y = 10, therefore are (0; 10) 3 (1) b) Assuming that (2; 0) is the x-intercept, then x –2 is a factor of f(x) f(x) = − x3 − x2 + x + 10 = (x − 2)( − x2 − 3x − 5) 33 ∴ x − 2 = 0 or − x2 − 3x − = 0 3 x = 2 but − x2 − 3x − 5 = 0 has no real solution. Hence (x – 2) is the only x-intercept 33 (5) c) At the turning point f ´(x) = − 3x2 − 2x + 1 = 0 3 (–3x + 1)(x +1) = 0 x = ​ 1_ ​ or x = – 1 33 3 1 1 1 1 270 − 3 + 9 − 1 275 5 When x = ​ __3 ​, y = − __ ​  27 ​− __ ​  9 ​+ __ ​  3 ​+ 10 = ____________ ​   27 ​= ___ ​  27 ​= 10​ __ 27 ​ Therefore turning point is ​__ ​  3 ​; ___(1 275 ) (1 ​  27 ​  ​= ​__ 5 ​  3 ​; 10​ __ ) 27 ​  ​3 When x = –1, y = 1 – 1 – 1 +10 = 9 Therefore turning point is − (1; 9) 3 (5) d) Y 3 __ 1 5 ​  3 ​;10 __ ​  27 ​3 10 3 (–1; 9) 3 2 X (4) e) At the point of inflection f ´´(x) =  − 6x − 2 = 0 3 2 1 ∴ at x = − ​ __6 ​=  − ​ __3 ​3 (2) 2. a) Since A and B are the x-intercepts of g they are solutions of –(2x – 5)(x + 2)2 = 0 3 5 i.e. x = –2 and x = ​ __2 ​The distance between –2 and 5 __ __5 ​  2 ​is ​  2 ​– (–2) = 4,5 units 3 (2) b) T is a turning point. g ´(x) = –6x2 –6x + 12 = 0. 3 –6(x2 + x – 2) = 0 –6(x + 2)(x – 1) = 0 When x = –2 or x = 1. 33 So the x-coordinate of T is 1. (3) 138 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 c) g ´(3) = –6(–3)2 –6(–3) + 12 = –24 3 So the equation of the tangent line is y – 11 = –24(x + 3) 3 which simplifies to y = –24x – 61 3(3) d) The graph of y = k is shown together with g(x) below. Using these graphs we can observe that, provided the line lies above the y-value of A and below that of T, the equation –2x3 – 3x2 + 12x + 20 = k will have 3 distinct roots. At T, g(1) = –2 – 3 + 12 + 20 = 27. So for 0 < k < 27 the equation has 3 distinct roots. 3333(4) y T y=k P(–3; 11) A B x e) g ´´(x) = –12x – 6 6 1 –12x – 6 = 0 when x = ____ ​  −12 ​=  − ​ __2 ​33(2) [31] 7.5.4 Finding the maximum or minimum f ´(x) = 0 shows us the local maximum or minimum points. We can use this to solve an applied problem that asks for a maximum or minimum value. This is revision of Grade 10 work that is needed in order to help you with some Grade 12 questions about measurement, volume, maximums and minimums. You need to know these formulae and use them to solve problems. 2-D shapes 3-D shapes 3-D shapes Right prisms Where the base is a polygon and the sides meet at one point, the apex. 1 Area V =Area of base × ⊥ height V = __ ​  3 ​Area of base × ⊥ Height & & 1 = __ ​  3 ​ A × H Perimeter Surface area = the sum of Where H is the perpendicular height (The distance around the outside) the areas of the flat shapes ​​​​​& 1 Surface area = Area of base + ​ __ 2 ​ ph where p is the perimeter of the base and h the slant height © Department of Basic Education 2015 UNIT 7 calculus 139 Mind the Gap Mathematics

7 Unit 1. Circle 1. Right cylinders 1.Cones r slan base perpendicular h t he r r height ight base radius h circular base A = ​πr​ ​ 2 V = ​πr​2​× h 1 Surface area = 2πr2 + 2πrh V = __ ​  3 ​πr2 × H Circumference = 2πr 1 Surface area = πr2 + ​ __ 2 ​(2πr × h) Circumference = 2πr = πr + πrh 2 2. Square 2. Square prism 2. Square base pyramid rectangular face side of height h pyramid a height base slant height side of base l b V = ​  1__3 ​ a2 × H A = length × length = a2 Surface area = area of square + Perimeter = 4a Note: l = b = h = a 4 × area of triangle V = a × a × a = a3 = a2 + 4 ​(__ ​  2 ​ ⋅ a ⋅ h )​ 1 Surface area = 6a2 = a2 + 2ah 3. Rectangle 3. Rectangular prism rectangular The slant height runs from the face middle of the edge of the base to the apex. [ a h We calculate the slant heights using the perpendicular height height and the dimensions of the b base by using the Theorem of Area: A = length × breadth Pythagoras. = ab Perimeter = 2a + 2b l b V=l×b×h Surface area = 2lb+2lh+2bh 140 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 4a. Right-angled triangle 4. Triangular prism 4. Triangular base pyramid rectangular face a c perpendicular height slant base t b heigh h Area: H 1 g ht A = ​ __ 2 ​ × base × height hei base b 1 = ​ __ 2​ × b × a Perimeter = a + b + c V = ​(__ 1 ) ​  2 ​ × b × h ​× H 1 V = ​ __ 3 ​ area of base triangular × H Surface area of triangular prism Surface area = area of base triangular 4b. Triangle = 2 × area of triangle + (sum of areas of 3 triangles) + (sum of areas of 3 rectangles) a h c b Area: 1 V = ​ __ 2 ​ × base × ⊥ height 1 = ​ __ 2​ × b × h Perimeter = a + b + c 2-D shapes 3-D shapes CONVERSIONS 1. Circle 1. Spheres 1 millilitre = 1cm3 radius circumference r 1 m3 = 1000 litres 4 A = πr2 V = ​ __ 3 ​ πr 3 Circumference = 2πr Surface area = 4πr2 © Department of Basic Education 2015 UNIT 7 calculus 141 Mind the Gap Mathematics

7 Unit Activity 6 1. A drinking glass, in the shape of a cylinder (shown here), must hold 200 ml of liquid when full. Find the value of r for which the total surface area of the glass is a minimum. (6) 2. A rectangular box is constructed Height = h in such a way that the length (l) of the base is three times as long as its width. The material used to construct the top and the bottom of the box r costs R100 per square metre. The material used to construct the sides of the box costs R50 per square metre. The box must have a volume of 9 m3. Let the width of the box be x metres. 2.1 Determine an expression for the height (h) of the box in terms of x. (3) 2.2 Show that the cost to h construct the box can be expressed as 1200 C = ____ ​  x ​+ 600x2 (3) x 2.3 Calculate the width of the l box (that is the value of x) if the cost is to be a minimum.  (4) 3. A tourist travels in a car over a mountainous pass during his trip. The height above sea level of the car, after t minutes, is given as s(t) = 5t3 − 65t2 + 200t + 100 metres. The journey lasts 8 minutes. 3.1 How high is the car above sea level when it starts its journey on the mountainous pass? (2) 3.2 Calculate the car’s rate of change of height above sea level with respect to time, 4 minutes after starting the journey on the mountainous pass. (3) 3.3 Interpret your answer to QUESTION 3.2. (2) 3.4 How many minutes after the journey has started will the rate of change of height with respect to time be a minimum?  (3) [26] Solutions 1. Find an equation for what you want to minimise: Surface area of glass = area of base + area of curved surface So S = πr2 + 2πrh 3 Because you cannot take the derivative if there are two different variables in the equation (r and h), you must use other information to help you get the equation for what you want to minimise in terms of one variable only. We know the glass holds 200 ml = 200 cm3. The volume of the glass is πr2h 200 So πr2h = 200 so h = ___ ​  πr2 ​3 142 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 7 And so we can say S = πr2 + 2πr ​___ (200 ) 400 ​  πr2 ​  ​= πr2 + ___ ​  r ​3 Now the only variable is r, because π is a constant. Write S in a way that is easy to find the derivative: S = πr2 + 400r−1 Take the derivative of the function you want to minimise: S´ = 2πr − 400r−2 3 Put the derivative equal to 0: 2πr − 400r−2 = 0 2πr = 400r−2 2πr3 = 400 r≠0 r3 = ​  400 ___ ​ 3 2π ___ √ 2π so r = 3​ ​  400 ___ ​ ​ ≈ 3.99 cm 3(6) 2.1 Volume = l × b × h 3 9 = 3x.x.h 9 = 3x2h 3 3 h = __ ​  x2 ​3 (3) 2.2 C = [​ 2(3xh) + 2xh ]​× 50 + (​2 × 3x2 )​ × 100 (2(3xh) + 2xh) × 50 + (2 × 3x2) × 100 3 = 8x​__(3 ) ​  ​x2​ ​​  ​× 50 + 600x2 3 1200 = ____ ​  x ​600x2 3(3) 1200 2.3 C = ____ ​  x ​+ 600x2 = 1200x−1 + 600x2 3 dC ___ ​  dx ​= − 1200x−2 + 1200x 3 1200 0 = − ____ ​  ​x​2​ ​+ 1200x 3 ∴ 1200x3 = 1200 x3 = 1 x = 1 3(4) 3.1 s(t) = 5t3 − 65t2 + 200t + 100 t = 0 Therefore it is ​5(0)​3​ – ​65(0)​2​+ 200(0) + 100 = 100 metres 33 (2) 3.2 s´(0) = 15t2 − 130t + 200 3 s´(4) = 15(4)2 − 130(4) + 200 3 = – 80 metres per minute 3 (3) 3.3  The height of the car above sea level is decreasing at 80 metres per minute and the car is travelling downwards hence it is a negative rate of change. 33 (2) 3.4 s´(t) = 15t2 − 130t + 200 s´´(t) = 30t − 130 3 30t = 130 3 ∴t = ​  130 ___ ​ 3 30    • t = 4.3​ 3​ (3) [26] © Department of Basic Education 2015 UNIT 7 calculus 143 Mind the Gap Mathematics

7 Unit What you need to be able to do: • Determine the average gradient between two points on a curve • Differentiate from first principles • Differentiate using the rules • Determine the equation of tangents • Use the Remainder and Factor theorem to find factors of equations in the third degree • Solve equations in the third degree • Draw a sketch graph of a cubic function using the x- and y-intercepts, turning points and/or stationary points • Determine the coordinates of the point of inflection • Discuss the nature of stationery points including local minimum, local maximum and points of inflection • Use differentiation to maximise or minimise an equation February/March 2014 Questions 10, 11 and 12 November 2013 Questions 8, 9 and 10 February/March 2012 Questions 8, 9, 10 and 11 November 2012 Questions 8, 9 and 10 February/March 2011 Questions 9, 10 and 11 November 2011 Questions 8, 9, 10 and 11 February/March 2010 Questions 10, 11 and 12 November 2010 Questions 8, 9, and 10 February/March 2009 Questions 11, 12 and 13 November 2009 Questions 10, 11, and 12 Unused Paper: Questions 9, 10, and 11 November 2008 Questions 8, 9, and 10 Keep going! 144 UNIT 7 calculus  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Probability 8.1 Revision Probability is the study of how likely it is that an event will happen. The following questions are typical probability questions: • What is the chance that it will rain tomorrow? • If I buy a Lotto ticket, what is the chance that I will win the Lotto? We can use a probability scale to decide what chance there is of an event happening. 1 1 3 0 ​ __ __ __ 4 ​ ​  2 ​ ​  4 ​ 1 Impossible Poor chance; Fair chance Good chance Certain; 0% unlikely, but very likely; definite possible; 75% 100% 25% AL ATTENTION NB • We can work out the probability using the formula: number of favourable outcomes ____________________ Probability = ​      ​ number of possible outcomes • This ratio can be expressed as a common fraction, a decimal fraction or a percentage. So a probability of 5 out of 8 can be written as _​ 5 ​or as 0,625 or as 62,5%. 8 • Probability always lies between 0 and 1, measured as a fraction or as a decimal. ed examples If probability is shown as a percentage, then it lies between 0% and 100%. ACTIVITIES boy and girl-left and right of page © Department of Basic Education 2015 UNIT 8 probability 145 Mind the Gap Mathematics

8 Unit 8.2 Theoretical probability and relative frequency If you flip a coin: • The possible outcomes are H (heads) or T (tails). • There are two possible outcomes. Each has a 50% chance of happening. • We say that there is a theoretical probability 1 of ​ __ 2 ​for each outcome. The theoretical probability of getting the outcome heads (H), is written as P(H). 1 P(H)NB= ​ __ PAY SPECIAL ATTENTION 2​ NB Relative frequency PAY SPECIAL ATTENTION NB Try this NB experiment: HINT! hint • Flip a coin 10 times. Did it land on heads exactly 5 out of 10 times? Relative frequency is called Empirical probability or Experimental probabilityHINT! e.g. 1 EG - worked examples hint Although the theoretical • Mantse flipped a coin 10 times and it landed on heads 7 times. So for probability of getting heads exams 7 her experiment, the relative frequency of heads is __ ​  10 ​. is ½, your experiment often does not show this • Jake flips a coin 100 times and records his results. His record shows e.g. EG - worked examples exactly. The results of your that he flipped heads 55 times. So the relative frequency of heads 55 45 experiment give you the is ​ ___ 100 ​. Therefore, the relative frequency of tails is ___ ​  100 ​. activity relative frequency of getting • Jake flips the coin 1 000 ACTIVITIES boy times. Now it is likely that heads and tails heads in that particular and girl-left and will come up about the same number of times. He is likely to get exams right of page experiment. heads 499 to 501 times. activity Now the relative frequency is equal to or close to the theoretical 1 probability of __ ​  2 ​. Step by step comment comment NB activity AL ATTENTION ACTIVITIES boy and girl-left and • An EVENT is a happening or an activity that has outcomes or results. right of page Example: activity Rolling an even number is an event with given outcomes. • An OUTCOME is the possible result of an event. Example: The possible outcomes Step byof rolling a die arecomment step comment 1, 2, 3, 4, 5 and 6. ed examples • The SAMPLE SPACE is the set of all possible outcomes. 146 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 8.3 Venn diagrams We use Venn diagrams to help us to represent different events. Venn Diagram consists of circles and a rectangle The rectangle S represents the sample space (all of the possible outcomes). Each circle inside S represents a different event. If the two circles intersect, the intersection shows which outcomes belong to both events. S (sample space) Event A Event A AND Event B event B PAY SPECIAL ATTENTION NB NB HINT! hint e.g. 2 EG - worked examples PAY SPECIAL ATTENTION 1. Draw a Venn diagram to show the sample space NB exams S = {1; 2; 3; 4; 5; 6; 7; 8; 9; 10} Indicate the following events in the sample space: Event A is the set of prime numbers. HINT! ∴A = {2; 3; 5; 7} activity Event B is the set ACTIVITIES of even numbers. hint boy ∴B = {2; 4; 6; 8; right 10}of page and girl-left and • Both sets have 2 in them, activity so A and B must intersect. e.g. EG - worked examples • Write in 2 first in the Solution intersection. Step by step comment comment S • Then write in the remaining numbers in exams each event. A • Check if there are any B 3 8 numbers that are not in 2 6 Event A or Event B. 5 4 • 1 and 9 is part of the 7 10 sample space, but not in 1 9 A oractivity in B. Write it in the rectangle, but not in A ACTIVITIES boy and girl-left and or B. right of page activity Step by step comment © Department of Basic Education 2015 UNIT 8 probability 147 Mind the Gap Mathematics

HINT! 8 Unit hint e.g. 3 EG - worked examples Use the Venn diagram in the previous example to determine: exams 1. P(A) 2. P(B) 3. P(A and B) activity 4. P(A or B) ACTIVITIES boy and girl-left and right of page Solutions activity 4 1. P(A) = __ ​  10 ​= ​ _25 ​ 5 __by step 1 comment ​  10 ​ = ​ _2 ​ 2. P(B) = Step comment 1 3. P(A and B) = __ ​  10 ​ 8 4 4. P(A or B) = __ ​  10 ​= __ ​  5 ​ PAY SPECIAL ATTENTION NB NB • P(A)+P(B) – P(A and B) 4 5 1 8 4 = __ ​  10 ​+ __ ​  10 ​– __ ​  10 ​= __ ​  10 ​= ​ __5 ​ HINT! 8 4 • hint P(A or B) = ​ __ __ 10 ​= ​  5 ​ ∴P(A or B) = P(A)+P(B) – P(A and B) e.g. EG - worked examples exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 148 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 8.4 Mutually exclusive events Mutually exclusive events are events that cannot happen at the same time. There is no PAY SPECIAL intersectionATTENTION between the events. NB NB • Mutual: applies to two or more people or events. • Exclude: to keep out, not allow a person in. • HINT!exclusive: Both events keep the other out. So there is no Mutually hint outcome that can happen in both events at the same time. e.g. 4 EG - worked examples If you roll a die, it is impossible for it to land on a 1 and a 6 at the same exams time. So P(1) and P(6) are mutually exclusive. When you roll a die, what are the chances of getting a 6 or a 1? 1 1 2 1 So P(1 or 6) = P(1) + P(6) = __ ​  6 ​+ __ ​  6 ​= __ ​  6 ​= __ ​  3 ​ 1 chance of rolling either a 1 or a 6 is __ So the activity ​  3 ​or 33,3% ACTIVITIES boy S: Possible outcomes and girl-left and right of page for rolling a die activity Roll a 5 PAY SPECIAL ATTENTION Step by step Roll a comment comment 1 Roll a 6 NB Roll a 4 Roll a 2 Roll a 3 HINT! hint When two events are mutually exclusive, P(A and B) = 0 If P(A and B) = 0 or if P(A or B) = P(A) ∴P(A or B) = P(A) + P(B) for mutually exclusive events e.g. + P(B), then theEGevents mutually exclusive are - worked examples We can also use this rule for the number of elements or outcomes in each event, if the events are mutually exclusive: exams n(A or B) = n(A) + n(B) When the two events are mutually exclusive, then they do not overlap. Therefore the intersection of A and B is empty and we write A∩B = ∅ (empty set) and P(A∪B) = 0 activity ACTIVITIES boy and girl-left and right of page activity Step by step comment © Department of Basic Education 2015 UNIT 8 probability 149 Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB 8 Unit HINT! 8.5 Complementary events hint Events that are mutually exclusive and make up the whole sample • Complement: space are called complementary events. There is no intersection and no (noun) something that elements from the sample set are outside the two sets. e.g. completes anEG -event; workeditexamples adds what is missing to The possible events when you roll a die are 1; 2; 3; 4; 5 or 6. make up the whole. 1 The probability of rolling a 4 is ​ __ 6 ​. • Complementary: 5 exams (adjective) an event that The probability of not rolling a 4 is __ ​ 6 ​. S: Possible outcomes completes or adds to So the event not rolling a 4 is for rolling a die other events to make up the complement of the event Roll a 5 the whole sample space. rolling a 4. NOTE: A complement is not 1 5 the same as a compliment! So P(4) + P(4´ ) = ​ __ __ 6 ​ + ​  6 ​= 1 Roll a 1 Roll a 6 activity A compliment is a positive The complementary rule: Roll a 4 comment made to a person Roll a 2 Roll a 3 P(A´) + P(A) = 1 or P(A´) = 1 – P(A) ACTIVITIES boy or a group of people. and girl-left and P(A´) means probability of ‘not A’. right of page activity In the example, n(not rolling a 4) + n(rolling a 4) = 5 + 1 = 6 Activity 1 Step by step comment comment 1. If S = {1; 2; 3; 4; 5; 6; 7}, A = {1; 3; 5; 7} and B = {2; 4; 6}, what is the probability of choosing a number that is not in set A? (2) 2. S = {a; b; c; d; e; f; g; h; i; j} and A = {a; e; i}, B = {b; c; d; f; g; h}, C = {b; h; j}. a) Draw a Venn diagram to represent S. (4) b) Give a description of set A. (1) c) Are there any complementary sets? Explain. (2) d) Which sets are mutually exclusive, but they are not complementary? Give a reason for your answer.  (2) 3. A DVD shop has 180 comedies, 250 drama films, 230 science fiction movies and 120 thrillers. If you select a DVD at random, what is the probability that this movie is a comedy OR a thriller?(3) [14] 150 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Solutions 4 3 1. P(A´ ) = 1 – P(A) = 1 – ​ __7 ​ = __ ​  7 ​ 33 (2) 2. a) S A C e g j bh a d f i c  B 3333(4) b) Set A is the set of vowels from a to j; or the set of the first three vowels of the alphabet. 3 (1) c) Sets A and B are not complementary because they do not include element j. Sets A and C are also not complementary. Sets B and C share elements b and h, so they are not mutually exclusive or complementary. 33(2) d) Set A and set B are mutually exclusive, but they are not complementary. They do not share any elements, but they do not make up the whole sample space. Set A and Set C are also mutually exclusive, but not complementary. 33(2) 3. No DVD is marked as both a comedy and a thriller, so there is no overlap in events. These are mutually exclusive (but not complementary). There are 250 + 230 + 120 = 600 DVDs in the sample space. Use P(A or B) = P(A) + P(B). P(comedy or thriller) = P(comedy) + P(thriller) 180 120 300 5 = ___ ​  780 ​3 + ___ ​  780 ​3 = ___ ​  780 ​= __ ​  13 ​3 (3) [14] © Department of Basic Education 2015 UNIT 8 probability 151 Mind the Gap Mathematics

8 Unit 8.6 NB Events which NB are not PAY SPECIAL ATTENTION mutually exclusive HINT! hint Sometimes two events have some outcomes that are the same. e.g. 5 EG - worked examples S exams 4 8 10 1 3 6 9 5 activity 12 ACTIVITIES boy 7 and girl-left and right of page 11 activity A B 2 Step by step comment comment The sample space S = {1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12} In the sample space, event A is the set of multiples of 3. So Set A = {3; 6; 9; 12} Event B is the set of odd numbers. So B = {1; 3; 5; 7; 9; 11} favourable outcomes 4 1 P(A) = ________________ ​   __ __   ​= ​  12 ​= ​  3 ​ possible outcomes 6 1 P(B) = __ ​  12 ​= __ ​  2 ​ 4 6 10 So P(A) + P(B) = ​ __ __ __ 12 ​+ ​  12 ​= ​  12 ​ P(A or B) is the chance of getting the numbers in set A or in set B. We cannot count the 3 and the 9 for both sets. We cannot repeat the numbers in the intersection of set A and set B. 8 So P(A or B) = __ ​  12 ​. So P(A) + P(B) ≠ P(A or B) To make them equal, we need to subtract the probability of the intersection, P(A ∩ B) 4 6 2 8 P(A) + P(B) – P(A ∩ B) = ​ __ __ __ __ 12 ​ + ​  12 ​ – ​  12 ​ = ​  12 ​ This is the answer we found for P(A or B). The ADDITION rule for the probability of ANY two events in a sample space: P(A or B) = P(A) + P(B) – P(A ∩ B) We can also use this rule for the number of elements or outcomes in each set: n(A or B) = n(A) + n(B) – n(A ∩ B) 152 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Activity 2 In a group of 50 learners, 35 take Mathematics and 30 take History. 12 learners do not take Mathematics or History. 1. Draw a Venn diagram to represent this information.  (4) 2. If a learner is chosen at random from this group, what is the probability that he takes both Mathematics and History? (2) [6] Solutions 1. Use M for Mathematics and H for History. S M H 8 3 x 12 3333 (4) – Draw the sample space and sets for the events M and H. – We do not know yet how many learners (outcomes) are in the intersection of M and H. So let M ∩ H = x – We do know that 12 learners are not in M or H. 35 – x + x + 30 – x + 12 = 50 –x = –27 x = 27 So place 27 in the intersection of M and H. M = 35 – 27 = 8 H = 30 – 27 = 3. 27 2. P(M and H) = __ ​  50 ​ 33 (2) [6] © Department of Basic Education 2015 UNIT 8 probability 153 Mind the Gap Mathematics

8 Unit 8.7 Summary of symbols and sets used in probability There are some symbols you need to use when describing probability. We have used some of them already. To explain the use of each symbol, we will use these sets again: S = {a; b; c; d; e; f; g; h; i} and A = {a; e; i}, B = {a; b; c; d; f; g;}, where S= Sample Space, A and B are two sets is the sample space P(A) (A) means the probability that an element from set A will occur. 3 1 P(A) = ​ __ __ 9 ​ = ​  3 ​ n(A) n(A) means the number of elements in set A. n(A) = 3 A´ A´ means all the elements of the sample space that are NOT in set A. This is the complement of set A. A´ = {b; c; d; f; g; h} ∪ A ∪ B means the same as A OR B. It means the union of the two sets and represents the total of all the elements that are in set A or set B. No elements are repeated. A ∪ B = {a; b; c; d; e; f; g; i} ∩ A ∩ B is the same as A and B. It means the intersection of sets A and B and represents all the elements that they share. (All the elements that are in both set A and set B at the same time). This is where the sets overlap. A ∩ B = {a} P(A ∩ B) P(A ∩ B) means the probability that an element from (A ∩ B) 1 will occur. P(A ∩ B) = ​ __ 9​ P(A ∪ B) P(A ∪ B) means the probability that an element from (A ∪ B) 8 will occur. P(A ∪ B) = ​ __ 9​ n(A ∪ B) n(A ∪ B) means the number of elements in set A or set B. n(A ∪ B) = 8 n(A ∩ B) n(A ∩ B) means the number of elements in set A and set B at the same time (the elements they share). n(A ∩ B) = 1 (A ∩ B)´ (A ∩ B)´ means all the elements of the sample space that are NOT in (A ∩ B), the complement of A ∩ B. (A ∩ B)´ = {b; c; d; e; f; g; h; i} (A ∪ B)´ (A ∪ B)´ means all the elements of the sample space that are NOT in (A ∪ B). (A B)´ = {h} 154 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 AB AB AB OR The shaded areas represent: The shaded area represents: (A or B) or (A ∪ B) (A and B) or (A ∩ B) You also need to be able to work with three sets in probability, using a Venn diagram and the formulae. A B A B A B C C C The shaded areas represent: (A or B or C) or (A ∪ B ∪ C) A B A B C C The shaded area represents: The shaded area represents: (A and B and C) or (A ∩ B ∩ C) (A and C) or (A ∩ C) © Department of Basic Education 2015 UNIT 8 probability 155 Mind the Gap Mathematics

HINT! 8 Unit hint e.g. 6 EG - worked examples A survey is conducted with a group of 50 learners to find out what is more exams popular at the school tuckshop. They are asked if they usually buy toasted sandwiches (T), salads (S) or burgers (B). They can choose none, one, two or three of the meals. The survey results are shown with this Venn diagram: activity ACTIVITIES boy and girl-left and right of page S activity T 2 10 15 1 comment 3 5 Step by step comment 4 B a) How many people did not buy salads, toasted sandwiches or burgers? b) Calculate the probability that a learner selected at random from this survey: i) buys salad, but not toasted sandwiches or burgers. ii) buys toasted sandwiches and salad, but not burgers. iii) buys salad or burgers or both, but not toasted sandwiches. Solutions: a) 50 – (15 + 2 + 10 + 1 + 3 + 5 + 4) = 50 – 40 = 10 10 learners bought none of the items listed. 10 1 2 1 10+5+4 19 b) (i) __ ​  50 ​= __ ​  5 ​ (ii) __ ​  50 ​= __ ​  25 ​ (iii) ______ ​  50 ​= __ ​  50 ​ 156 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Activity 3 A school organised a camp for 103 Grade 12 learners. The learners were asked which food they prefer for the camp. They had to choose from chicken (C), vegetables (V) and fish (F). The following information was collected: • 2 learners do not eat chicken, fish or vegetables • 5 learners eat only vegetables • 2 learners only eat chicken • 21 learners do not eat fish • 3 learners eat only fish • 66 learners eat chicken and fish • 75 learners eat vegetables and fish Let the number of learners who eat chicken, vegetables and fish be x. a) Draw a Venn diagram to represent the information. (6) b) Calculate x. (3) c) Calculate the probability that a learner, chosen at random: i) Eats only chicken and fish, and no vegetables. (2) ii) Eats any TWO of the given food choices: chicken, vegetables and fish. (2) [13] Solutions: a) Fill in any given information that C V you can. y We do not know where these 5 2 belong yet: x – 21 learners do not eat fish 66–x – 66 learners eat chicken and fish 2 75–x So let x be learners who eat 3 chicken, fish and vegetables. F Then 66 – x is learners who eat only chicken and fish. Introduce y, the learners who do not eat fish, but do eat chicken and vegetables. Then 2 + y + 5 + 2 = 21 ∴y = 12 333333(6) b) 2 + 12 + 5 + 66 – x + x + 3 + 75 –x + 2 = 103 33 –x + 165 = 103 –x = –62 x = 62 3 (3) 66–x ___ ____ 4 c) (i) ​  103 ​= ​  103 ​ 33 (2) 4+12+13 29 (ii) _______ ​  103 ​= ___ ​  103 ​ 33 (2) [13] © Department of Basic Education 2015 UNIT 8 probability 157 Mind the Gap Mathematics

8 Unit 8.8 Tree diagrams and NB contingency tables PAY SPECIAL ATTENTION NB 1. Independent events Two successive events are independent if the outcomes of the one HINT! hint event do not influence the outcomes of the other event. e.g. 7 EG - worked examples 1 The probability of flipping a coin and it lands on heads is P(H) = __ ​  2 ​. exams What is the probability of flipping two coins and they both land on heads? Solution Thereactivity are four possible outcomes: H and H; H and T; TACTIVITIES and boy H; T and T. and girl-left and 1 So H and H is 1 out of 4 outcomes and P(H and H) = __ right of page ​  4 ​. activity A tree diagram is a picture that helps you to list all possible outcomes of the events. Step by step comment comment Here is the tree diagram for P(H and H) if you flip a coin twice: 1st flip 2nd flip outcomes 1 ​ __ 2​ H HH 1 ​ __ H 2​ 1 __ HT ​  2 ​ T 1 1 ​ __ ​ __ ​ H TH 2​ T 2 1 ​ __ 2​ T TT The tree diagram shows 4 outcomes. Each time you flip the coin, the outcomes (heads or tails) do not depend on the outcomes of the last flip. So these events are independent of each other. 158 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 8 e.g. 8 EG - worked examples You have a pack of cards (no jokers). exams What is the probability of these two events? – Event A: Drawing a heart card from a pack of cards and activity putting it back. ACTIVITIES boy and girl-left and – Event B: Drawingrightaofheart page card There are 52 cards in a pack (or deck). activity from the pack again. There are 4 suits: A and B are independent events. No hearts; spades; diamonds and clubs. matter what card is drawn in Event To draw a card, means to take a A, it is put Step by back step in the comment pack. So the comment card out of the pack. outcome of event B does not depend on the outcome of event A. Tree diagram Here is the tree diagram for all possible outcomes of the two events. Event A Event B Outcomes for heart then heart 13 ___ ___ 13 13 ___ 1 __ 1 __ ___1 heart ​  heart ​  ​ 52 52 ​ × ​  52 ​ = ​  4 ​ × ​  4 ​ = ​ 16 ​ 13 ___ ​  52 ​ not a ___ ​  39 heart 52 ​ 1 __​ heart ​  13 ___ P(A and B) = ​  16 not a 52 ​ Compare this with: heart 39 ___ not a 39 ___ ​  1 ​ – __ P(A) × P(B) = – __ ​  1 ​ = __ ​  1 ​ ​  52 ​ heart ​  52 ​ 4 4 16 ∴ P(A and B) = P(A) × P (B) Events are independent if the probability of one event happening is not influenced by another event happening. P(A and B) = P(A) × P(B)……..if the events are independant 2. Dependent events Two successive events are dependent if the outcomes of the one event influence the outcomes of the other event. © Department of Basic Education 2015 UNIT 8 probability 159 Mind the Gap Mathematics

HINT! 8 Unit hint e.g. 9 EG - worked examples What is the probability of these two events? exams – Event A: Drawing a heart from a regular pack of cards and not putting it back. – Event B: Drawing a heart again, from the rest of the pack (51 cards left). activity ACTIVITIES boy and girl-left and Solution right of page activity A and B are dependent events, because event B depends on the outcomes of event A. Here is a tree diagram for Event A and Event B. comment Step by step comment Event A Event B Outcomes for heart then heart 13 12 156 1 heart __​ ​  12 heart ​ ___ ___ _____ ___ 52 ​ × ​  51 ​ = ​  2 652 ​ × ​  17 ​ 51 13 ___ ​  52 ​ not a 39 __ ​  51​ heart 1 __​ heart ​ __ 13 P(A and B) = ​  17 not a 51 ​ Compare this with: heart __ 39 not a 38 __ ​  1 ​ × __ P(A) × P(B) = __ ​  1 ​ = ​  1 __​ ​  52 ​ heart ​  51 ​ 4 4 16 ∴ P(A and B) ≠ P(A) × P (B) When the events are dependent. Events are dependent if the probability of one event happening is influenced by how another event happens. P(A and B) ≠ P(A) × P(B)…………. for dependant events 160 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 8.9 Contingency tables We can also use a contingency table to represent all possible outcomes of events. Look at the same example we used for the tree diagram above: What is the probability of these two events? – Event A: Drawing a heart card from a pack of cards and putting it back. – Event B: Drawing a heart card from the pack again. We can make a table of possible outcomes using columns for the type of card drawn and rows for the events: Numbers in each heart not heart Total row add up to totals NB A Event PAY SPECIAL ATTENTION 13 NB 39 52 on the right. Numbers in each Event B 13 39 52 column add up Total HINT! 26 78 104 to totals below the hint table. e.g. 10 EG - worked examples The hair colour of 50 learners was recorded. The table below represents exams the information Girls Boys Total Black 10 12 22 activity Brown 8 9 17 ACTIVITIES boy and girl-left and Blond right of page6 5 11 activity Total 24 26 50 Calculate the probability that learner chosen at random: 1) will have brown Step by step haircomment comment 2) will have blond hair 3) will have black hair or brown hair 4) will have blond hair or brown hair or black hair Solutions 17 1) 17 learners have brown hair out of a total of 50 ∴ P (brown hair) = __ ​  50 ​ 11 2) 11 learners have blond hair out of a total of 50 ∴ P (blond hair) = __ ​  50 ​ 3) 22+17=39 learners have black or brown hair out of a total of 50 39 ∴ P (black or brown hair) = __ ​  50 ​ 4) 22+17+11=50 learners have black or brown or blond hair out of a total of 50 50 ∴ P (black or brown or blond hair) = __ ​  50 ​= 1 © Department of Basic Education 2015 UNIT 8 probability 161 Mind the Gap Mathematics

8 Unit PAY SPECIAL ATTENTION NB NB Activity 4 HINT! 1. P(A) = 0,45; P(B) = 0,3 and P(A or B) = 0,165. hint Are the events A and B: a) mutually exclusive b) independent  (7) One die 2. What is the probability of throwing at least one six in four rolls of a e.g. EG - worked examples regular die? (3) 3. What is the probability of throwing four 6s in a row with four rolls of a regular die? (3) exams Two dice 4. If two dice are rolled at the same time, what is the probability that the sum of the two numbers is 9?  (3) [16] Solutions 1. a) P(A or B) = P(A) + P (B) – P(A and B) activity Events A and B are mutually exclusive if P(A and B) = 0 ACTIVITIES boy and girl-left and ∴ if events are mutually exclusive, then P(A or B) = P(A) + P(B) right of page P(A or B) = 0,165 activity P(A) + P(B) = 0,45 + 0,3 = 0,75 33 ∴ P(A or B) ≠ P(A) + P(B) 3 (3) Events A and B are not mutually exclusive. b) Events A and B are independent if P(A and B) = P(A) × P(B) Step by step comment comment To work out P(A and B), use the rule for P(A or B): P(A or B) = P(A) + P(B) – P(A and B) for all events A and B 0,165 = 0,45 + 0,3 – P(A and B) 33 ∴ P(A and B) = 0,75 – 0,165 = 0,585 3 But P(A) = 0,45 and P(B) = 0,3 ∴ P(A) × P(B) = 0,45 × 0,3 = 0,135 3 ∴ P(A and B) ≠ P(A) × P(B) ∴ Events A and B are not independent. (4) 2. The probability of not throwing a six in 4 rolls of a die is: (5 ) 4 625 ​  6 ​  ​​ ​= ____ P(6) = ​​ __ ​  1296 ​ 625 671 ∴ P(at least one 6) = 1 − ____ ​  1296 ​= ____ ​  1296 ​333(3) 3. Each roll of the die is independent of the previous one. 1 1 1 1 1 P(four 6s in a row) = __ ​  6 ​× __ ​  6 ​× ​ __6 ​× __ ​  6 ​= _____ ​  1 296 ​ 333(3) 162 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Solutions (continued) 4. Use a table: Let the columns represent die 1 and the rows represent die 2. Die 1 1 2 3 4 5 6 6 1;6 2;6 3;6 4;6 5;6 6;6 5 1;5 2;5 3;5 4;5 5;5 6;5 Die 2 4 1;4 2;4 3;4 4;4 5;4 6;4 3 1;3 2;3 3;3 4;3 5;3 6;3 2 1;2 2;2 3;2 4;2 5;2 6;2 1 1;1 2;1 3;1 4;1 5;1 6;1 There are 4 throws of both dice that give numbers with a sum of 9. 4 1 ∴ P(sum of 9) = __ ​  36 ​= __ ​  9 ​333 (3) [16] © Department of Basic Education 2015 UNIT 8 probability 163 Mind the Gap Mathematics

8 Unit 8.10 NB Counting principles PAY SPECIAL ATTENTION NB Statistics has many applications in everyday life. The tree diagrams and contingency HINT! tables used so far are useful if there are not too many hint outcomes or possibilities. Look at these examples. e.g. 11 EG - worked examples 1. How many different outfits can be combined using a shirt and a pair exams of pants from 3 shirts (red, white or blue) and 2 pairs of pants (black or brown)? SHIRTS: red; white and blue PANTS: black and brown activity ACTIVITIES boy Solution and girl-left and right of page red activity black white brown Step by step comment comment blue Outcomes: red/black red/brown white/black blue white/brown ​ ______​ blue/black ∴6 different outfits (3 × 2 = 6) brown 2. How many different meals could you have if the menu at a restaurant offered: Dinner Drink Dessert fried chicken orange juice ice cream fish and chips Coca-Cola apple pie hamburger coffee tea Solution We can use 3 × 4 × 2 = 24 to find the number of different meals. We need a more effective way of counting and keeping track of all possibilities. 164 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

NB PAY SPECIAL ATTENTION NB Unit 8 1. Counting permutations Permutation HINT! hint A way of grouping a) The number of permutations of n different items elements of a group in a specific order. e.g. 12 EG - worked examples How many different ways could you arrange 4 books on a exams shelf ? R P Q Call them P, Q, R and S. S } There are 6 possible outcomes that start with P: P Q activity R S ACTIVITIES boy and girl-left and P Q S R right of page activity P R S Q P R Q S 24 different ways to arrange P S R QStep by step comment comment 4 books P S Q R Now start with Q (6 possibilities) Now start with R (6 possibilities) Now start with S (6 possibilities) Instead of writing down all the possibilities, we can find the answer by using the factorial (!) key on a calculator. 4! = 4 × 3 × 2 × 1 = 24 Using n factorial (n!): NBexclamation mark ! is called The NBthe factorial symbol. PAY SPECIAL ATTENTION 4! reads as ‘four factorial’ and means 4 × 3 × 2 × 1 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5 040 HINT! hint b) Permutations of outcomes that are not all distinct (some are the same) e.g. 13 Consider the word TAN EG - worked examples How many word arrangements can be made with the word TAN exams Solution There are three letters. Possible words are: TAN TNA ANT ATN NTA NAT 3!=3 activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 8 probability 165 Mind the Gap Mathematics

HINT! 8 Unit hint e.g. 14 Consider the word MOM EG - worked examples How many word arrangements can be made with the word MOM if the repeated examsare treated as different letters? letters How many word arrangements can be made with the word MOM if the repeated letters are treated as the same letters. activity Solutions ACTIVITIES boy and girl-left and 1. There are three letters. Let us write the first M as M1 and the second M and M2 right of page activity M1OM 2 M1M2O OM1M2 OM2M1 M2OM1 M2M1O ∴3! = 3 × 2 × 1 = 6 2. If we remove the 1 and the 2 from the letter M, we will obtain the following words: Step by step comment comment MOM MMO OMM Therefore, there are only three possible words from the word MOM 3! 3 = ​ __ 2! ​. The 3! (the numerator) indicates the total number of words formed with three letters and the 2! (the denominator) indicates the number of times a letter is repeated. Activity 5 1. Determine the number of permutations that can be formed from all the letters of the word ABRACADABRA.(4) 2. Determine the number of permutations that can be formed from all the letters of the word ABRACADABRA. This time, the first and last letters must be A. (4) 3. Determine the number of permutations that can be formed from all the letters of the word ABRACADABRA. This time, all the As have to be next to each other. (4)  [12] Solutions 1. There are 11 letters (so n = 11), but some letters are repeated. There are 5 As; 2 Bs; 2 Rs; 1 C and 1 D. 11! The number of permutations will be ​ _______ 5!2!2!1!1! ​333 = 83 160 3(4) On a calculator, use the multiplication sign between factorial factors. 2. The first and last letters are ‘fixed’, so there are 9 letters that can change positions (n = 9). There are 3 As; 2 Bs; 2 Rs; 1 C and 1 D. 9! The number of permutations will be _______ ​  3!2!2!1!1! ​333= 15 120 3 (4) 3. Treat ‘AAAAA’ as one possible outcome, so we have n = 7. There are one AAAAA; 2 Bs; 2 Rs; 1 C and 1 D. 7! The number of permutations will be _______ ​  1!2!2!1!1! ​333= 1 260 3 (4) [12] 166 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB Unit 8 HINT! hint number of permutations of m distinct objects taken n at a time C) The e.g. 15 EG - worked examples There are 6 people in a room. Call them A, B, C, D, E and F. exams How many different groups of 2 people are possible? Solution The question activityis really – how many permutations of 2 people (A to F) are possible? ACTIVITIES boy and girl-left and We can list them: right of page AB ACactivity AD AE AF (5) or BA CA DA EA FA (5) BC BD BE BF (4) or CB DB EB FB (4) CD CE CF (3) or DC EC FC (3) Step by step comment comment DE DF (2) or EC FD (2) EF (1) or FE (1) There are 5 + 4 + 3 + 2 + 1 + 5 + 4 + 3 + 2 + 1 = 30 different groups of 2 people. To find this answer without writing all the possibilities out, we can use the formula: m! Permutations: m Pn = _____ ​  (m–n)! ​ where m = total number of possibilities n = number of items chosen in a group 6! 6! 6×5×4×3×2×1 So 6P2 = _____ ​  (6–2)! ​= __ ​  4! ​= ___________ ​   4×3×2×1 ​= 6 × 5 = 30 PAY SPECIAL ATTENTION UseNB NB or work it out as shown here. the factorial key on the calculator, D) The number HINT! of permutations of m items taken n at a time (where hint the items may be repeated any number of times) e.g. 16 EG - worked examples In a multiple choice test there are 5 questions, each with 4 multiple choice exams answers. How many possible ways are there of answering the questions if you guess the answers? Solution activity Since you can choose ACTIVITIES fromboy4 answers for each question, you can represent the answers with 5 right ‘boxes’ of 4 solutions: and girl-left and of page 4×4×4 × 4 × 4 = 45 = 1 024 activity Step by step comment comment © Department of Basic Education 2015 UNIT 8 probability 167 Mind the Gap Mathematics

HINT! 8 Unit hint e.g. 17 EG - worked examples How many three digit numbers can be formed with the digits exams 0 – 9, if numbers can be repeated? Solution 10 ‘boxes’ of 3 numbers: activity 10 × 10 × 10 = 103 = 1ACTIVITIES 000boy and girl-left and right of page activity SUMMARY The basic counting principle: The numberStepof ways of comment by step making comment several decisions in succession (call them m1; m2 and m3 etc…) is determined by multiplying the numbers of choices that can be made in each decision. m1 × m2 × m3 … Permutations • The number of permutations of m different items is m! • The number of permutations of m items of which: m! a are alike, b are alike, c are alike is: ​ ________ a! × b! × c! ​ • The number of permutations of m items taken n at a time, when each of the items may be repeated any number of times, is: m × m × m × m × … to n factors = mn times. • The number of ways that m items taken n at a time can be arranged, m! is mPn = ​ _____ (m–n)! ​ 168 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 Activity 6 1. At Angelo’s pizza place you can choose from 6 different types of pasta and 28 different sauces. How many different meals of 1 type of pasta and 1 type of sauce can you have? (2) 2. In how many different ways can we arrange 7 books on a shelf ? (2) 3. In how many different ways can 9 girls sit on one side of a table? (2) 4. In how many ways can a three-letter word be made from the letters c; d; e; f without repeating any letters? (3) 5. How many possible choices can be made in a multiple choice quiz if there are 4 questions each with 3 answers? (3) 6. How many different words can be made using the letters from LIMPOPO?(4) 7. How many 3-digit numbers can be made with the digits 1 – 5 if: a) repetitions are allowed (2) b) repetitions are not allowed (3) 8. A code is made using the format XYY, where the X is any letter in the alphabet and Y represents any digit from 0 to 9. a) How many possible codes can be formed if the letters and digits are repeated? (3) b) How many possible codes can be formed if the letters and digits are not repeated? (3) [27] Solutions 1. 6 × 28 = 168 different meals. 33(2) 2. 7! = 5 040 different ways 7 books can be arranged on a shelf. 33(2) 3. 9! = 362 880 different ways for 9 girls to sit on one side of a table. 33 (2) 4! 4! 4. 4P3 = _____ ​  (4–3)! ​= __ ​  1! ​= 24 ways a 3-letter word can be made from c; d; e; f with no repetition. 333(3) 5. 4 ‘boxes’ of 3 ∴ 3 × 3 × 3 × 3 = 34 = 81 possible choices. 333(3) 6. LIMPOPO m = 7; one L; one I; one M; two Ps; two Os. 7! ​ _______________    1! × 1! × 1! × 2! × 2! ​333 = 1 260 3(4) 7. a) 5 ‘boxes’ of 3 = 53 = 125 3-digit numbers (repetitions allowed) 33 (2) 5! 5! b) 5P3 = _____ ​  (5–3)! ​= __ ​  2! ​= 60 3-digit numbers (no repetition allowed) 333(3) 8. a) In first slot, we have 26 possible options (26 letters in the alphabet) In the second slot, we have 10 possible options (digits 0 to 10) In the third slot, we have 10 possible options (digits 0 to 10 – the digits may be repeated) ∴ 26 × 10 × 10 = 2 600 possible codes 333(3) b) In first slot, we have 26 possible options (26 letters in the alphabet) In the second slot, we have 10 possible options (digits 0 to 10) In the third slot, we have 9 possible options (the digits may be repeated) ∴ 26 × 10 × 9 = 2 340 possible codes 333(3) [27] © Department of Basic Education 2015 UNIT 8 probability 169 Mind the Gap Mathematics

8 Unit PAY SPECIAL ATTENTION NB NB 8.11 Use of counting principles in probability HINT! hint e.g. 18 EG - worked examples 1. What is the probability that a random arrangement of the letters of exams BAFANA starts and ends with an ‘A’? 2. A drawer contains 20 envelopes. 8 of the envelopes each contain 5 blue and 3 red sheets of paper. The other 12 envelopes each contain 6 blue and 2 red sheets of paper. One envelope is chosen at random. A sheet of paper is activity chosen at random from it. What is the probability that this sheet of paper is red? ACTIVITIES boy and girl-left and right of page activity Solutions 1. There are 6 letters: one B; 3 As; one F and one N. 6! Total number of arrangements of BAFANA = ____________ ​  1!    × 3! × 1! × 1! ​= 120 Step by step comment comment Word starts and ends with A: (A _ _ _ _ A): one B; one A; one F; one N (4 letters in the middle) 4! Number of middle arrangements = ____________ ​  1!    × 1! × 1! × 1! ​= 24 24 1 Probability of starting and ending with A = ___ ​  120 ​= __ ​  5 ​= 20% 2. Use a tree diagram: Red being chosen: ​ 8 __ ​ × ​  3 ___ ​ = ​  3 _ ​ = ​  24 __ ​ 20 8 160 20 3 ​ __​ 8 __ ​  20 ​ red 8 5 __ Envelope A ​  8 ​ Envelopes blue 2 24 3 __ ​ × __ Red being chosen: ​ 12 ​  8 ​= ___ ​  160 ​= __ ​  20 ​ blue __6 20 Envelope B ​  ​ 12 12 red 2 ​ __ 20 ​ ​ __ 12 ​ 3 3 6 Probability of a red sheet of paper being chosen = __ ​  20 ​+ __ ​  20 ​= __ ​  20 ​= 0,30 = 30% 170 UNIT 8 probability  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 8 What you need to be able to do: • Revise the addition rule for mutually exclusive events: P (A or B) = P (A) + P (B) • Revise the complementary rule: P(A′) = 1 – P (A) • Revise the identity P (A or B) = P (A) + P (B) – P (A and B) for all probability events. • Identify dependent and independent events and use the product rule • Use Venn diagrams to solve problems for up to three events. • Introduce an x for an event to solve problems. • Use tree diagrams and contingency tables for probability of consecutive events or simultaneous events which are not necessarily independent. • Understand and use Counting principles in probability. NOV 2013 P3 Q3, Q4, Q6 FEB/March 201 P3 Q4, Q5 and Q6 FEB/March 2012 P3 Q5, Q6 and Q7 NOV 2011 P3 Q3, Q5, Q6 FEB/March 2011 P3 Q3, Q5, Q6 NOV 2010 P3 Q1, Q5 Keep going! © Department of Basic Education 2015 UNIT 8 probability 171 Mind the Gap Mathematics

9 Unit Analytical Geometry Analytical geometry works with the Cartesian plane and with algebra to define points, lines and shapes. 9.1 Revise: Analytical Geometry This topic is also called Coordinate Geometry 1. Gradient of a line The gradient is the slope of a straight line. It shows y how steep the line is. The steeper the gradient, the bigger the angle it makes with the ground or the positive side of the x-axis. x y2 − y1 gradient m = ​ _____ x2 − x1 ​ OR change in y ​ _________ change in x ​ where (x1; y1) and (x2; y2) are two points on the line. m < o (negative gradient) m=o m > o (positive gradient) m is undefined y y y y x x x x 172 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 Parallel lines have equal gradients. The product of the gradients of lines that are perpendicular is –1. AB ∥ CD and mAB = mCD This means that the gradient of one line is the negative reciprocal of the gradient of the second line: y C AB ⊥ PQ A mAB × mPQ = − 1 y A x Q PAY SPECIAL ATTENTION D NB B NB x P HINT! B hint Note: The equation must always be in form y = mx + c e.g. 1 EG - worked examples 1. The graphs of y = 2x + 1 and y = 2x + 5 are parallel because they both have m = 2. exams 1 The graphs of y = 2x + 1 and y = − __ ​  2 ​x + 5 are perpendicular 1 __ because 2x − ​  2 ​= − 1 activity 2. The distance formula ACTIVITIES boy PAY SPECIAL ATTENTIONand girl-left and Learn the formularight of NB forpagedistance: NB ________________ activity Length of AB = ​√(x2 − x1)2 +    (y2 − y1)2 ​ You can also find the coordinates for a point on the line using the HINT! hint distance formula. comment comment Step by step e.g. 2 EG - worked examples 1. L(-5;-2) and M (-1;-6) are two sets of co-ordinates on the same exams straight line. Determine the length of LM _________________ ​ (x2 − x1)2 + LM = √    (y2 − y1)2 ​ ___________________ LM = √​ (− 5 + 1)2 +    (− 2 + 6)2 ​ _______ activity = ​√16 + 16 ​ ___ ACTIVITIES boy = ​√32 ​ and girl-left and right of page __ = 4 ​√2 ​ activity __ 2. The length of the straight line PQ is given as 2​√5 ​. The co-ordinates of P (5;2) and Q(3;t) are given. Find the value(s) of t. _________________ comment PQ = ​√(x2 − x1)2 + (y2 − y1)2 ​ Step by step comment    __ _______________ 2​√5 ​= ​√(5 − 3)2    + (2 − t)2 ___ ______________ ​ ​√20 ​= ​√4 + (4 −    4t + t2) ​ square both sides 20 = 8 − 4t + t 2 t2 − 4t − 12 = 0 (t − 6)(t + 2) = 0 t = 6 or t = −2 © Department of Basic Education 2015 UNIT 9 analytical geometry 173 Mind the Gap Mathematics

9 Unit Activity 1 1. For a line passing through the two points A(6; 6) and B(3; 2), calculate the length of AB. (3) 2. If PQ = 5 units; P (5; t) and Q (1; –3) determine the possible value(s) of t.(3)  [6] Solutions _________________ 1. Length AB = ​√(x2 − x1​)2​ ​+    (y2 − y1)2 ​3 _______________ = ​√(6 − 3)2 +    (6 − 2)2 ​ 3 ______ = ​√32 + 42 ​ ___ = ​√25 ​ =5 The length of AB is 5 units. 3 (3) __________________ 2. PQ = ​ (xQ – xP)2 + √   (yQ − yP)2 ​ _________________ 5 = ​√(1 − 5)2 +    (− 3 − t)2 ​ ________________ = ​√(− 4)2 + 9    + 6t + t2 ​ _____________ = √ ​ 16 + 9    + 6t + t2 ​ __________ = ​√t2 + 6t + 25 ​ 3 25 = t2 + 6t + 25 (square both sides) 0 = t2 + 6t 3 0 = t (t + 6) (factorise by taking out the HCF) t = 0 or t = −6 3 (both solutions are correct – plot the points to see why!) (3) [6] 3. The midpoint of a line A(x1;y1) B(x1;y1) 0 If you know the coordinates of the two endpoints of a line, you can find the point that is halfway between them. This is called the midpoint. The midpoint of a line has the coordinates (x + x y + y ) ​_____ ​  2 ​; _____ 1 2 1 2 ​  2 ​  ​ where (x1; y1) and (x2; y2) are the endpoints of the line. 174 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 9 e.g. 3 EG - worked examples For a line passing through the two points A(6; 6) and B(3; 2), find the exams coordinates of the midpoint of AB. PAY SPECIAL ATTENTION NB NB (x + x y + y ) Midpoint of AB = ​_____ ​  2 ​; _____ ​  2 ​  ​ 1 2 1 2 = ​(____ ​  2 ​  )​= ​(4​ __2 ​;4 )​ 6+3 6+2 1 HINT! ​  2 ​; ____ hint activity the midpoint has the coordinates ​(4​ __​; 4 )​ 1 So ACTIVITIES boy 2 and girl-left and right of page e.g. activity 4 EG - worked examples The coordinates of the midpoint of the line AB are (1;-4). Find the exams coordinates ofby step Step A if the comment coordinates of B are (4;-3). comment When midpoint is given: xA + xB yA + yB x as the midpoint of AB = ______ ​  2 ​and y as midpoint of AB = ______ ​  2 ​ xA + 4 activity yA − 3 1 = ​ _____ _____ 2 ​and –4 = ​  2 ​ ACTIVITIES boy 2 = xA + 4 and –8 = yAright and girl-left and −of3page –2 = xA and –5 = yA activity Coordinates of A are (-2;-5) comment Activity 2 Step by step comment K (–1; –6) and L (5; 4) are two coordinates on the same straight line. Determine the coordinates of the midpoint. (2) If M (–1; 4) is the midpoint of line segment AB, and the coordinates of A (3;6) are given, find the coordinates of the endpoint B. (3) [5] Solutions 1. Midpoint of KL = ​_____ ​  2 ​; _____ ​  2 ​  ​ (x + x y + y ) 1 2 1 2 (− 1 + 5 − 6 + 4 ) = ​______ ​  2 ​; ______ ​  2 ​  ​3 = (2 ; -1) 3 (2) 2. Let B have coordinates (xB; yB). (3 + x 6 + y ) (–1; 4) = ​_____ ​  2 ​; ​ _____ B 2 ​  ​3 B 3 + xB 6 + yB –1 = _____ ​  2 ​ and 4 = _____ ​  2 ​ (–1)(2) = 3 + xB (4)(2) = 6 + yB –2 = 3 + xB 8 = 6 + yB –5 = xB 3 2 = yB 3 ∴ the coordinates of B are(–5; 2). We can use coordinate geometry to identify the properties of geometric shapes on the Cartesian plane. (3) [5] © Department of Basic Education 2015 UNIT 9 analytical geometry 175 Mind the Gap Mathematics

9 Unit Activity 3 A. (–4; 7), B (4; 5), C (0; –1) and D (a; b) are the vertices of parallelogram ABCD. a) Draw the parallelogram on squared paper. (2) b) Find the midpoint of the diagonal AC. (2) c) Use the information that you have to find the coordinates of point D. (3) [7] Solutions a) A y 6 5 B 3 4 3 3 D 2 1 –8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 –1 –2 C (2) b) A (–4; 7) and C (0; –1) (x + x y + y ) (−4 + 0 7 − 1 ) Midpoint ​_____ ​  2 ​; _____ 1 ​  2 ​  ​= ​_____ 2 1 ​  2 ​; ____ 2 ​  2 ​  ​= (–2; 3) 3 3 So the midpoint of AC is (–2; 3) (2) c) Diagonals of parallelogram ABCD bisect each other ∴ midpoint of DB is (–2; 3). So midpoint (–2; 3) = ​____ (4 + a 5 + b ​  2 ​ )​3 ​  2 ​; ____ 4+a 5+b –2 = ____ ​  2 ​ and 3 = ​ ____ 2 ​ –4 = 4 + a and 6=5+b –8 = a 3 and 1=b3 ∴ Point D has coordinates (–8; 1) (3) [7] 176 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 9.2 The equation of a line You can find the equation of a straight line using y = m x + c, if you know the gradientPAYmSPECIAL and the y-intercept c. ATTENTION NB NB You can also find the equation of a straight line using y − y1 = m (x − x1), if you know the gradient m and any point (x1; y1) on the line, or if two points given. HINT! hint y and x are the coordinates of a specific point on the line. NOTE: 1 1 PAY SPECIAL ATTENTION NB NB e.g. 5 EG - worked examples If the gradient HINT! of a line is –2 and the line cuts the y-axis at 1, then the hint exams equation of the line is y = –2x + 1. e.g. 6 EG - worked examples activity of a line is –2 and the point (4; –1) lies on the line, find the If the gradient exams −y 1 andm (x − x1) ACTIVITIES boy equation of the line.y and girl-left= right of page y – (–1) = –2(x – 4) activity substitute (4; –1) into the equation y + 1 = –2x + 8 simplify y = –2xactivity +7 Step by step We usually comment put the answer in the form y = mx + c. comment ACTIVITIES boy and girl-left and right of page Summary activity If you know Formulae to use The gradient and the y-intercept y = mx + c Step by step comment comment The gradient and the coordinates of at least y − y1 = m (x − x1) one point on the graph. or y = mx + c y2 − y1 Two points on the line: first calculate the m = ​ _____ x2 − x1 ​and y = mx + c gradient and then substitute into y = mx + c. © Department of Basic Education 2015 UNIT 9 analytical geometry 177 Mind the Gap Mathematics

9 Unit Activity 4 1. Determine the equation of the straight line that passes through the points P(1; 2) and Q(3; 8) in the form y = ....(3) 2. Line AB is perpendicular to CD, which has a gradient of –2. The point (3; 4) lies on AB. Determine the equation of line AB. (2) [5] Solutions 1. First calculate the gradient of PQ: y2 − y1 8−2 ____ 6 m = ​ _____ __ x2 − x1 ​= ​  3 − 1 ​= ​  2 ​= 3 3 Then use the form y − y1 = m (x − x1) y − y1 = 3(x − x1) 3 Substituting P(1; 2) y − 2 = 3 (x − 1) y − 2 = 3x − 3 ∴ The equation of linePQ is y = 3x − 1. 3 (3) 2. mCD = –2 and CD ⊥AB. 1 ∴ mAB = __ ​  2 ​ 1 So now we have y = __ ​  2 ​x + c Substitute (3; 4) to find the value of c. 1 4 = ​ __2 ​(3) + c 3 c = 4 – 1​ __12 ​ 1 ∴ c = 2​ __2 ​ 1 1 ∴ equation of line AB is y = __ ​  2 ​x + 2​ __2 ​ 3 (2) [5] 178 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 9.3 The inclination of a line In trigonometry, you used the ratios tan θ, sin θ and cosθ. To find the inclination of a line, or the angle it makes with the x-axis, we use tan θ. opp BC In triangle ABC, tan θ = ​ ___ ___ adj ​ = ​  AC ​. BC change in y y ​  AC ​is also ​ _________ ___ change in x ​which is the gradient of AB. B We write gradient of AB as mAB. So we can say mAB = tan θ PAY SPECIAL ATTENTION NB NB Angle θ shows the slope or inclination of the line AB. Ɵ A θ is called the angle of inclination. C x HINT! hint θ ∈ (0 ;180 ) NOTE: 0 0 e.g. 7 EG - worked examples 1 __ 1 If tan θ = __ 2​  ​, then θ = 26,56505 …° (Press: shift tan ​  2 ​on your calculator) exams θ = 26,57 (rounded off to two decimal places) activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 9 analytical geometry 179 Mind the Gap Mathematics

9 Unit Activity 5 Give your answers correct to two decimal places. 1. Line AB is perpendicular to CD, which has a gradient of –4. Find the inclination θ of AB.  (2) 2. Determine the inclination of the straight line that passes through the points P (–6; 2) and Q (3; 10). (2) 3. Given the points A (–2; –1), B (5; 6) and C (7; –2), calculate the size     ^ of A​ B​C.(6) [10] Solutions 1 1. mCD = –4 and mAB⊥mCD, – 4 × __ ​  4 ​= –1 1 So mAB =​ __4 ​3 1 So tanθ = __ ​  4 ​= 0,25 and θ = 14,04° 3. (2) 2. P (–6; 2) and Q (3; 10) y2 − y1 10 − 2 8 So mPQ = _____ ​  x − x ​= ​ _______​= ​ __​3 3 2 1 3 − ( −6) 9 8 So tan θ = __ ​  9 ​ [To find θ, use 8 ÷ 9 = shift tan on your calculator] Angle of inclination: θ = 41,63° NOTE: (round off to 2 decimal places) (2) 3. Draw a rough sketch first. Place the triangle on the Cartesian plane. Use angles α and β y mAB = tan α. B (5; 6) 6+1 7 ∴tan α = ____ ​  5 + 2 ​= ​ __7 ​= 1 3 ∴α = 45° 3 (special angles) mBC = tan 3 − 2 − 6 −8 α β ∴ tanβ = ______ ​  7 − 5 ​= ​ ___ 2 ​= −4 3 x ∴ β = −75,963°… + 180° A (–2; –1) C (7; –2) = 104,04° 3     ^ A​ B​C = β−α (ext angle of ∆) = 104,04°− 45° = 59,04° 3 (6) [10] 180 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 Activity 6 1. For a line passing through the two points A(6; 6) and B(3; 2): 1.1 Calculate the length of AB. 1.2 Find the coordinates of the midpoint of AB. 1.3 Calculate the angle of inclination of the line. 1.4 Determine the equation of the line passing through A and B. 1.5 Determine the equation of a line GH perpendicular to AB passing through the midpoint of AB.  (11) 2. F, G, H and I are the vertices of a quadrilateral shown below. What kind of quadrilateral is FGHI? y 7 G (4; 6) 6 F (1; 5) 5 4 3 2 H (8; 2) 1 0 1 2 3 4 5 6 7 8 I (2; 0) (5)  3. ABCD is a quadrilateral with vertices A(5 ; 1), B(−3 ; 5), C(−1 ; −5) and D(9 ; −7). y B(–3; 5) A(5; 1) x O M C(–1; –5) D(9; –7) 3.1 Calculate the gradient of AC. (2) 3.2 Determine the equation of AC in the form y = … (3) 3.3 Hence, or otherwise, show that the midpoint M of BD lies on AC. (3)      ^ 3.4 Show that A​ M​B = 90°.(2) 3.5 Calculate the area of ∆ ABC. (5) [31] © Department of Basic Education 2015 UNIT 9 analytical geometry 181 Mind the Gap Mathematics

9 Unit Solutions _________________ _______________ 1. 1.1 Length AB = ​√(x2 − x1)2 +    (y2 − y1)2 ​= ​√(6 − 3)2 +    (6 − 2)2 ​ 3 ______ ___ ​√32 + 42 ​= ​√25 ​= 5 units 1.2 Midpoint coordinates: x1 + x2 3+6 1 _____ ​  2 ​= ____ ​  2 ​= 4​ __2 ​3 y1 + y2 6+2 1 _____ ​  2 ​= ​ ____ 2 ​= 4. So the midpoint has the coordinates (4​ __2 ​; 4) 2−6 −4 4 1.3 tan θ = mAB = ____ ​  3 − 6 ​= ___ ​  −3 ​= __ ​  3 ​3 ∴θ = 53,13° 3 4 1.4 mAB = __ ​  3 ​and you know the coordinates of A and B. Use y – y1 = m(x – x1) 4 y – y1 = __ ​  3 ​(x – x1) now substitute either point A or point B 3 4 y – 2 = __ ​  3 ​(x – 3) here point B has been substituted for (x1; y1) 3 4 ∴y = ​  34__​x – 2 3 y – 2 = __ ​  3 ​x – 4 1.5 AB⊥ GH ∴ mAB × mGH = –1 3 ∴ mAB = ​  4__3 ​so mGH = −​  43__​3 1 The midpoint of AB is (4​ __2 ​; 4) y – y1 = m(x – x1) 3 9 y – 4 = − ​ __4 ​(x − __ ​  2 ​) 3 3 27 y – 4 = − ​ __4 ​x + __ ​  8 ​ 3 3 y = − ​ __4 ​x + 3​ __8 ​+ 4 3 3 y = − ​ __4 ​x + 7​ __8 ​3 (11) 2. 6–5 mFG = ____ ​  4 – 1 ​= __ ​  3 ​3 1 mH1 = ​  82____ –0 2 __ 1 __ – 2 ​= ​  6 ​= ​  3 ​3 ∴ FG and HI are parallel. 0–5 –5 mF1 = ____ ​  4 – 1 ​= __ ​  –1 ​= 5 3 and mGH = ​  28 − ____ 6 − 4 ___ − 4 ​= ​  4 ​= − 1 3 so FI is not parallel to GH. ∴ FGHI is a trapezium (one pair of opp sides∥) 3 (5) 3. yC − yA substitution 3.1 mAC = ​ ______ xC − xA ​ 3 answer 3 −5−1 = ______ ​  − 1 − 5 ​(2) −6 = ___ ​  − 6 ​ Answer only: full marks =1 3.2 33 substitution y − y1 = m(x − x1) y − 1 = 1(x − 5) y = x − 4 3 equation  (3) 182 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 3.3 Midpoint of BD = ​_____ (x + x y + y ) ​  2 ​; _____ 2 1 ​  2 ​  ​ 2 1 (− 3 + 9 5 − 7 ) = ​______ ​  2 ​; ____ ​  2 ​  ​ = (3; − 1) 3 midpoint (3;-1) line AC is y = x − 4 y=3−4 y =  −1 3 substitution of M in the equation of ∴ M lies on AC. line AC 3 conclusion (3) 3.4 y2 − y1 MAM = ​ _____ x2 − x1 ​ 3 gradient of AM −1−5 = ​ ______ 3+3 ​ = −1 −1−1 and MMB = ​ ______ 3−5 ​ 3 gradient of BM =1 MAM × MMB = −1 MAM × MMB = −1       ^ ∴ A​ M​B = 90°.(2) _________________ 3.5 BM = ​√(5 + 1)2 +    (− 3 − 3)2 ​ 3 substitution into distance formula ___ BM = ​√72 ​ _______________ ___ AC = ​√(5 + 1)2 +    (1 + 5)2 ​ 3 BM = ​√72 ​ ___ ___ AC = ​√72 ​ 3 AC = ​√72 ​ ___ ___ 1 Area of ∆ABC = ​ __2 (​​ √ ​ 72 ​  )(​​ √ ​ 72 ​  )​ 3 formula for area of ∆ = 36 square units 3 answer (5) [31] © Department of Basic Education 2015 UNIT 9 analytical geometry 183 Mind the Gap Mathematics

9 Unit 9.4 Circles in analytical geometry A circle is made up of a set of points that are radius equidistant from its centre. The circumference is the distance around the whole circle. The distance from the centre to any point on the circumference of the circle is called the radius of the circle. 9.4.1 The equation of a circle CIRCLE WITH CENTRE AT THE ORIGIN y We can use the distance formula to work out P(x; y) the equation of a circle with centre (0; 0). r NB y) is any point on the circle C(0; b) x PAY SPECIAL ATTENTION If P(x; NB with radius r, then _______________ • r = ​√(x − 0)2 +    (y − 0)2 ​ HINT! hint r2 = x2 + y2 e.g. 8 EG - worked examples Find the equation of a circle centre 0 with the point P(5; 2) on its exams circumference. x2 + y2 = r2 This is the general equation. We just need a value for r2. (5)2 + (2)2 = r2 At the point (5; 2) r2 = 25 + 4 activity = 29 ∴ x2 + y2 = 29 ACTIVITIES boy and girl-left and right of page activity CIRCLES NOT CENTRED AT THE ORIGIN y P(x; y) If we move the centre of the circle to any point r on the Cartesian Step by step plane C(a; comment b),comment C(a; b) then (x – a)2 + (y – b)2 = r2 ______________ x and r = ​√(x − a)2 +    (y − b)2 ​ Note: Recap on completing of the square on unit 2 184 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 9 e.g. 9 EG - worked examples The equation of a circle is (x + 1)2 + (y – 3)2 = 16. exams Determine the coordinates of the centre and the length of the radius. The equation is already in the form (x – a)2 + (y – b)2 = r2, with a = –1, b = 3 and r2 = 16 ___ So the activity centre is (–1; 3) and the radius is ​√16 ​= 4. ACTIVITIES boy and girl-left and Remember that the right radius of page can only be a positive number because it is a length.activity Activity 7 Step by step comment comment 1. Determine the coordinates of the centre and the length of the radius if a circle has the equation: x2– 2x + y2+ 10y = –14 (3) 2. Determine the equation of a circle with centre C(–1; –2) and passing through the point B(1; –6).  (3) [6] Solutions 1. To get the equation in the form (x – a)2 + (y – b)2 = r2, we need to add in numbers to complete the squares using x2 with –2x and y2 with 10y. (x2– 2x) + (y2+ 10y) = –14 (x2– 2x + 1) + (y2+ 10y + 25) = –14 + 1 + 25 3 (x – 1)2 + (y + 5)2= 12 3 ___ ____ __ ​ 12 ​= √ So the centre is the point (1; –5) and the radius is √ ​ 22.3 ​= 2​√3 ​3 (3) 2. First find the value of r2: r2 = (x – a)2 + (y – b)2 r2 = (x + 1)2 + (y + 2)2 3 Substitute B(1; –6) r2 = (1+ 1)2 + (–6 + 2)2 3 r2 = (2)2 + (–4)2 r2 = 4 + 16 =20 ∴20 = (x + 1)2 + (y + 2)2 3 (3) [6] © Department of Basic Education 2015 UNIT 9 analytical geometry 185 Mind the Gap Mathematics

9 Unit 9.4.2 The equation of a tangent to a circle A tangent is a straight line which touches a A circle at one point only. So ADB is a tangent, but AP is not a tangent. D A tangent to a circle at any point on the circumference is perpendicular to the radius at that point.PAY So AB ⊥ CD. SPECIAL ATTENTION NB NB P B We can use all the formulae we know from analytical geometry to solve problems involving a tangent to a circle (distance, midpoint, HINT! gradient, angle of inclination, the equation of a line and the equation of a circle). hint e.g. 10 EG - worked examples Find the equation of the tangent APB which touches a circle centre C with exams equation (x – 3)2 + (y + 1)2= 20 at P(5; 3). Solution Draw a sketch to help you. activity A Centre of circle is C(3; –1) so the gradient ACTIVITIES boy of the radius CP (mCPand )rightgirl-left of page and 3 − ( −1) activity is ​ _______ 5 − 3 ​= 2. P (5; 3) radius ⊥ tangent, so mAPB × mCP = –1 and so C mAPB = – __​  12 ​Step by step comment comment Equation of tangent: y – y1 = m(x – x1) B y – 3= – ​  1__2 ​(x – 5) P is a point on the tangent y – 3= – ​  1__2 ​x + 2​  1__2 ​ y = – ​  1__2 ​x + 5​  1__2 ​ 186 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 Activity 8 1. ABCD is a quadrilateral with vertices A(–3; 0), B(–1; –3), C(2; –1) and D(0; 2). y D(0; 2) A(–3; 0) θ x 0 C(2; –1) B(–1; –3) 1.1 Determine the coordinates of M, the midpoint of AC. (2) 1.2 Show that AC and BD bisect each other. (3) ^ 1.3 Prove that A​ |D​C = 90°. (4) 1.4 Show that ABCD is a square. (4) 1.5 Determine the size of θ, the angle of inclination of DC, correct to ONE decimal place. (3) 1.6 Does C lie inside or outside the circle with centre (0; 0) and radius 2? Justify your answer. (2) [18] Solutions 2−3 −1+0 ____ ______ 1 1 1.1 Midpoint M of AC: ​  2 ​; ​  2 ​= ( − ​ __2 ​; − ​ __2 ​) 33 (2) 1.2 Midpoint M of BD: (− 1 + 0 − 3 + 2 ) ​______ ​  2 ​; ______ 1 1 ​  2 ​  ​= ( − ​ __2 ​; − ​ __2 ​) 33 ∴ So the midpoint of AC and the midpoint of BD are the same point, so they bisect each other. 3 (3) 2−0 2 −3 1.3 mAD= _______ ​  0 − ( − 3) ​= ​ __3 ​3 and mDC= ​  −21 − 2 ___ ______ − 0 ​= ​  2 ​3 2 −3 PAY SPECIAL ATTENTION mADNB × mDC = __ ​  3 ​× ​ ___ 2 ​= –1 3 NB ∴AD ⊥ DC      ^ ∴A​ D​C = 90° 3 (4) 1.4 There HINT! are several ways to prove that ABCD is a square: hint • Prove that diagonals are equal and bisect each other at 90° • Prove that ABCD is a rectangle and has a pair of adjacent sides equal. e.g. EG - worked examples • Prove that all four sides are equal and that one internal angle is 90°. Here is one possible answer: exams The diagonals AC and BD bisect each other (proved in 1.2)      ^ A​ D​C = 90° (proved in 1.3) 3 AD2 = (2 – 0)2 + (0 – (–3))2 = 4 + 9 = 13 3 © Department of Basic Education 2015 UNIT 9 analytical geometry 187 activity Mind the Gap Mathematics ACTIVITIES boy

9 Unit ___ AD = ​√13 ​ CD2 = (– 1 – 2)2 + (2 – 0)2 = 9 + 4 = 13 3 ___ CD = ​√13 ​ So adjacent sides are equal in length 3 ∴ ABCD is a square. (4) −1−2 3 1.5 tan θ = mDC = ​ ______ __ 2 − 0 ​= − ​  2 ​3 θ = –56,3099324... + 180° 3 θ = 123,7° 3 (3) 1.6 OC2 = (2 – 0)2 + (– 1 – 0)2 OC2 = 4 + 1 = 5 3 __ OC = ​√5 ​ This is longer than the radius of 2 of the given circle, so C(2; –1) lies outside the circle 3. (2) [18] 2. O is the centre of the circle in the figure below. P(x; y) and Q(12; 5) are two points on the circle. POQ is a straight line. The point R(t; –1) lies on the tangent to the circle at Q. y Q(12; 5) x 0 R(t; –1) P(x; y) 2.1 Determine the equation of the circle. (3) 2.2 Determine the equation of the straight line through P and Q. (2) 2.3 Determine x and y, the coordinates of P. (4) 12 __ 2.4 Show that the gradient of QR is –​  5 ​.(2) 2.5 Determine the equation of the tangent QR in the form y = … (3) 2.6 Calculate the value of t.(2) 2.7 Determine an equation of the circle with centre Q(12; 5) and passing through the origin. (3) [19] 188 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 9 Solutions The centre is at the origin, so x2 + y2 = r2. 2.1 OQ2 = (5)2 + (12)2 = 25 + 144 = 169 33 So the equation of the circle is x2 + y2 = 169 3 (3) 0–5 5 2.2 mPQ = mOQ = _____ ​  0 – 12 ​= __ ​  12 ​3 PQ has y-intercept of 0. 3 (2) 5 y = __ ​  12 ​x 2.3 By symmetry, P is the point (–12; –5). 33 OR 5 Substitute y = __ ​  12 ​x into x2 + y2 = 169 x2 + ​​(​  12 5 ​x )​​ = 169 2 __ ​ 25 x2 + ___ ​  144 ​x2 = 169 144x2 + 25x2 = 169 × 144 169x2 = 24 336 x2 = 144 x = 12 or x = –12 x = –12 according to given diagram 3 5 ​x = ___ 5 y = ​  12 __ ​  ​× (–12) = –5 3 12 (4) So P is the point (–12; –5). 2.4 tangent ⊥ radius so QR ⊥ PQ 3 mPQ = ​  00−− _____ 5 __5 12 ​= ​  12 ​ − 12 ∴mQR = ​ ____ 5 ​3 (2) –12 2.5 y = ___ ​  5 ​x + c 3 y – y1 = ​  –12 OR ___ 5 ​(x – x1) 3 Substitute Q(12; 5) into equation to find c: –12 ​  5 ​(12) + c 3 5 = ___ y – 5 = ​  –12 ___ 5 ​(x – 12) 3 144 ​  5 ​= c 5 + ___ y = ​  –12 ___ 144 5 ​x + ​  5 ​+ 5 ___ 169 169 c = ___ ​  5 ​3 y = –​  12 __ 5 ​x + ​  5 ​3 ___ –12 169 y = ___ ​  5 ​x + ___ ​  5 ​ (3) −12 169 2.6 R(t; –1) lies on line with equation y = ____ ​  5 ​x + ___ ​  5 ​ −12 169 ​  5 ​t + ___ ∴ –1 = ____ ​  5 ​3 –5 = –12t + 169 12t = 174 t = 14,5 3 (2) 2.7 OQ2 = (x – 12)2 + (y – 5)2 3 3 Q(12; 5) is centre of circle Substitute (0; 0) into equation: OQ2 = (0 – 12)2 + (0 – 5)2 OQ2 = 144 + 25 = 169 3 ∴ (x – 12)2 + (y – 5)2 = 169 (3) [19] © Department of Basic Education 2015 UNIT 9 analytical geometry 189 Mind the Gap Mathematics

9 Unit What you need to be able to do: From Grade 10 and 11: • Find the distance between any two points on the Cartesian plane using the distance formula: _________________ • Distance = ​√​(​x​2​− ​x​1​)​2​+    ​(​y​2​− ​y​1​)​2​ ​ • Find the midpoint between two points on a line, using the (​x​ ​+ ​x​ ​ ​y​ ​+ ​y​ ​) formula ​_____ ​  2 ​; _____ 1 2 1 ​  2 ​  ​. 2 ​y​2​− ​y​1​ • Find the gradient of line using m = _____ ​  ​x​2​− ​x​1​​ • Find the equation of a line given: • The gradient and the y-intercept using y = mx + c • The gradient and the coordinates of at least one point on the graph. • You can use y – y1 = m(x – x1) • Two points on the line: first calculate the gradient, then substitute one of the points into y = mx + c. • Find the inclination θ of a line, where m = tan θ. • Find other angles, using geometry. From Grade 12: • Determine the equation of a circle with radius r and centre (a; b). • Determine the equation of a tangent to a circle centre (a; b) • Know the properties of triangles (isosceles, scalene, equilateral, right- angled triangle);square, rectangle, trapezium, rhombus and parallelogram. Keep going! 190 UNIT 9 analytical geometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 Trigonometry 10.1 Revise: Trig ratios Trigonometry is the study of the relationship between the sides and angles of triangles. The word trigonometry means ‘measurement of triangles’. The trigonometric ratios Using θ as the reference angle in ∆ABO A • The side opposite the 90° is the hypotenuse side, therefore side AO is the hypotenuse side. hypotenuse • The side opposite θ is the opposite side, therefore AB is the opposite side. opposite • The side adjacent to θ is called the adjacent side, therefore OB θ is the adjacent side. O B adjacent We work with the ratios of the sides of the triangle: opposite • The ratio _________ ​ hypotenuse ​is called sine θ (abbreviated to sin θ) adjacent • The ratio _________ ​ hypotenuse ​is called cosine θ (abbreviated to cos θ) opposite • The ratio _______ ​ adjacent ​is called tangent θ (abbreviated to tan θ) adjacent: the side adjacent opposite side Therefore sin θ = __________ ​  hypotenuse ​= AB/AO opposite: hypotenuse: opposite: to the angle θ the side the side opposite the side opposite θ the right angle opposite θ θ adjacent side cos θ = __________ ​  hypotenuse ​= OB/AO θ hypotenuse: opposite side tan θ = ​ __________ adjacent: the side opposite adjacent side ​= AB/OB the side adjacent to the angle θ the right angle © Department of Basic Education 2015 UNIT 10 trigonometry 191 Mind the Gap Mathematics

10 Unit y We can also place the same triangle on the Cartesian plane in standard position, with a vertex at the origin and one side on A(x; y) the x-axis like this: r • On the Cartesian plane, A is the point (x; y).     ^ • The angle A​ O​B or θ is positive (we rotate in an anti-clockwise θ direction) 0 B (x; 0) x • The length of OB is x units and the length of AB is y units. • We can find the length of AO, using the Theorem of Pythagoras. In ∆ABO, AO2 = AB2 + OB2 PAY SPECIAL ATTENTION NB AO = x + y 2 2 2 NB r2 = x 2 + y 2 IAL ATTENTION HINT! NB hint NOTE: The Theorem of Pythagoras Look at the circle centre O. AO is also a radius of this circle. In any right-angled triangle, y the square on the hypotenuse e.g. Now we can name the examples EG - worked trigonometric ratios A(x; y) is equal to the sum of the in terms of x, y and r. squares on the other two y r sides. • The ratio _​ r ​is called sin θ θ x ed examples •exams The ratio _​ r ​is called cos θ 0 B (x; 0) x y • The ratio _​ x ​is called tan θ PAY SPECIAL ATTENTION NB activity NB PAYLearn SPECIALthese ratios: ATTENTION NBy opposite ACTIVITIES boy NB y θ = _​ x adjacent opposite sin θ = ​ _r ​= ________ ​  ​ cos and ​= ________ girl-left and right ofrpage ​  ​ tan θ = _​ x ​= ______ ​  ​ hypotenuse hypotenuse adjacent activity HINT! hint HINT! y A(x; y) ACTIVITIES boy and girl-left and hint Remember the word SOHCAHTOA: right of page r SOH C A HEG - worked examples TOA e.g.​ O ​ sin θ = __ Step by step cos θ = ​ A__​ comment O comment tan θ = ​ __ ​ θ B (x; 0) x e.g. H EG - worked H examples A exams ep comment comment exams activity activity ACTIVITIES boy and girl-left and right of page ACTIVITIES boy and girl-left and activity right of page activity 192 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 Activity 1 1. ∆MNP is a right-angled triangle. Write down the trig ratios for: a) sin α b) sin β N (4) c) tan β d) cos α (3) 2. If MP = 13 and NP = 5, calculate cos β. α β P M [7] Solutions MN NP 1. a) sin α = ____ ​  MP ​3 (1) b) sin β = ___ ​  MP ​3 (1) NP NP c) tan β = ____ ​  MN ​3 (1) d) cos α = ​ ___ MP ​3 (1) (4) 2. MP = 13 and NP = 5, so we can find MP, MP2 = MN2 + NP2 ……….Pythagoras 3 13 = MN + 5 2 2 2 169 = MN2 + 25 MN2 = 169 – 25 MN2 = 144 3 ∴MN = 12 MN 12 cos β = ____ ​  MP ​= __ ​  13 ​3 (3) [7] A O B PAY SPECIAL ATTENTION –45° PAY SPECIAL ATTENTION NB NB NB Angles measured in an anti- Angles measured in an anti- 45° clockwise direction from the A clockwise direction from the O HINT! B x-axis are positive x-axis are negative. HINT! hint hint ∴angle is negative e.g. EG - worked examples e.g. EG - worked examples exams exams activity activity ACTIVITIES boy ACTIVITIE and girl-left and and girl-l right of page right of p activity activity © Department of Basic Education 2015 UNIT 10 trigonometry 193 Mind the Gap Mathematics

10 Unit 10.2 Trig ratios in all the quadrants of the Cartesian plane PAY SPECIAL ATTENTION NB NB plane has four quadrants (quarters). We label them 1, 2, The Cartesian 3 and 4 starting from the quadrant with positive x- and y-values. We can calculate trig ratios for any angle size in the Cartesian plane. HINT! 90° 90° hint y CAST 2 1 2 S A 1 sin θ +ve All ratios +ve For trig ratios that are positive All other ratios –ve e.g. in the 4 quadrants.EG - worked examples 180° 0° 180° 0° x 360° x tan θ +ve cos θ +ve All other ratios –ve 3 All other ratios –ve 4 3 T C 4 exams 270° 270° • In the first quadrant x, y and r are positive. Therefore, all the trig functions are positive. • In the second quadrant, y and r are positive, therefore sin θ is positive. In the second quadrant, x is negative, therefore cos θ and activity tan θ are negative. • boy ACTIVITIES In the third quadrant, x and y are negative, therefore tan θ is positive. andthe third quadrant, r is positive, therefore cos θ and sin θ are and girl-leftIn right of page negative activity • In the fourth quadrant, x and r are positive, therefore cos θ is positive. In the fourth quadrant, y is negative, therefore sin θ and tan θ are negative. Step by step comment comment 194 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 Activity 2 1. If sin θ is negative and cos θ is positive, then which statement is true? A. 0° < θ < 90° B. 90° < θ < 180° C. 180° < θ < 270° D. 270° < θ < 360°  (1) 2. If tan θ < 0 and cos θ < 0, then which statement is true? A. 0° < θ < 90° B. 90° < θ < 180° C. 180° < θ < 270° D. 270° < θ < 360° (1) 3. Will the following trig ratios be positive or negative? a) sin 315° b) cos (–215°) c) tan 215° d) cos 390° (4) [6] Solutions 1. Sin θ is negative in 3rd and 4th quadrants; cos θ is positive in 1st and 4th quadrants. So θ is in the 4th quadrant. D. 270° < θ < 360° 3 (1) 2. tan θ < 0 in 2nd and 4th quadrants; cos θ < 0 in 2nd and 3rd quadrants. So θ is in the 2nd quadrant. B. 90° < θ < 180° 3 (1) 3. a) sin 315° is in 4th quadrant so it is negative. 3 (1) b) cos (–215°) is in 2nd quadrant so it is negative. 3 (1) c) tan 215° is in 3rd quadrant, so it is positive. 3 (1) d) cos 390° is the same as cos 30° in the 1st quadrant, so it is positive. 3 (1) [6] © Department of Basic Education 2015 UNIT 10 trigonometry 195 Mind the Gap Mathematics

10 Unit 10.3  NB Solving NB triangles with PAY SPECIAL ATTENTION trig HINT! For some trigonometry problems, it is helpful to draw a diagram showing hint the angle involved and the x, y and r values. e.g. 1 EG - worked examples __ If tan θ = –​√3 ​and 180° < θ < 360°, determine, using a diagram, the value exams of: a) sin θ b) 3 cos θ Solutions activity __ y –​√3 ​ a) tan θ = _​  x ​= ___ ​  1 ​. ACTIVITIES boy and girl-left and right of page 180° < θ < 360° and tan θ is negative in the 4th quadrant activity By Pythagoras, r2 = x2 + y2 __ r2 = (1)2 + (–​√3 ​)2 r2 = 1 +Step3by=step4 comment comment r=23 __ θ y –​√3 ​ ∴ sin θ = _​  r ​= ___ ​  2 ​3(4) b) 3 cos θ __ x 1 3 (1; –​√3 ​) 33 = 3(​ _r ​) = 3(​ __2 ​) 3 = __ ​  2 ​= 1,5 3 (2) [6 ] Activity 3 p If cos β = ___ ​  __ ​where p < 0 and β ∈ [180°; 360°], determine, using a ​√5 ​ diagram, an expression in terms of p for: a) tan β b) 2 cos2β – 1 [6] Solutions p x __ ​  √__ ​= _​  r ​; a) cos β = ___ so x = p and r = ​√5 ​ p ​ 5​ By Pythagoras, y2 = r2 – x2 β __ ∴ y2 = ​​(​√5 ​  )2​​ ​– p2 __ = 5 – p2 ​√5 ​ 33 _____ ∴ y = ± ​√5 – p2 ​. _____ ∴ y = –​√5 – p2 ​3 since β is in quadrant 3, y is negative _____ –​√5 – p2 ​ ​  p ​3 (4) ∴ tan β = ______ p (​ 5 ​) 2 b) 2 cos2β – 1 = 2 ​​ ___ ​  √__ ​  ​​ ​– 1 3 2p2 = ___ ​  5 ​– 1 3 (2) [6] 196 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.4 Using a calculator to PAY SPECIAL ATTENTION find trig ratios NB NB The scientific HINT! calculator calculates trigonometric ratios as decimal hint fractions. e.g. 2 EG - worked examples 1. sin 58° = 0,8480480962…. [Press: sin 58 =] exams 2. cos 222° = –0,7431448255… [Press: cos 222 =] 3. Calculate (correct to 2 decimal places): cos 238° tan 132° = 0,5885349 … ≈ 0,59 (to 2 decimal places) activity [Press: cos 238 × tan 132 =] sin _______2 327 4. ​  5+tan 37 ​= 0,05155and… ACTIVITIES ≈ 0,052 boy girl-left and [NOTE: sin2 327° = (sin 327°)2] right of page 1 5. sin 30° = ​ __2 ​ activity Step by step comment comment © Department of Basic Education 2015 UNIT 10 trigonometry 197 Mind the Gap Mathematics

10 Unit 10.5 The trig ratios of special angles 1. Special trig ratios using the unit circle Consider a circle on the Cartesian plane that has a radius of one unit. We can find the trig ratios for 0° (or 360°), 90°, 180° and 270° using the unit circle. Label the (x; y) coordinates on each axis. Label the angles on each axis. y (0; 1) 90° (–1; 0) r=1 (1; 0) 180° 360° 270° (0; –1) From the unit circle: • At 0 or 360°: x = 1, y = 0 and r = 1 • At 90°: x = 0, y = 1 and r = 1 • At 180°: x = –1, y = 0 and r = 1 • At 27°: x = 0, y = –1 and r = 1 0 1 sin 0° = __ ​ 1 ​= 0 sin 90° = ​ __ 1 ​= 1 1 0 cos 0° = ​ __ 1 ​= 1 cos 90° = ​ __ 1 ​= 0 0 1 tan 0° = ​ __ 1 ​= 0 tan 90° = ​ __ 0 ​is undefined 0 –1 sin 180° = ​ __ 1 ​= 0 sin 270° = ​ ___ 1 ​= –1 –1 0 cos 180° = ​ ___ 1 ​= –1 cos 270° = ​ __ 1 ​= 0 0 −1 tan 180° = ​ ___ –1 ​= 0 tan 270° = ​ ___ 0 ​is undefined CIAL ATTENTION NB Summary θ 0° 90° 180° 270° 360° sin θ 0 1 0 –1 0 cos θ 1 0 –1 0 1 tan θ 0 undefined 0 undefined 0 ked examples 198 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 2. Special trig ratios using an equilateral triangle We use an equilateral triangle that has sides of 2 units to find the trig ratios for the special angles 30° and 60°. The perpendicular bisector of one side creates two triangles. The angles of an equilateral triangle are P P equal, so angles P, Q 30° and R are each 60°. 30° 30° P is bisected, so ​  ^ QPS​ 2 2 2 _ ^ ➡ ​√3 ​ = ​ RPS​= 30°. By Pythagoras, Q 60° 60° R Q 60° S S 1 PR2 = PS2 + RS2 2 22 = PS2 + 12 PS2 = 4 –__1 = 3 ∴ PS = ​√3 ​ Now we can use ∆PQS to work out trig ratios of 30° and 60°. PAY SPECIAL ATTENTION NB √ __ NB ​ ​ 3 ​​ sin 60° = ___ sin 30° = _​ 1 ​ 2 2 __ √ cos 30° = ​  3 ​​ ​___ cos 60° = ​ _1 ​ 2 2 __ __ √ 1__ ​ tan 60° = ​ ​___ 3 ​ √ HINT! tan 30° = ​ ___ ​= ​ 3 ​ hint √ ​ 3 ​ 1 3. Special trig ratios using a right-angled isosceles triangle e.g. EG - worked examples Use a right-angled isosceles triangle with A sides of one unit to work out the trig ratios for 45°. The angles opposite the equal 45° sides are equal, so they are each 45° (sum __ examsin ∆). of angles 1 ​√2 ​ By Pythagoras, 45° AC2 = AB2 + BC2 B C 1 AC2 = 12 + 12 AC2 = 1 +__1 = 2 ∴ AC = ​√2 ​activity __ The hypotenuse will be ​√2 ​ units.ACTIVITIES boy and girl-left and right of page NB activity ECIAL ATTENTION __ __ √ √ ​  1__ ​ = ​ ​___ sin 45° = ___ 2 ​ 1__ ​ = ​  ​___ cos 45° = ​ ___ 2 ​ tan 45° = ​ _1 ​= 1 ​√2 ​ 2 ​√2 ​ 2 1 Step by step comment comment orked examples © Department of Basic Education 2015 UNIT 10 trigonometry 199 Mind the Gap Mathematics

10 Unit Summary of special angles y You should memorise the special angles, (0; 1) 90° as you will use them often. You will be asked exam questions where you are not allowed to use a calculator, and must show that you have used the special angles. (–1; 0) r=1 (1; 0) If you just remember these three diagrams, 180° 360° x you can work out all of the special angles. 270° (0; –1) P A 30° 45° 2 _ 1 __ ​√3 ​ ​√2 ​ 60° 45° Q S B C 1 1 If you find it difficult to remember the diagrams, then learn this summary of the special angles. θ 30° 45° 60° __ __ sin θ 1 ​√2 ​ ​√3 ​ ​ __ 2​ ___ ​  2 ​ ​ ___ 2​ __ __ cos θ ​√3 ​ ___ ​√2 ​ 1 ​  2 ​ ​ ___ 2​ __ ​  2 ​ __ tan θ ​√3 ​ __ ___ ​  3 ​ 1 ​√3 ​ You can also use a scientific calculator to find these special angle ratios. 200 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.6 Using reduction formulae Look at the angles here. If θ < 90°, it is in the first quadrant, therefore θ is an acute angle. Quadrant I Quadrant II Therefore y y • angle (180° – θ) in quadrant II • angle (180° + θ) in the quadrant III r • angle (360° – θ) in quadrant IV. r y y 180 – θ You can work out which trig ratios will θ θ be positive and which will be negative, x x x x according to the quadrants they are in. Quadrant III Quadrant IV y y 360 – θ 180 + θ x x θ x θ x y y r r a) Reduction formulae y III Quadrant II: 180° – θ Quadrant III: 180°+ θ Quadrant IV: 360° – θ S A sin(180° – θ) = sin θ sin(180° + θ)= –sin θ sin(360° – θ) = –sin θ sin all cos(180° – θ) = –cos θ cos(180° + θ) = –cos θ cos(360° – θ) = cos θ x PAY SPECIAL ATTENTION NB T C tan(180° – θ) = –tan θ tan(180° + θ) = tan θ tan(360° – θ) = –tan θ tan cos III IV b) Angles greater than 360° We can add or subtract 360° (or multiples of 360°) and will always HINT! end up with an angle in the first revolution. For example, 390° can hint be written as (30° + 360°), so 390° has the same terminal arm as When you divide, you 30°. sometimes need to round off c) Negative angles: • (–θ) lies in quadrant IV and is the same as 360° – θ. e.g.to the closest EG numbers that - worked are easier to divide. examples sin(–θ) = –sin θ cos(–θ) = cos θ tan(–θ) = –tan θ PAY SPECIAL ATTENTION NB • (θ–180)lies in the third quadrant exams sin(θ –180) = -sin θ cos(θ–180) = –cos θ tan(θ–180) = tan θ • (–θ–180)lies in the second quadrant HINT! sin(–θ –180) = sin θ cos(–θ–180) = –cos θ tan(–θ–180) = –tan θ hint • (θ–360) lies in the first quadrant When youactivity divide, you sometimes need to round off ACTIVITIES boy sin(θ– 360) = sin θ cos(θ–360) = –cos θ tan(θ–360) = –tan θ e.g.to the closest numbers that EG - worked are easier to divide. examples and girl-left and right of page activity exams © Department of Basic Education 2015 UNIT 10 trigonometry 201 Mind the Gap Mathematics Step by step comment

HINT! 10 Unit hint e.g. 3 EG - worked examples sin (360˚+ θ) = sin θ cos (360˚+ θ) = cos θ tan (360˚+ θ) = tan θ exams Activity 4 activity Without using a calculator, determine the value of: 1. cos 150° 2. and sin (–45°) ACTIVITIES boy and girl-left 3. tan 480° right of page [7] activity Solutions 1. cos 150°Step by step comment rewrite as (180 – ?) comment = cos(180° – 30°) quadrant II, cos θ negative = –cos __ 30° 3 special ratios ​√3 ​ = – ​ ___ 2 ​3 (2) 2. sin(–45°) sin(–θ) = –sin θ; quadrant IV, sin θ negative = –sin 45° 3 special ratios 1__ = – ​ ___ √ ​3 (2) ​ 2​ 3. tan 480° write as an angle in the first rotation of 360° = tan (480° – 360°) = tan 120° 3 quadrant II, rewrite as (180 – ?) = tan (180° – 60°) tan θ negative = –tan __ 60° 3 special ratios = –​√3 ​3 (3) [7] d) Co-functions     ^ In this right-angled triangle, the sides are a, b and c and ​ B​ = θ. A = 90° and angles of a triangle are supplementary     ^ ∴​ C​= (90° – θ). Look at the sine and cosine ratios for the triangle: b b sin θ = ​ __a ​ and cos (90° – θ) = ​ __a ​ 90° – θ ∴ cos (90° – θ) = sin θ c c cos θ = ​ __a ​ and sin (90° – θ) = ​ __a ​ a b ∴ sin (90° – θ) = cos θ Trig ratios of angles that add up to 90°, are θ called co-functions. c 202 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 PAY SPECIAL ATTENTION NB NB sin (90˚– θ) = cos θ (quadrant I) sin (90˚+ θ) = cos θ HINT! (sin θ positive in quadrant II) sin (θhint – 90˚) = sin[–(90° – θ)] (common factor of –1) = –sin(90° – θ) (sin θ negative in quadrant IV) = –cos θ e.g. cos (90˚– θ) = sin θ (quadrant I) EG - worked examples cos (90˚+ θ) = –sin θ (cos θ negative in quadrant II) cos (θ – 90˚) = cos[–(90° – θ)] (common factor of –1) = +cos(90° – θ) exams (cos θ positive in quadrant IV) = +sin θ Activity 5 activity Write the trig ratios as the trig ratios of their co-functions: ACTIVITIES boy and girl-left and 1. sin 50° 2. cos 70° 3.right ofsin page100° 4. cos 140° [4] activity Solutions 1. sin 50° = sin(90° – 40°) = cos 40° 3 2. cos 70° = cos(90°Step– by20°) step = sin 20° 3 comment comment 3. sin 100° = sin(90° + 10) = cos 10° 3 4. cos 140° = cos(90° + 50°) = –sin 50° 3 [4] Summary Any angle (obtuse or reflex) can be reduced to an acute angle by using: • Convert negative angles to positive angles • Reduce angles greater than 360° • Use reduction formulae • Use co-functions © Department of Basic Education 2015 UNIT 10 trigonometry 203 Mind the Gap Mathematics

10 Unit Activity 6 Simplify without using a calculator: sin(180° + x). cos 330°.tan 150° _______________________ 1. ​      ​ (4) sin x cos 750°.tan 315°.cos(–θ) 2. ​ _________________________    ​     cos(360°– θ).sin 300°.sin(180°– θ) (8) tan 480°.sin 300°.cos 14°.sin(–135°) 3. ​ __________________________       ​ sin104°.cos 225° (9) cos 260°.cos 170° 4. ​ __________________   ​    sin10°.sin 190°.cos 350° (7) [28] Solutions sin(180° + x).cos 330°. tan 150° 1. _______________________ ​       sin x ​ reduction formulae in numerator 3 3 3 (–sin x)(+ cos 30°)(–tan 30°) = _____________________ ​      sin x ​ (use brackets to separate ratios) __ __ ​√3 ​ ​√3 ​ + sin x. ___ ​  2 ​3. ___ ​  3 ​ _____________ = ​   sin x ​ special angles __ __ ​√3 ​ ​√3 ​ = ___ ​  2 ​.___ ​  3 ​ 3 1 = __ ​  6 ​= __ ​  2 ​ (4) cos 750°.tan 315°. cos(–θ) 2. __________________________ ​        cos(360°– θ). sin 300°. sin(180° – θ) ​ use reduction formulae cos30° 3. (–tan 45°) 3. cosθ 3 = _______________________ ​  cosθ        3. (–sin60°) 3. sinθ 3 ​ use special angles __ ​√3 ___ ​ ​  2 ​.(–1)cos θ ____________ = ​     __ ​3 cosθ.​– ___ ( ​√3 ​ ​  2 ​  ​sinθ ) –1 1 = ____ ​  –sinθ ​= ____ ​  sinθ ​3 (8) tan 480°. sin 300°. cos 14°. sin(–135°) cos 260°. cos 170° 3. ___________________________ ​       sin 104°. cos 225° ​ 4. ___________________ ​      sin10°. sin 190°. cos 350° ​ tan 120°.(–sin 60) 3. cos 14°. sin 225° –cos 80°3. (–cos 10°) = ____________________________ ​       ​ = ​    sin76° 3. (–cos 45°) 3 _____________________    sin10°. (–sin 10°)3.cos 10°3 ​ __ __ (–​√3 ​) (–​√2 ​) __ cos(180° + 80°). cos(180° – 10°) (–​√3 ​). ​___ ​  2 ​  ​. sin 76. ​___ ​  2 ​  ​ = ____________________________ ​        sin(180° + 10°). cos(360° – 10°) ​ = ___________________ ​       __ ​33 sin10°. ( ) –​√ ___ sin76°. ​​  2 ​  ​ 2​ (–tan60°)3.(–sin 60°). sin 76°3. (–sin 45°)3 –sin 10° 3. (–cos 10°) = ________________________________ ​          sin76°. (–cos 45°) ​ = ​ ___________________       sin10°. (–sin 10°). cos 10° ​ __ __ (–​√3 ​) (–​√2 ​) __ (–​√3 ​). ​___ ​  2 ​  ​. sin 76. ​___ ​  2 ​  ​ –1 = ___________________ ​       __ ​33 = _____ ​  sin10° ​3 (7) √ –​ ___ sin76°. ​​  2 ​  ​ 2 ​ ( ) 3 = __ ​  2 ​3 (9) [28] 204 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.7 Trigonometric identities sin θ sin2θ =1 – cos2θ 1. tan θ = ​ ____ cos θ ​; (cos θ ≠ θ) 2. sin2θ + cos2θ = 1 (the quotient identity) (the square idedtity) cos2θ = 1 – sin2θ AL ATTENTION NB PROOF OF IDENTITIES Proof of the identities are examinable with y the RHS and break it down into its x, y and r values. r sin ____ θ Proof: ​  cos θ ​ y y θ = _​ r ​÷ _​ xr ​ 0 d examples y = _​  r ​× ​  xr_​ x x y = _​  x ​= tan θ Proof: sin2θ + cos2θ y 2 = ​​(_​  r ​  )​​​ + ​​(_​  xr ​  )​​​ 2 y2 ​  x2 ​ Use LCD r2 2 = __ ​  2 ​ + __ r r x2 + y2 = _____ ​  2 ​ x2 + y2 = r2 (Pythagoras) r __ r 2 = ​  2 ​ =1 r ACTIVITIES boy and girl-left and We can use the identities and the reduction formulae to help us simplify trig expressions. right of page p comment comment © Department of Basic Education 2015 UNIT 10 trigonometry 205 Mind the Gap Mathematics

10 Unit Activity 7 Simplify the following expressions. cos (180°–x) sin (x –90°) – 1 __________________________ 1. ​        tan2(540° + x) sin (90° + x) cos (– x) ​ (8) 2. [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)](3) 3. cos2θ (1 + tan2θ)(3) 1 – cos2θ 4. ​ _______ 1– sin2θ ​ (3)  [17] Solutions cos (180° – x) sin (x–90°) – 1 1. __________________________ ​  ​tan​2       ​ ​(540° + x) sin (90° + x) cos (–x) – use reduction formulae and co-functions (–cos x)3 (–cos x) 3 –1 = ___________________________ ​  tan2    ​    (540° – 360° + x) cos x3. cos x3 – multiply out numerator and denominator reduction of angle > 360° cos2x–1 = ​ _______________   ​ tan2(180° + x). cos2x – use trig identity format for cos2x – 1 reduction formula –(1 – cos2x) ​  tan2x3.cos2x ​ = __________ – use trig identities for 1 – cos2x and for tan x  –sin2x3 = ​ _________ sin2x cos2 ​ x – simplify ​  2 ​3.​  1 ​ ____ ____ cos x –sin2x = ​ _____ sin2x ​= –1 3 (8) 2. [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)] – reduce to angle < 90° =[–sin θ + cos θ][cos (–(90° – θ))+ (–cos θ)] – simplify; use co-functions =(–sin θ + cos θ)(sin θ – cos θ) – multiply out using FOIL = –sin θ + sin θ cos θ + cos θ sin θ – cos θ 33 2 2 = –(sin2θ + cos2θ) + 2 sin θ cos θ – use trig identity = –1 + 2 sin θ cos θ – use double angle identity = –1 + sin2θ 3 (3) 3. cos2 θ (1 + tan2 θ) – multiply out the bracket = cos θ + cos θ.tan θ 3 2 2 2 – use trig identity for tan θ cos2θ sin2θ = cos2θ + ____ ​  1 ​.____ ​  cos2θ ​ – simplify = cos2θ + sin2θ 3 = 1 3 – use trig identity sin2θ + cos2θ = 1 (3) 1 – cos2θ 4. ​ _______ 1– sin2θ ​ – use trig identity sin2θ + cos2θ = 1 3 sin2θ = ​ ____ cos2θ3 ​ – use trig identity for tan θ = tan²θ 3 (3) [17] 206 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 PAY SPECIAL ATTENTION NB 10.8 More trig identities You need to be able to use all the information you have about trig ratios and ways to simplify them in order to solve more complicated trig identities. HINT! hint Hints for solving trig identities: Activity 8 • Choose either the left- hand side or the right- Prove the following identities: e.g. EG - worked examples hand side and simplify it to look like the other side. 1 1. sin x ∙tan x + cos x = ____ ​  cos x ​ (4) • If both sides look difficult, you can try to simplify on 2. ( sin x ) (sin x + tan x) ​_______ ​  1 + cos x ​  ​= sin x. tan x (7) exams both sides until you reach ____ 1 cos x a point where both sides 3. ​  cos x ​= _______ ​  1 + sin x ​+ tan x (6) are the same. ____ 1 tan x • It is usually helpful to 4. ​  tan x ​+ tan x = ____ ​  sin2x ​ (5) sin θ write tan θ as ​ ____ ​. [22] cosθ • Sometimes you need to activitysin ____ simplify ​  cos θ ​to tan θ. Solutions θ ACTIVITIES boy • If you have sin2x or cos2x 1. LHS: sin x. tan x + cos x and girl-left and right of page sin x with +1 or –1, use the = sin x . ____ ​  cos x ​+ cos x 3 + cos x squares identities sin2 x cos x activity (sin2θ + cos2θ = 1). cos x ​+ ​  1 ​ = ​ ____ ____ • Find a common sin2 x + cos2 x 3 1 cos x 3   ​= ​  cos x ​3 = RHS (4) = ​ ____________ ____ denominator when fractions are added or 1 ∴ sin x. tan x + cos x = ____ ​  cos x ​ (4) subtracted.Step by step comment • Factorise if necessary – sin x 2. LHS: (sin x + tan x) ​_______ ( ​  1+ cos x ​  ​ ) RHS: sin x. tan x specify with examples i.e. common factor, DOPS, sin x = ​(sin x + ____ ( sin x ) ​  cos x ​3 )​​_______ ​  1+ cos x ​ ​ sin x = sin x. ____ ​  cos x ​3 Trinomial, sum/diff of two cubes sin x cos x + sin x 3 x sin2 x cos x 3   ​ )​​(​ 1 + cos x ​  )​ = ​(_______________ sin cos x ​3 = ​ ____ ​  _______ sin x (cos x + 1) 3 = ​(______________cos x    ​  )​​(​  1 + cos x ​  )​ sin _______x ​  sin2 x cos x ​3 (7) = ​ ____ ∴ LHS = RHS (7) cos x 1 + sin x ​+ tan x 3. RHS: ​ _______ cos x sin x 1 + sin x ​+ ​  cos x ​3 = ​ _______ ____ cos2 x + sin x (1 + sin x) 3 = ​ ___________________    cos x (1 + sin    x) 3 ​ cos2 x + sin x + sin2 x 3 = ​ _________________    cos    x (1 + sin x) ​ trig identity: cos²x + sin²x = 1 1 + sin x 3 = ​ ____________    cos x (1 + sin x) ​ 1 cos x ​3 = LHS = ​ ____ 1 cos x ​  cos x ​= _______ ∴ ____ ​  1 + sin x ​+ tan x (6) © Department of Basic Education 2015 UNIT 10 trigonometry 207 Mind the Gap Mathematics

10 Unit 1 tan x ​  tan x ​+ tan x = ____ 4. ____ ​  sin2 x ​ 1 tan x tan x ​+ tan x LHS: ​ ____ RHS: ​ ____ sin2 x ​ 1 sin x sin x 1 = ​ ____x ​+ ​  cos x ​3 cos x ​3. ​  sin2 x ​ ____ = ​ ____ ____ sin ____ ​  cos x ​ cos x sin x 1 ​  sin x ​3+ ____ = ____ ​  cos x ​ sin x.cos x ​ = ​ ________ cos2 x + sin2 x 3 = ​ ____________ x3 ​ sin x.cos    1 sin x.cos x ​ = ​ ________ ∴ LHS = RHS (5) [22] 208 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.9 Solving trigonometric equations To solve a trig equation where the angle is unknown, you need to find all the possible values of the angle. 1 For example, if sin θ = ​ __ 2 ,​ we know that θ could be 30°. However, there are other values for θ in the other quadrants. Have a look at the graph for sin 1 θ = ​ __ 2 ​, θ ∈ [–360°; 720°]. There are six values for θ between –360° and 720°. y y = sin x 1 y = 0,5 0,5 –360° –270° –180° –90° 90° 180° 270° 360° 450° 540° 630° 720° x –0,5 –1 If 30° is our reference angle in quadrant I. 1 In quadrant II: sin (180° – 30°) = sin 30° = ​ __2 ​ So θ is 150° In quadrant III and IV, the sine ratio is negative, so there is no solution for θ. The angle could be greater than 360°. 1 In quadrant I: sin (360° + 30°) = sin 30° = ​ __ 2 ​ So θ is 390° In quadrant II: sin (540° – 30°) = sin ((540° – 360°) – 30°) 1 = sin (180° – 30°) = sin 30 = ​ __ 2​  So θ is 510° You can also work out that θ = –210° or θ = –330° You do not need to draw a graph to solve these equations. PAY SPECIAL ATTENTION NB NB A method to find the general solution of trig equations: 1. Isolate the trig function on one side of the equation. 2. Determine the reference angle: put the positive number for the angle in the calculator and press the trig key and the inverse key: HINT! hint shift sin / shift cos /shift tan. Use special angles if the question does not allow you to use a calculator. 3. For sin x and cos x, place the reference angle in the two possible quadrants where they are positive or negative e.g. (according to the question). Always put k ∈핑. The period of the sine and cosine graphs is 360°, so add k 360° to each solution. EG - worked examples 4. For tan x, place the reference angle in one correct quadrant where it is positive or negative (according to the question). The period of the tan graph is 180°, so add k 180°. Always put k ∈핑. exams 5. If x must be solved for a given interval: a) Find the general solution b) Substitute k with –1; 0; 1; 2, etc. to find the solutions in the correct interval. activity © Department of Basic Education 2015 UNIT 10 trigonometry 209 Mind the Gap Mathematics ACTIVITIES boy and girl-left and right of page

HINT! 10 Unit hint e.g. 4 EG - worked examples Solve for x: sin x = 0,7 [On your calculator, press: sin–1 0,7 =] 1. exams The calculator answer is 44,42……° We call this the reference angle, as it is not the only solution to the equation. sin x is positive, so angle x must be in quadrant I or quadrant II in activity the first revolution. In quadrant I: x =ACTIVITIES 44,42……. boy and girl-left and ° right of page AND In activity quadrant II: x = 180° – 44,42….° = 135,57…….. ° The period of the sin graph is 360°, so the other points of intersection occur 360° to the right or left of these solutions. We addStepkbyrevolutions step to the comment comment two angles in the first revolution. k is an integer (…–1; 0; 1; …). We call this the general solution of the equation. So we can say the solution to sin x = 0,7 is x = 44,42° + k360° or x = 135,57° + k360°; k ∈핑. (Correct to two decimal place) 2. Solve for x: sin x = –0,7 This time, place the reference angle in quadrants III and IV (sin x is negative) x = 180°+ 44,42…..°+ k360° or x = 360°– 44,42…..°+ k360° k ∈핑 x = 224,42° + k360° or x = 315,57° + k360; k ∈핑 (Correct to two decimal place) 3. Solve for x: cos x = –0,7 Reference angle = 134,427….° cos x is negative in quadrants II and III. x = 360° – 134,43° = 225,57° x = 134,43° + k360° or x = 225,57° + k360°; k ∈핑 (Correct to two decimal place) 4. Solve for x: cos x = 0,7 Reference angle = 45,57……° This time, place the reference angle in quadrants I and IV where cos x is positive: x = 45,57….° + k360° or x = 360° – 45,57……° + k360° PAY SPECIAL ATTENTION x = 45,57° + k360° or x = 314,43° + k360°; k ∈핑. NB NB (Correct to two decimal place) 5. Solve for x: tan x = 0,7 tan x is positive in quadrants I and III. HINT! Reference angle = 34,99° (correct to 2 dec places) hint x = 34,99….° or 180° + 34,99……° = 214,99…….° Now the period of the tan graph is 180°, so the other points You do not need to write of intersection occur 180° to the right or left of the solutions. down the solution of 215°. x = 34,99° + k180°; k ∈핑 e.g. EG - worked examples This solution is already there (Correct to two decimal place) because 6. Solve for x: tan x = –0,7 34,99° + (1)180° = 215° tan x is negative in quadrants II and IV. exams The reference angle is –34,99…...° 180° – 34,99…….° = 145,01…..° x = 145,01° + k180°; k ∈핑. 210 UNIT 10 trigonometry  © Department of Basic Education 2015  activity Mind the Gap Mathematics

Unit 10 Activity 9 1. If cos 20° = p, determine the following ratios in terms of p: a) cos 380° b) sin 110° c) sin 200° (6) 2. Determine the general solution for x in the following equations: a) 5 sin x = cos__ 320° (correct to 2 decimal places) b) 3 tan x + ​√3 ​= 0 (without using a calculator) tan x–1 c) ​ ______ 2 ​= –3 (correct to one decimal place)  (10) 3. Determine x for x ∈[–180°; 180°] if 2 + cos (2x – 10°) = 2,537 (6) [22] Solutions p ​  1 ​so x = p and r = 1 1. cos 20° = __ By Pythagoras, y2 = r2 – x2 ​y2​ ​= ​1​2​– ​p2​ ​= 1– ​p2​ ​ _____ y = √​ 1 – ​p2​ ​ ​ first quadrant, so y is positive a) cos 380° = cos (360° + 20°) = cos 20° 3 = p 3 (2) b) sin 110° reduction formula = sin (180° – 70°) = sin 70° 3 co-function = sin (90° – 20°) = cos 20° 3 = p 3 (3) c) sin 200° = sin (180° + 20°) = –sin20° 3 _____ –​√1 –​p2​ ​ ​ _____ ​  1 ​ = –​√1 – ​p2​ ​ ​ (1) = ______ (6) Calculator keys: 2. a) 5 sin x = cos 320° 3 cos 320 = 5 sin x = 0,766044 ÷5= sin x = 0,15320... 3 SHIFT sin ANS = Ref angle = 8,81° x = 8,81° + k360° OR x = 180° – 8,81° + k360° 3 x = 171,19° + k360° 3 k ∈핑 (4) __ b) 3 tan x + ​ 3 ​= 0 √ __ 3 tan x = –__ ​√3 ​ __ –​√3 ​ ​√3 ​ tan x = ___ ​  3 ​3 [special angle: tan 30° tan 30° = ___ ​  3 ​] Ref angle = 30° x = 180° – 30° + k180° 3 x = 150° + k180° 3 k ∈핑 (3) © Department of Basic Education 2015 UNIT 10 trigonometry 211 Mind the Gap Mathematics

10 Unit tan x–1 c) ______ ​  2 ​= –3 multiply both sides by 2 tan x – 1 = –6 tan x = –5 3 reference angle is 78,69…° ∴x = 180° –78,69…° + k180° 3 x = 101,31° + k180°; k ∈핑 3 (3) (10) 3. 2 + cos (2x – 10°) = 2,537 cos (2x – 10°) = 0,537 Ref angle = 57,52…..° 2x – 10° = 57,52…..° + k360° or 2x – 10° = 360° – 57,52° + k360° [solve equations] 2x = 67,52….° + k360° or 2x = 312,48…° + k360° 3 [divide all terms on both sides by 2] x = 33,76° + k180° or x = 156,24° + k180° 3 k ∈핑 x ∈ [–180°; 180°] So for k = –1: x = 33,76° –180° = –146,24° or x = 156,24° – 180° = –23,76° 3 For k = 0: x = 33,76° or x = 156,24° 3 (For k = 1, x will be > 180°, so it is too big) Solution: x ∈ {–146,24°; –28,76°; 33,76°; 156,24°} 33 (6) [22] 212 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.10 More solving trig equations using identities CIAL ATTENTION NB a sin θ =NBb cos θ: single sin and cos function with the same angle PAY SPECIAL ATTENTION NB • 1) Divide by the cos function 2) Change HINT! sin ​ cosθ ____​to tan θ θ hint e.g. 5 EG - worked examples PAY SPECIAL ATTENTION rked examples Solve for x (give general solution) and round off your answer to 2 decimal NB exams places. • a sin θ = b cos β: single 1. 3 sin x = 4 cos x sin and cos function with 2. 4 cos2 x + 4 sinx cos x + 1 = 0 the different angles HINT! 1.hint [6] activity Use co-functions ACTIVITIES boy to get the same and girl-left and Solutions right of page function i.e. change 1. 3 sin x = 4 cos x activity Divide both sides by cos x to create tan x on LHS the sin function to a 3 sin x ______ 4 cos x e.g. EG - worked examples cos function or the cos x ​= ​  cos x ​3 ​ _____ Trig identity for tan x cos function to a sin 3 tan x Step = by4step comment function. ACTIVITIES boy comment 3 tan x = __ and girl-left and ​  4 ​3 2. exams If sinθ = sinβ, we right of page Ref angle = 53,13° equate the angles then θ = β and θ = x = 53,13° + k180° k ∈핑 3 (3) 180° – β. If cosθ = cosβ we 2. 4 cos2 x + 4sinx cos x + 1 = 0 use 1 = sin2x + cos2x equate the angles 4 cos2 x + 4sinx cos x + (sin2x + cos2x) 3 = 0 then θ = β and θ = tep comment comment 360° – βactivity 5 cos2 x + 4sinx cos x + sin2x = 0 ACTIVITI (5 cos x + sinx)(cos x + sinx) = 0 and girl- right of p 5 cos x + sinx = 0 or cos x + sinx = 0 5 cos x –sin x cos x –sin x activity cos x ​= ​  cos x ​ ​ ______ cos x ​= ​  cos x ​ _____ or ​ ____ _____ 5 = –tan x ∴ tanx = –5 1 = –tan x ∴ tanx = –1 Reference angle = 78,69° Reference angle = –45° x = 180° – 78,69° + k180° or ∴ x = 180°–45° + k180° ∴ x = 101,3° + k180° 3 ∴ x = 135° + k180° 3 k ∈핑 (3) Step by step com [6] © Department of Basic Education 2015 UNIT 10 trigonometry 213 Mind the Gap Mathematics

HINT! 10 Unit hint e.g. 6 EG - worked examples Solve for x (give general solution) and round off your answer to 2 decimal places. exams sin (x + 20°) = cos 3x [7] Solution sin (x activity + 20°) = cos 3x Use co-functions sin (x + 20°) = sin (90°ACTIVITIES – 3xboy) 3 and girl-left and Choose one angle to be the reference angle Ref angle = (90° – 3x) PAY SPECIAL ATTENTION right of page NB activity NB x + 20° = 90° – 3x + k360° 3 or x + 20° = 180° – ( 90° – 3x) + k360 3 4x = 70° + k360° 3 x + 20° = 180° – 90° + 3x + k360° x = 17,5° +Step kby90° HINT! 3 –2 comment x = 70° + k360° 3 hint step comment x = –35° – k180° 3 k ∈핑 [7] e.g. 7 EG - worked examples PAY SPECIAL ATTENTION NB NB 1. sin2 A − sinA cosA = 0 exams Trigonometric equations 2. cos2A − 2 cosA − 3 = 0 that leads to quadratic equations 3. cos2x + 3 sinx = − 3 [16] HINT! Solutions hint activity 1. sin2A − sinA cosA ACTIVITIES = 0 boy and girl-left and sinA(sinA − cosA) = 0......3.......factorise by means of a HCF right of page activity ∴ sinA = 0 or sinA − cosA = 0 3 e.g. EG - worked examples ∴ sinA = 0 or sinA = cosA ∴ A = 0° + 360°n 3 or tanA =1 3 comment Step by step comment ∴ A = 45° + 180°n.......n ∈핑 3 exams (5) 2. cos2A − 2 cosA − 3 = 0 (cosA + 1) (cosA − 3) = 0 3 ∴ cosA + 1 = 03 or cosA − 3 = 0 ∴ cosA = − 1 3 or cosA = 3 3 ∴ A = − 180° + 360°n .....n ∈핑 activity if cosA = 3....no solution 3 ACTIVITIES boy (5) and girl-left and right of page 3. cos²x + 3 sin x = –3 Use cos²x = 1 – sin²x to make activity a quadratic equation in sin x 1 – sin²x + 3 sin x + 3 = 0 –sin²x + 3 sin x + 4 = 0 sin²x – 3 sin x – 4 = 0 3 (sin x – 4) (sin x + 1) = 0 3 Step by step comment commen sin x – 4 = 0 or sin x + 1 = 0 sin x = 4 3 sin x = –1 3 No solution 3 ref angle = –90° (–1 ≤ sin x ≤ 1) x = – 90° + k360° or x = 360° – 90°+ k360° x = 270° + k360° 3 (6) [16] 214 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 10.11 Compound and double angle identities sin (20° + 30°) ≠ sin 20° + sin 30° When two angles are added or subtracted to form a new angle, then a compound or a double angle is formed. Sketch 1 Sketch 2 Sketch 3 A D F E P S β β α a a a B C G H T R     ^ Sketch 1 : The compound angle A​ B​C is equal to the sum of α and β eg. 75° = 45° + 30°     ^ Sketch 2 : The compound angle E​ G​H is equal to the difference between α and β NOTE: eg. 15° = 60° – 45° or 15° = 45° – 30°     ^ sin (α + β) ≠ sinα + sinβ and Sketch 3 : The double angle P​ T​R is equal to the sum of α and α cos (α – β) ≠ cosα – cosβ eg. 45° = 22,5° + 22,5° Using the same methods as we did to establish the reduction formulae, we can also establish the compound angle identities. Given any angles a and β, we can find the values of the sine and cosine ratios of the angles a + β, a – β and 2a. AL ATTENTION NB sin (α + β) = sin α cos β + cos α sin β sin (α – β) = sin α cos β – cos α sin β cos (α + β) = cos α cos β – sin α sin β These formulae are provided on cos (α – β) = cos α cos β + sin α sin β the information sheet in the final exam. sin 2α = 2sin α cos α You should learn these formulae, ed examples cos 2α = cos α – sin α 2 2 as you will use them often. = 2cos2 α – 1 = 1 – 2 sin2 α ACTIVITIES boy and girl-left and right of © Department of Basic Education 2015 page UNIT 10 trigonometry 215 Mind the Gap Mathematics

10 Unit Accept: cos(α – β) = cosα.cosβ + sinα.sinβ, and derive the other compound angle identities PAY SPECIAL ATTENTION NB NB These are examinable, learn them well. Proof: cos(α + β) = cos[α – (–β)] = cosα.cos(–β) + sinα.sin(–β) HINT! = cosα.cosβ + sinα.(–sinβ) hint = cosα.cosβ – sinα.sinβ Proof: sin(α + β) = cos[90° – (α + β)] = cos[90° – α – β] = cos[(90° – α) – β] e.g. = cos(90° – α). cos(β) + sin(90° – α). sin(β) EG - worked examples = sinα.cosβ + cosα.sinβ Proof: β) = cos[90° – (α – β)] = cos[90° – α + β] = cos[(90° + β) – α] sin(α –exams = cos(90° + β). cosα + sin(90° + β). sinα = –sinβ.cosα + cosβ.sinα = sinα.cosβ – cosα.sinβ activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 216 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 10 e.g. 8 EG - worked examples Simplify without the use of a calculator: exams 1. cos70° cos10° + cos20° cos80° 2. 2 sin15° cos 15° 3. sin 15° [10] activity Solutions ACTIVITIES boy and girl-left and right of page 1. cos70° cos10° + cos20° cos80° activity = cos70° cos10° + sin70° sin10° = cos(70° – 10°) 3 = cos60° 3step Step by comment comment = ​  _1 ​ 3 (3) 2 2. 2 sin15° cos15° = sin 2(15°) 3 = sin 30° 3 = ​  1_ ​ 3 (3) 2 3. sin 15° = sin (45° – 30°) n = sin 45°.__ cos 30° – cos 45°. sin 30° n 1 ​√3 ​ 1 1 = ___ ​  √__ ​× ___ ​  2 ​– ___ ​  √__ ​× __ ​  2 ​33 ​ 2​ ​ 2​ __ √ ​ 3​ 1 = ___ ​  √ ​_ – ___ ​  √ ​_ 2​ 2 ​ 2​ 2 ​ __ __ √ ​ 3 ​–1 √ ​2__​​ = ____ ​  √__ ​3 × ​ ___ 2​ 2 ​ ​√2 ​ __ __ ​√2 ​(​√3 ​–1) = _______ ​  4 ​3 (4) [10] © Department of Basic Education 2015 UNIT 10 trigonometry 217 Mind the Gap Mathematics

10 Unit Activity 10 Do NOT use a calculator __ __ to answer this question. Show ALL calculations. Prove that: ​√2 ​(​√3 ​–1) 1. cos 75° = _______ ​  4 ​ (5) 2. Prove that cos(90° – 2x).tan(180° + x) + sin2(360° – x) = 3sin2x (7) 3. Prove that (tan x – 1)(sin 2x – 2cos x) = 2(1 – 2sin x cos x)(7) 2 [19] Solutions 1. LHS = cos 75° = cos(45° + 30°) 3 = cos45°.cos30° __ __ __ – sin45°.sin30° 3 ​ 2 ​ ​√3 ​ √ √ ​ 2​ 1 = ___ ​  2 ​.___ ​  2 ​– ___ ​  2 ​.__ ​  2 ​33 __ __ __ ​√2 ​.​√3 ​ ​√2 ​ = _____ ​  4 ​– ___ ​  4 ​ __ __ ​√2 ​(​√3 ​–1 = _______ ​  4 ​3 = RHS (5) 2. LHS = cos(90° – 2x).tan(180° + x) + sin2(360° – x) co-functions and reductions = sin2 x3. tan x3 + sin x3 2 double angle for sin 2x trig identity for tan x sin x = 2sin x.cos x3. ____ ​  cos x ​3+ sin x 2 simplify = 2 sin2 x + sin2 x3 = 3 sin2x3 = RHS (7) 3. There are several ways to prove this. Here is one solution. LHS = (tan x – 1)(sin 2x – 2cos2x) sin x = ​(____ ​  cos x ​3 – 1 )​(2sin x. cos x3 – 2cos2x) double angle identity for sin 2x = 2sin x – 2sin x. cos x – 2sin x. cos x + 2cos2x33 2 multiply out = 2 sin x – 4 sin x cos x + 2 cos x3 2 2 = 2(sin2 x – 2sin x. cos x + cos2x)3 trig identity sin2 x + cos2 x = 1 = 2(1 – 2sin x. cos x)3 = RHS (7) [19] 218 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 Activity 11 Determine the general solution for x in the following: a) sin 2x. cos 10° – cos 2x. sin 10° = cos 3x  (8) b) cos2 x = 3 sin 2x (11) c) 2sinx = sin(x + 30°) (5) [24] Solutions a) sin 2x. cos 10° – cos 2x. sin 10° = cos 3x use compound angle identity ∴ sin (2x – 10°)3 = cos 3x use co-functions ∴ sin (2x – 10°) = sin (90° – 3x)3 ∴ 2x – 10° = 90°–3x + k360°3or 2x–10° = 180°–(90°–3x) + k360°3 k ∈핑 ∴ 5x = 100° + k360°3 2x – 10° = 90° + 3x + k360° ∴ x = 20° + k72°3 –x = 100 + k360°3 x = –100 – k360 3 k ∈핑 (8) b) cos2 x = 3 sin 2x use double angles for sin 2x cos2 x = 3(2 sin x.cos x)3 make LHS = 0 cos2 x – 3(2 sin x.cos x) = 0 multiply out cos2 x – 6 sin x.cos x = 03 common factor cos x (cos x – 6 sin x)3 = 0 ∴ cos x = 03 or cos x – 6 sin x = 03 cos x 6 sin x cos x = 0 or ____ ​  cos x ​= _____ ​  cos x ​ cos x = 0 or 1 = 6 tan x 1 cos x = 0 or tan x = __ ​  6 ​3 Reference angle = 90° or reference angle = 9,46° ∴ x = 90° + k360°3 or x = 360°–90° + k360° or x = 9,46° + k180° 3k ∈핑 x = 270° + k360°3 or x = 180° + 9,46° + k360°3k ∈핑 = 189,46° + k360° 3 k ∈핑 (11) c) 2 sin x = sin ​(x + 30° )​ 2 sin x = sin x. cos30° + cosx.sin30°3 __ ​√3 ​ 1 2 sin x = sin x.​ ___ 2 ​+ cos x.​  2 ​3 __ multiply by 2 __ 4 sin x = ​√3 ​sin x + cos x divide by cos x __ 4 tan x = ​√3 ​tan x + 1 __ 4 tan x − √ ​ 3 ​tan x = 13 1 tan x =​ _____ √ ​_ 3 4 − ​ 3​ x = 23,79° + k180°; k ∈핑 3 (5) [24] © Department of Basic Education 2015 UNIT 10 trigonometry 219 Mind the Gap Mathematics

10 Unit 10.12 Determining x for NB whichNB the identity is PAY SPECIAL ATTENTION HINT! undefined hint AL ATTENTION e.g. 9 EG - worked examples NB 1 tan x For which values of x is this identity undefined? ____ ​  tan x ​+ tan x = ____ ​  sin2 x ​ exams any number [4] • ________ ​  ​is undefined. 0 Therefore if the denominator of an identity Solution tan x = 0, then the identity is ​  tan x ​+activity 1 ____ tan x = _____ ​  sin2 x ​is undefined if tan x = 0 or if sin²x = 0 or if tan x is undefined undefined ACTIVITIES boy and girl-left and • y = tanx is undefined right of page d examples for certain value of x. activity [division by 0 is undefined] Therefore if a tan function if tan x = 0 OR if sin²x = 0 OR tan x is undefined is in an identity then x = 0° + k180°3 OR sin x = 0 33 x = 90° + k180° 3 (4) the identity is undefined comment x = 0° + k360° OR x = 180° + k360° Step by step comment where the tan function is undefined. So the identity is undefined for x = 0° + k360° or x = 180° + k360° or x = 90° + k180° All these solutions are the same as x = 0° + k90° for k ∈핑. [4] ACTIVITIES boy and girl-left and right of page p comment comment 220 UNIT 10 trigonometry  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 10 What you should be able to do: • Simplify expression, without a calculator by using a sketch. • Use reduction formulae and/or co-functions • Use special angles • Derive and use the trig identities: (Quotient, square, compound and double angle identities). • Determine for which values an identity is undefined • Determine the general solution of trigonometric equations • Solve trigonometric equations with a given interval • Use identities to prove identities and solve equations Feb/March 2014 Q8 & Q9.1 & 9.2 Nov 2013 Q10 & Q11 Feb/March 2013 QQ8 & Q9 Nov 2012 Q8 & Q9 Feb/March 2012 Q11 & Q12 Nov 2011 Q9.1 & 9.2 & Q12 Feb/March 2011 Q10 Keep going! © Department of Basic Education 2015 UNIT 10 trigonometry 221 Mind the Gap Mathematics

11 Unit Trigonometry: Sine, cosine and area rules We use these three rules to find the lengths of sides, sizes of angles and the area of any kind of triangle. To ‘solve a triangle’ means you must calculate the unknown sides and angles. 11.1 Right-angled triangles You can use the trig ratios to find angles and lengths of a right-angled triangle. Activity 1     A ^ In triangle ABC,     ​ B​ = 37° and ^ AC = 16 cm. ​ C​= 90°. Calculate the length of AB and BC (correct to one decimal place). (3) 16 cm 37° B C [3] Solution To calculate the length of AB, use 37° as the reference angle, then AC = 16 cm is the opposite side and AB is the hypotenuse. Use the sine ratio. opp 16 sin 37° = ___ ​  hyp ​= ___ ​  AB ​ AB sin 37° = 16 16 AB = ​ _____ sin37° ​= 26,6 cm 3 To find the length of BC, you can use adj cos 37° = ___ ​  hyp ​= BC/26.6 26,6 cos 37° = BC 3 BC = 21,2 cm (to one decimal place) 3 You can also use Pythagoras’ theorem: AB2 = AC2 + BC2 [3] 222 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 11 Activity 2 In triangle PQR, PQ = 12,3 m and P      ^ PR = 13 m. Calculate the size of ​ Q​.  (2) (adj) (opp) 12,3 m 13 m θ Q R (hyp) [2] Solution Use PQ and PR. opp 13 ​  adj ​= ____ tan θ = ___ ​  12.3 ​ 3 ( 13 ) θ = tan–1 ​____ ​  12.3 ​  ​= 46,58° 3 [2] © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 223 Mind the Gap Mathematics

11 Unit 11.2 Area rule Area of a right-angled triangle: 1 Area D = __ ​ 2 ​ base × perpendicular height 1 Area D = ​ __ 2 ​ bh perpendicular perpendicular height height base base Proof of Area rule [STUDY FOR EXAM PURPOSE]     ^     ^ If ​ A​is acute If ​ A​is obtuse B B c a h h 180°–A A b C A D C D b 1 2 ​ bh……………(1) Area of ∆ABC = ​ __ 1 h 2 ​ bh……………(1) Area of ∆ABC = ​ __ But sin A = ​ __c ​∴ h = c sin A h But sin (180° – A) = ​ __c ​∴ h = c sin A Substituting into (1) 1 Substituting into (1) 2 ​ bc sin A Area of ∆ABC = ​ __ 1 Similarly it can be shown that 2 ​ bc sin A Area of ∆ABC = ​ __ 1 2 ​ ab sin C Area of ∆ABC = ​ __ Similarly it can be shown that 1 1 2 ​ ab sin C Area of ∆ABC = ​ __ 2 ​ ac sin B = ​ __ 1 2 ​ ac sin B = ​ __ 224 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 11 ECIAL ATTENTION NB If a base or height is unknown, you can use trig ratios to work them out. If the perpendicular height is not given and cannot be worked out, then we need a different area formula. There is a formula that works to find the area of any triangle, even if we do not know the perpendicular height. The area of any ∆ABC is half the product of two sides and sine of the included angle. A orked examples b c C B a So if you choose to use angle A, then Area ∆ABC = _​  1 ​ bc sin A 2 If you choose to use angle B, then Area ∆ABC = _​  1 ​ ac sin B 2 If you choose to use angle C, then Area ∆ABC = _​  1 ​ ab sin C 2 PAY SPECIAL ATTENTION ACTIVITIESNBboy NBcan work out the others from that. Learn one form of the formula – you and girl-left and right of page To find the area of any triangle, you need to know the lengths of two sides and the size of the angle between the two sides. HINT! hint e.g. 1comment EG - worked examples step comment Calculate the area of ∆MNK with m = 3,5 cm; exams      ^ n = 4,8 cm and ​ K​= 112°. M Choose the version of the formula that uses the sides m and n and the angle K becauseactivity these are known values. k 1 ACTIVITIES n = 4,8 Area ∆ MNK = ​ __2 ​mnand boy sin K girl-left and right of page 1 = __ activity ​  2 ​(3,5)(4,8) sin 112° K N = 8,4 sin 112° m = 3,5 (correct = 7,788 cmcomment Step by step 2 to 3 decimal places) comment © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 225 Mind the Gap Mathematics

11 Unit 11.3 Sine rule If you have enough information about the sides and angles of any triangle, you can use the sine rule to find the other sides and angles. Sine rule A The ratio of sine of the angle divided c by the side opposite that angle is b the same for all three pairs of sides B and angles. C a So … In any triangle ABC: sin A sin B sin C ​  a ​= ​ ____ ____ ____ b ​= ​  c ​ We can also use the ratios with the sides in the numerator: a b ____ ____ c ​ ____ sin A ​ = ​  sin B ​ = ​  sin C ​ The formula will be provided on the information sheet Proof of Sine rule [STUDY FOR EXAM PURPOSE]     ^     ^ If ​ A​is acute If ​ A​is obtuse B B c a a c A b C A b C Using Area rule for ΔABC: 1 1 1 2 ​ bc sin A = ​  2 ​ ab sin C = ​  2 ​ ac sin B ​ __ __ __ 1 sin A sin C sin B Dividing each by __ ​ 2 ​ abc results: ____ ​  a ​= ​ ____c ​= ​  b ​ ____ PAY SPECIAL ATTENTION NB NB To use the sine rule you need to know at least one side and its matching opposite angle and one more side or angle. A HINT! hint b c B C a e.g. EG - worked examples The sine rule can be used to solve many problems if the right information about the triangle is given. exams 226 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 11 Activity 3      ^ Solve     ∆XYZ in which z = 7,3 m, ​ X​ = 43° X ^ and ​ Y​ = 96°. Give your solutions correct to 43° 3 decimal places. (4) y z Z 96° x Y [4] Solution The angle opposite the known side is not given, but you can work it out.     ^ ​ Z​= 180° – (43° + 96°) (sum angles of ∆) Using the sine rule again to     ^ find x: ​ Z​= 41° 3 x 7,3 y ​  sin 43° ​= ______ ______ ​  sin 41° ​ 7,3 To find y: ​ ______ ​= ______ ​  sin 96° sin 41° ​ 3 7,3 sin 43° 7,3 sin 96° x = ​ ________ sin 41° ​ y = ________ ​  sin 41° ​ x = 7,589 m 3 y = 11,066 m 3 [4] © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 227 Mind the Gap Mathematics

11 Unit 11.4 Cosine rule You apply the cosine rule If you are given the values of: • two sides and the included angle OR • three sides of a triangle, Cosine rule: In any triangle ABC: If you choose to use angle A, then A a2 = b2 + c2 – 2bc cos A b c If you choose to use angle B, then B b2 = a2 + c2 – 2ac cos B C a If you choose to use angle C, then c2 = a2 + b2 – 2ab cos C Proof of Cosine rule [STUDY FOR EXAM PURPOSE]     ^     ^ If ​ A​is acute If ​ A​is obtuse B B h h c a 180°–A A D C b D A C b In Δ BDC: a2 = BD2 + CD2 (Pythagoras Theorem) In Δ BDC: a2 = BD2 + CD2 (Pythagoras Theorem) = BD2 + (b – AD)2 = BD2 + (b + AD)2 = BD2 + b2 – 2bAD + AD2 = BD2 + b2 +2bAD + AD2 But BD + AD2 = c2 2 (Pythagoras Theorem) But BD2 + AD2 = c2 (Pythagoras Theorem) Thus ​a​​= ​b​​+ ​c​​+ 2bAD …………………..(1) 2 2 2 Thus ​a ​= ​b​​+ ​c​​− 2bAD ……………………..(1) ​ 2 2 2 InΔABD:. AD AD In Δ ABD: cosA = ​ ___ c ​∴ AD = c cos A ………..(2) cos(180°- A) = ___ ​  c ​∴ AD = − c cosA……….(2) Substituting (2) into (1) Substituting (2) into (1) ∴ ​a​​= ​b​ ​+ ​c​​− 2bc cosA 2 2 2 ∴ a2 = b2 + c2 − 2bc cosA Similarly it can be shown that: Similarly it can be shown that: ​b​​= ​a​​+ ​c​​− 2ac cos B and 2 2 2 b2 = a2 + c2 − 2ac cosB and ​c​​= ​a​ ​+ ​b​ ​− 2ab cos C 2 2 2 c2 = a2 + b2 − 2ab cosC 228 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 11 e.g. 2 EG - worked examples      P ^ 1. Solve ∆PQR if q = 462 mm, p = 378 mm and ​ R​= 87°. Q exams Using the cosine rule 462 (two sides and the included angle are given so you can 378 find the side opposite the given angle) 87° = p2 + q2 – 2pq cos R PQactivity 2 R PQ2 = (378)2 + (462)ACTIVITIES– 2 boy2(378)(462).cos 87° and girl-left and PQ = 338 048,5159 2 right of page activity PQ = 581,42 mm [take square root] Using the sine rule: 378 ____ 581,42 ​  sin P ​= _____ ​  sin Step by step ​ 87° comment comment sin P ____ _____ sin 87°     ^ ​  378 ​= ​  581,42 ​ (it is easier to have ​ P​in the numerator) 378 × sin 87° sin P = __________ ​  581,42 ​ sin P = 0,649     ^ ​ P​= ​sin​–1​(0,649) = 40,48°      ^ ∴ ​ Q​= 180° – (87° + 40,48°) = 52,52° [sum angles of ∆] 2. Determine the biggest angle in ΔABC if a = 7 cm; b = 9 cm and c = 15 cm. You are given three sides, so use the cosine rule. ^     The biggest angle will be ​  C​(opposite the longest side). c2 = a2 + b2 – 2 ab cos C rearrange the formula to get cos C on its own 2 ab cos C = a2 + b2 – c2 a2 + b 2 − c 2 cos C = ________ ​  2ab ​ 72 + 92 − 152 cos C = ​ _________ 2(7)(9) ​     ^ cos C = –0,753968… cos θ is negative in quad 2, so ​ C​is obtuse. reference angle is 41,06°     ^ ​ C​= 180° – 41,064…° = 138,94° (correct to two decimal places) © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 229 Mind the Gap Mathematics

11 Unit 11.5 Problems in two and three dimensions Activity 4 1. PQRS is a trapezium with PQ // SR, PQ = PS, SR = 10 cm,      ^ QR = 7 cm, ​ R​= 63°. Calculate: P Q a) SQ (2) b) PS 7 cm (6) c) area of quadrilateral PQRS. (correct to 2 63° S R IAL ATTENTION decimal places) 10 cm (5) NB [13] When solving triangles, Solutions start with the triangle a) In ∆ QSR, you know two sides and the included angle, so use the cosine which has most rule. information (i.e. triangle SQ² = 7² + 10² – 2(7)(10)cos 63° 3 with three sides or two sides and an angle or two SQ² = 85,44… find the square root angles and a side given) SQ = 9,24 cm 3 (2) ked examples b) In Δ PQS, you know that PQ = PS and you worked out that SQ = 9,24 cm. Think about the question first     ^ If you can find ​ P​then you can use the sine rule to find PS.     ^      ^     ^ To find ​ P​, you need to first find P​ Q​S or P​ S​Q.      ^     ^ P​ Q​S = P​ S​Q (alternate angles, PQ // SR)     ^ Now can you work out a value for Q​ S​R?      ^ In ∆QSR, you know three sides and ​ R​.     ^ So it is easiest to use the sine rule to find Q​ S​R.     ^ sin Q​ S​R ______ sin 63° ACTIVITIES boy ​  7 ​=  ​ _____ 9,24 ​ 3 and girl-left and     ^ 7 sin 63° sin Q​ S​R = ______ right of page ​  9,24 ​= 0.675004     ^ ∴ Q​ S​R = 42,45° 3      ^     ^ P​ Q​S= Q​ S​R = 42,45° (alternate angles, PQ // SR)      ^     ^ P​ Q​S= P​ S​Q = 42,45° (base angles of isosceles ∆)     ^ comment ∴ ​ P​= (180° – (42,45° + 42,45°) 3 ep comment = 95,1° 3 (sum angles in ∆)     ^ Now we can find PS using the sine rule and ​ P​. PS 9,24 In ∆PQS ​ _______ ______ sin 42,45° ​= ​  sin 95,1 ​ 3 9,24 sin 42,45° PS = ​ ___________ sin 95,1 ​ PS = 6,26 cm 3 (6) 230 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 11 c) To find the areas of PQRS, find the area of the two triangles and add them together.     ^ To find the area of ∆PQS, use ​ P​= 95,1° and PS = PQ = 6,26 cm. 1 Area ∆PQS = ​ __2 ​qs sin P 3 1 Area ∆PQS = __ ​  2 ​(6,26)(6,26)sin95,1° Area ∆PQS = 19,52 m² 3      ^ To find the area of ∆RQS, use ​ R​= 63°, QR = 7 cm and SR = 10 cm. 1 Area ∆RQS = ​ __2 ​(7)(10)sin63 3 Area ∆RQS = 31,19 m² 3 ∴Area PQRS = 19,52 + 31,19 = 50,71 m² 3 (5) [13] Activity 5 A In the diagram alongside,   AC = 7 cm, ^ DC = 3 cm, AB = AD, D​ C​A = 60°,      ^     ^ β D​ A​B = β and A​ B​D = θ. ___ ​√37 ​sin β Show that BD = ​ _______ sin θ ​ 7cm θ 60° B D 3cm C [3] Solution A​D2​ ​   = ​AC​2​+ ​CD​2​− 2AC.CD cos  6​0°​​= ​(7)​2​+ ​(3)​2​– 2 × 7 × 3 × 0,5 3 AD2 =58 – 21 AD2 = 37 ___ AD = ​√37 ​P Applying sine rule: AD sin β ___ BD AD ​ ____ sin β ​= ​ ____ _______ sin θ ​⇒ BD = ​  sin θ ​   but AD = ​ 37 ​ 3 √ ___ ​√37 ​sin β ∴BD = ​ _______ sin θ ​3 [3] © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 231 Mind the Gap Mathematics

11 Unit Activity 6 1. In the diagram alongside, ABC is a right angled triangle. C     ^     ^ KC is the bisector of A​ C​B. AC = r units and B​ C​K = x x x 1.1 Write down AB in terms of x(2) r      ^ 1.2 Give the size of A​ K​C in terms of x(2) AK 2 1.3 If it is given that ​ ___ AB ​= ​ __3 ​, calculate the value of x(7) B K A 2. A, B and L are points in the same horizontal plane, HL is a vertical pole of length 3 metres, H     ^ AL = 5,2 m, the angle A​ L​B = 113° and the angle of elevation of H from B is 40°. 2.1 Calculate the length of LB. (3) 3m 2.2 Hence, or otherwise, calculate the length of AB. (3) 2.3 Determine the area of ∆ABL. (3) 40° L 113° B 5,2 A 3. The angle of elevation from a point C on the ground, at the centre of the goalpost, to the highest point A of the arc, directly above the centre of the Moses Mabhida soccer stadium, is 64,75°. The soccer pitch is 100 metres long and 64 metres wide as prescribed by FIFA for world cup stadiums. Also AC ⊥ PC. In the figure below PQ = 100 metres and PC = 32 metres 3.1 Determine AC (2)      ^ 3.2 Calculate P​ A​C(2) 3.3 A camera is placed at D, 40 m directly below A point A, calculate the distance from D to C (4) [28] 40 m D P Q 64.75° C M 232 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 11 Solutions AB 1.1 sin 2x = ___ ​  r ​∴ AB = r sin 2x 33 (2)      ^ 1.2 A​ K​C = 90° + x [ext. angle of ∆CBK] 33 (2) AK r r sin x sin x ​= ​  sin(90° + x) ​∴ AK = ​  cos x ​ 33 _________ _____ 1.3 ​ ____ r sin x AK ​  cos x ​ _____ r sin x 1 2 ​  AB ​= ​  r sin 2x ​= ​  r   ___ ______ _____________ _____ __ ​= ​  ​= ​  ​33 cos x.2 cos x sin x 2cos2x 3 3 ∴ cos2x = ​ __4 ​3 __ ​√3 ​ cos x = ​ ___ 2 ​3 Hence x =  3​0​0​3 (7) 3 2.1 In ∆HLB, tan 40° = ___ ​  LB ​3 [∆HLB is right-angled, so use a trig ratio] 3 LB = _____ ​  tan 4​00​ ​​3 LB = 3,5752… ≈ 3,58 metres 3 (3) 2.2 In ∆ABL, [∆ABL not right-angled. You have two sides and included angle, so use the Cosine Rule] AB2 = AL2 + BL2 – 2(AL)(BL).cos L 3 AB2 = (5,2)2 + (3,58)2 – 2(5,2)(3,58).cos 113° 3 AB2 = 54,40410… m2 AB = 7,38 m 3 (3) 1     ^ 2.3 Area ∆ABL = __ ​  2 ​AL × BL × sin A​ L​B 3 1 = __ ​  2 ​(5,2) × (3,58) × sin 113° 3 = 8,56805… 3 ≈ 8,57 m2 (3) CM CM 50m 3.1 cos 64,750° = ____ ​  AC ​∴ AC = ​ _______ ​= ​ _______​= 117,21 33 cos 64,7​50​ ​ 0,426569 (2)      ^PC 3.2 tanP​ A​C = ​ ___ AC ​      ^ P​ A​C = ​tan​−1 ​​ ___32 ​  AC ​  ​ 3 ( ) = 15,2​7​ ​ 3 0 (2) 3.3 D​C​2​= A​C​2​+ A​D​2​− 2AC.ADcos(9​0​0​− 64,7​5​0​) 3 D​C2​ ​= (117,21​)​2​+ (40​)​2​− 2(117,21).40cos(25,2​5​0​) 3 = 6857,289 3 DC = 82,81 m 3 (4) [28] © Department of Basic Education 2015 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RULE S 233 Mind the Gap Mathematics

11 Unit What you should be able to do: sin θ • Derive and use the trig identities: tan θ = ____ ​  cos θ ​and sin2θ + cos2θ = 1. • Derive and use reduction formulae to simplify expressions. • Determine for which values of a variable an identity is true. • Derive and use the sine, cosine and area rules. • Apply the sine, cosine and area rules to solve triangles in 2-D and 3-D problems. • Use the compound angle and double angle identities where necessary to prove and make necessary calculations. November 2013 Question 13 February/March 2013 Question 11 February/March 2012 Question 12 November 2012 Question 12 November 2011 Question 11 November 2010 Question 11 Keep going! 234 UNIT 11 TRIGONOMETRY: SINE, COSINE AND AREA RU LES  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 Euclidean Geometry 12.1  Revise: Proportion and area of triangles 1. Ratio and proportion Ratio compares two measurements of the same kind using the same units. Example If Line A is 2 units long and Line B is 6 units long, then the ratio of Line A : Line B is 2 : 6. This is the same ratio as 1 : 3. Line C is 1 unit long and Line D is 3 units long. So Line C : Line D is 1 : 3. So C and D are in the same proportion as A and B. A C B D A C So the two ratios are equal and we can say that __ ​  B ​= ​ __ D​ We say that A, B, C and D are in proportion. This proportion can be written in many ways: A C A B B D C D If ​ __ __ B ​ = ​  D ​, then 1. __ ​  C ​= ​ __ D​ 2. ​ __ __ A ​= ​  C ​ 3. ​ __ __ A ​ = ​  B ​ © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 235 Mind the Gap Mathematics

12 Unit Ratio and proportion of areas and sides of triangles 1. If two or more triangles have a common vertex (A) and lie between the same parallel lines, they also have a common perpendicular height (altitude). B C D E 2. The areas of triangles with equal altitudes D are in the same proportion as their bases. Remember: area ∆ = ½ base × perp height ∆ADB, ∆DBC and ∆ADC all have the same ⊥ height DE. A B E C So Area ∆ADB : Area ∆DBC : Area ∆ADC (½ AB × DE) : (½ BC × DE) : (½ AC × DE) AB : BC : AC 3. If two or more triangles lie CD between parallel lines, they have the same altitude. h 4. Triangles on the same base (or equal bases) and between AB parallel lines are equal in area. Area ∆ABC = ½(AB)h Area ∆ADB = ½(AB)h Area ∆ABC = Area ∆ADB 236 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 12.2 Proportion theorems Theorem 7 (Learn the proof for the examination) Proportional Theorem If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same proportion. (Prop theorem, DE ∣∣ BC) Given: Triangle ABC with D on AB and E on AC, DE ∣∣ BC A AD ​ = __ To prove: ​ __ ​  AE ​ DB EC Proof: Construction: Draw altitudes h and k in ∆ADE k h Join DC and BE h __ ​ Area of _________ ∆ADE ½.AD. Area of ∆BDE ​ = _____ ​  ½.DB.h ​= ​ AD ​(same altitude h) DB D E and ​ = ​  ½.AE.k ​= ​ __ Area of ∆ ADE _____ ​ _________ AE ​(same altitude k) Area of ∆ CED ½.EC.k EC but Area ∆ADE = Area ∆CED (same base DE; same altitude; DE ∣∣ BC) B C ∆ ADE Area of ∆ ADE ∴ ​  Area of _________ ​ = _________ ​  ​ Area of ∆ BDE Areaof ∆ CED AD ​ = __ ∴ ​ __ ​  AE ​ DB EC Solving problems using proportion Activity 1 1. Determine the value of x, in the diagram alongside, if PQ ∣∣ BC. (4) Solution AP AQ ​  PB ​= ___ ___ ​  QC ​ 3 (PQ ∣∣ BC, prop theorem) 3 5 4 ∴ __ ​  3 ​= __​  x ​ 3 ∴ 5x = (3)(4) 12 ∴ x = __ ​  5 ​= 2,4 cm 3 [4] 2. In ∆ABC, AB ∣∣ FD; AF ∣∣ DE and FE : EC = 3 : 4. A NOTE: Determine EC : BF 3 : 4 does not mean that D FE = 3 and EC = 4. For any a, we can say that FE = 3a and EC = 4a For every 3 of a in FE, there is 4 of a in EC. B F E C (7) © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 237 Mind the Gap Mathematics

12 Unit Solution Work with two different triangles: ∆ACF and ∆ABC A A D D B F E C B F E C In ∆ACF: ___AD FE ​  DC ​= ___ ​  EC ​ 3 (AF ∣∣ DE, prop intercept theorem) 3 In ∆ABC: AD ___ ___ BF ​  DC ​= ​  FC ​ (AB ∣∣ FD, prop intercept theorem) 3 FE BF AD ∴ ___ ​  EC ​= ___ ​  FC ​(both = ___ ​  DC ​) 3 ___FE 3a BF BF ​  EC ​= __ ​  4a ​ and ___ ​  FC ​= ___ ​  7a ​ 3a BF ∴ __ ​  4a ​= ___ ​  7a ​ 3 ∴ BF = 3​(__ ​  4 ​  )​= ​  7a 21a ___ 4​3 EC 21a ∴​ ___ BF ​= 4a ÷ ___ ​  4 ​ 4a 4 = __ ​  1 ​× ___ ​  21a ​ 16 = __ ​  21 ​3 ∴ EC : BF = 16 : 21 [7]  238 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 3. Determine the value of x if PQ ∣∣ BC. (4) Solution A ___AP AQ ​  PB ​= ​ ___ ( prop theorem , PQ ∣∣ BC ) 3 QC ​3 5 4 ​  3 ​= ​  x ​3 __ __ 5cm 4cm 5x = (3) (4) x = ​  12 __ 5 ​= 2,4cm 3 P Q 3cm x B C [4] 4. In the diagram, RF ∣∣ KG , ED ∣∣ KH, H RH = 3 units, RK = 9 units, HF = 2 units. GE: EK = 1:3 3 2 R F Calculate (stating reasons) the lengths of: 9 D 4.1 FG 4.2 FD (8) K G 3m E 1m Solutions 4.1 In ∆HKG FG 9 ___ ​  2 ​= ​ __3 ​ 3 S (line ∣∣ one side of a ∆) 3 R or (RF ∣∣ KG) FG = 6 units 3 S (3) GD GE 1 4.2 ​ ___ ___ __ GH ​= ​  GK ​= ​  4 ​3 S (line ∣∣ one side of a ∆) 3 R or (ED ∣∣ KH) 1 GD = ​ __4 ​.GH 1 GD = ​ __4 ​. (8) 3 S GD = 2 3 S ∴ FD = 6 − 2 = 4units 3 R OR In ∆HKG, HK ∣∣ DE GD EG ___ ___ __ 1 ​  DH ​= ​  EK ​= ​  3 ​3 S (line ∣∣ one side of a ∆) 3 R or (proportional theorem, HK//DE ) 6 − FD ______ __ 1 ​  2 + FD ​= ​  3 ​3 S 18 − 3FD = 2 + FD 3 ∴ FD = 4 units 3 (5) [8] © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 239 Mind the Gap Mathematics

12 Unit PAY SPECIAL ATTENTION NB NB 12.3 Similar polygons HINT! hint Similar polygons have the same shape, but not necessarily the same size. e.g. 1 EG - worked examples Every square is similar to every other square. exams activity ACTIVITIES boy and girl-left and right of page activity Polygons (with the same number of sides) are similar when: Step by step comment comment • All the pairs of corresponding angles are equal (They are equiangular) and PAY SPECIAL ATTENTION NB NB • All the pairs of corresponding sides are in the same proportion. Both of these conditions must hold at the same time. ||| is the symbol HINT! we use to say one polygon ‘is similar to’ another polygon. hint e.g. 2 EG - worked examples P Q exams A B T Corresponding sides are E sides in the same position (with respect to the angles) in each polygon. activity S R D E ACTIVITIES boy and girl-left and right of page activity Consider pentagon ABCDE and pentagon PQRST      ^     ^     ^      ^     ^      ^      ^     ^     ^     ^ ​ A​= ​ P​; ​ B​= ​ Q​; ​ C​= ​ R​; ​ D​= ​ S​; ​ E​= ​ T​ AND Step by step comment comment AB ___ BC DC ED ___ EA ​ ___ ___ ___ PQ ​= ​  QR ​= ​  SR ​= ​  TS ​= ​  TP ​ ∴ABCDE ||| PQRST (equiangular and corresponding sides in the same proportion) Triangles are special polygons: equiangular ∆s ➝ similar ∆s • If two triangles are equiangular, then their sides will always be in the same proportion, so the triangles are similar. corresponding sides ∆s in proportion ➝ ∆s are similar • If the sides of two triangles are in the same proportion, then the triangles will be equiangular, so the triangles are similar. 240 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 Theorem 9 (Learn the proof for the examination) If two triangles are equiangular, then the corresponding sides are in proportion and therefore the triangles are similar.     ^     ^     ^     ^     ^ Given: ∆ ABC and ∆ DEF with ​A ​ = ​ D​; ​B ​= ​E ​; = ​F ​ ​ DE ​ = __ To prove: __ ​  EF ​ = ​  DF __​ AB BC AC D Proof: On AB mark off AP = DE and on AC mark off AQ = DF Draw PQ In ∆ APQ and ∆ DEF AP = DE (Construction)     ^     ^ ​A ​ = ​ D​ (given) AQ = DF (Construction) ∴ ∆ APQ ≡ ∆ DEF (SAS) E F     ^     ^ ∴ ​P ​1 = ​E ​     ^     ^     ^     ^ ∴ ​P ​1 = ​B ​ (​E ​= ​B ​) A ∴ PQ ∣∣ BC (Corresponding ∠s equal) ​  AQ ​ (PQ ∣∣ BC in ∆ ABC ) ​  AP ​ = __ ∴ __ AB AC But AP = DE and AQ = DF P 1 Q DE ​ = __ ∴ ​ __ ​  DF ​ AB AC 2 Similarly, we can prove that B C DE ​ = __ ​ __ ​  EF ​ AB BC ​  DE ​ = __ ∴ __ ​  EF ​ = __ ​  DF ​ AB BC AC But the triangles are equiangular ∴ ∆ ABC ||| ∆ DEF NOTE: If two triangles have 2 corresponding angles equal, then the third angles will equal each other (sum angles of a triangle = 180°) and the triangles are therefore similar and their sides will be in proportion. The shortened reason you can use is (third angle) 40° 40° 80° 80° If two angles are the same, then the 3rd angle of both triangles is 180° – (40° + 80°) (sum angles in ∆) = 60° © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 241 Mind the Gap Mathematics

12 Unit Theorem 10 (Learn the proof for the examination) If two triangles have their sides in the same proportion, then the corresponding angles will be equal and the triangles are similar. Given: ​ DE ​ = __ ∆ ABC and ∆ DEF with __ ​  EF ​ = ​  DF __​ A AB BC AC     ^     ^ ^             ^ ^     ^ A ​ = ​ D​; ​B ​= ​E ​ ; ​ C​= ​F ​ To prove: ​     ^     ^     ^     ^ Proof: Draw ∆ PEF so that P​E ​F = ​B ​and E​F ​P = ​ C​ ∴ ∆PEF ||| ∆ABC (equiangular ∆s) PE ​ = __ ♦ ∴ ​ __ ​  EF ​ = __ ​  PF ​ B C AB BC AC DE ​ = __ But ​ __ ​  EF ​ = __ ​  DF ​ (Given) D AB BC AC ∴ PE = ED and PF = DF and EF is common ∴ DEF ≡ ∆ PEF (SSS) 1             E 1 ^ ^ ^ F ∴ ​F ​1 = ​F ​2 = ​ C​ 2 2 ♦     ^     ^     ^ and ​E ​1= ​E ​2 = ​B ​     ^     ^ ^         ^     ^     ^ i.e. ​A ​ = ​ D​; ​B ​= ​E ​1 ; ​ C​= ​F ​1 But the corresponding sides of the triangles are proportional P ∴ ∆ ABC ||| ∆DEF Theorem 11 (Learn the proof for the examination) Theorem of Pythagoras (proved using similar triangles) In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.     ^ Given: ∆ABC with ​A ​= 90° To Prove: BC2 = AB2 + AC2 Proof: Draw AD ⊥ BC A In ∆ABD and ∆CBA     ^ ​ B ​is common     ^     ^ A​ D​B = C​A ​B = 90° (given)     ^     ^ B​ A ​D = B​ C​A (3rd ∠ of ∆) ∴ ∆ABD ||| ∆CBA (AAA) ∴ ​  AB ​  BD ​ __​ = __ (ABD ||| CBA) BC AB ∴ AB2 = BC × BD B C D Similarly ∆ACD ||| ∆CBA and AC2 = DC × CB ∴ AB2 + AC2 = BC × BD + DC × CB AB2 + AC2 = BC (BD + DC) AB2 + AC2 = BC × BC AB2 + AC2 = BC2 242 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 Activity 2 1. C      ^ Diameter AME of circle with centre M bisects F​ A​B. MD is perpendicular to the chord AB. A D 4 B 2 1 3 ED produced meets the circle at C, and CB is joined. 1 2 a) Prove ∆AEF ||| ∆AMD (5) 1 2 b) Hence, find the numerical value M AF of ​ ___ AD ​. (5) 2 c) Prove ∆CDB ||| ∆ADE (4) 1 E d) Prove AD2 = CD. DE (3) [17] F Solution Solution AE EF AF a) b) ____ ​  AM ​= ____ ​  MD ​= ___ ​  AD ​ (||| ∆s) 3 C AM = ME (radii) 3 ∴ AE = 2AM 3 A D 4 B 2AM AF 1 2 1 3 2 ∴ _____ ​  AM ​= ___ ​  AD ​3 AF 1 ∴ ___ ​  AD ​= 2 3 (5) 2 M 2 1 E F     ^ ​ F​= 90° (∠ in semi-circle) 3      ^ ​ D​1 = 90° (given MD ⊥ AB) 3     ^      ^ ∴ ​ F​= ​ D​1 In ∆AEF and ∆AMD     ^      ^ ​ F​= ​ D​1 3 (proved)      ^      ^      ^ ​ A​1 = ​ A​2 (AM bisects F​ A​B) 3     ^       ^ ∴ ​ E​1 = ​ M​1 (third ∠ of ∆) ∴ ∆AEF ∣∣∣ ∆AMD (AAA) or ∠∠∠ 3 (5) © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 243 Mind the Gap Mathematics

12 Unit Solution Solution c) d) C C A D 4 D 4 2 1 3 B A B 1 2 1 3 2 1 2 1 1 2 2 M M 2 2 1 E 1 E F F CD DB In ∆CDB and ∆ADE ​  AD ​= ___ ___ ​  DE ​ (III ∆s)     ^      ^ ​ C​= ​ A​2 3 (∠s in same seg) 3 ∴ CD. DE = AD. DB 3     ^     ^ But AD = DB (MD ⊥ AB, M is centre) 3 ​ B​= ​ E​2 (∠s in same seg) 3      ^      ^      ^ ∴ CD. DE = AD 3 ​ D​4 = ​ D​1 + ​ D​2 (opp ∠) ∴ AD2 = CD. DE (3) ∴ ∆CDB ||| ∆ADE (AAA) 3 (4) [17] 244 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 2. A CD is a tangent to circle ABDEF at D. Chord AB is produced to C. Chord BE cuts chord AD in H and chord FD in G. F AC ∣∣ FD and FE = AB.           B 1 ^ ^ a) Prove that ​ D​4 = ​ D​2 (3) 3 2 3 H b) Prove that ∆BHD ||| ∆FED (5) G 2 1 AB FD 1 23 1 c) Hence ___ ​  BH ​= ___ ​  BD ​ (3) 3 2 1 2 [11] 4 E C D Solutions      ^      ^ a) ​ A​= ​ D​4 (tan-chord thm) 3      ^      ^ ​ D​2 = ​ A​ (alt ∠s CA ∣∣ DF) 3      ^      ^ ​ D​4 = ​ D​2 3 (3) A A F F B 1 B 1 3 2 3 2 3 H 3 H G G 2 1 2 1 1 23 1 1 23 1 3 2 1 2 3 2 1 2 4 E 4 E C C D D b) In ∆BHD and ∆FED     ^     ^ ​ B​2 = ​ F​ (∠s in same seg) 3      ^      ^ ​ D​3 = ​ D​1 3 (equal chords) 3      ^     ^ ​ H​2 = ​ E​2 (third ∠ of Δ) 3 ∴ ∆BHD ||| ∆FED ∠∠∠ 3 A (5) F FE FD c) ___ ​  BH ​= ___ ​  BD ​3 (||| ∆s) B 1 But FE = AB 3 (given) 3 2 3 H AB FD G ∴ ___ ​  BH ​= ___ ​  BD ​3 2 1 1 23 1 (3) 3 2 1 2 4 E C D [11] © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 245 Mind the Gap Mathematics

12 Unit 3. In the diagram ∆ABC is such that F is on AB and G is on AC. CB is produced to meet GF produced at E .DGFE is a straight line. BFA ∣∣ CD. AB = 20, BC = 10, EF = 8, EB = 5 and FB = 6. E F1 2 A 3 2 1 B 3 G12 2 1 D C EF 3.1 Determine the numerical value of ___ ​  ED ​ (3) 3.2 Calculate the length of ED (2) 3.3 Complete, without stating the reasons: ∆EFB III ∆ …. (1) 3.4 Hence, calculate the length of DC (3) AF FG 3.5 Prove that: ___ ​  CD ​= ___ ​  DG ​ (4) [13] Solutions BFA ∣∣ CD. AB = 20, BC = 10, EF = 8, EB = 5 and FB = 6 3.1 3.5 In ∆AFG and ∆CDG FB ll CD (Given)      ^     ^ ​ A​= ​ C​1 (alt ∠s. AF ∣∣ DC) 3 S/R EF EB ​ ___ ___ ED ​= ​  EC ​3 S (line ∣∣ one side of A) 3 R      ^      ^ ​ G​3 = ​ G​1 (vertically opp ∠s) 3 S/R EF 5 __ 1 ​ ___ __ ED ​= ​  15 ​= ​  3 ​3 S (3)     ^      ^ ​ F​3 = ​ D​ (alt ∠s. AF ∣∣ DC) ∆AFG III ∆CDG (∠∠∠) 3 R AF FG ​ ___ ___ CD ​= ​  DG ​ (∆AFG ∣∣∣ ∆CDG) 3 R (4) EF 1 3.2 ​ ___ __ ED ​= ​  3 ​from 3.1 and EF = 8 8 1 ∴ ___ ​  ED ​= __ ​  3 ​ 3 ED = 24 3 S (2) 3.3 ∆EFB ∣∣∣ ∆EDC 3 (1) DC ED ___ 3.4 ​  FB ​= ___ ​  EF ​(∆EFB ∣∣∣ ∆EDC) 3 R DC 24 ​ ___ __ 6 ​= ​  8 ​3 S DC = 18 3 S (3) [13] 246 UNIT 12 EUCLIDEAN GEOMETRY  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 12 4. In the diagram, PQCB is a cyclic quadrilateral. Chords BP and CQ are produced to meet at A such that AQ = BC. B P 1 2 A 2 1 Q C 4.1 Prove that: ΔAPQ III ΔACB (4) 4.2 Hence, prove that AQ² = AB.PQ (3) [7] Solutions 4.1 Proof: In ΔAPQ and ΔACB      ^      ^ ​ A​= ​ A​ (common) 3 S/R     ^     ^ ​ P​2 = ​ C​3 S (ext ∠ of a cyclic quad ) 3 R     ^     ^ ​ P​2 = ​ B​ (sum ∠s of ∆) or ( ext ∠ of cyclic quad ) ∆APQ ∣∣∣ ∆ACB ( ∠.∠.∠) 3 R (4) 4.2 ___AQ PQ ​  AB ​= ___ ​  BC ​3 S (∆APQ ∣∣∣ ∆ACB) 3 S AQ ___ PQ ​ ___AB ​= ​  AQ ​3 S ( AQ = BC ) AQ2 = AB.PQ (3) [7] Keep going! © Department of Basic Education 2015 UNIT 12 EUCLIDEAN GEOMETRY 247 Mind the Gap Mathematics

13 Unit Statistics Data handling is the study of statistics, or data. We collect, organise, analyse and interpret data. The data can inform students, researchers, advertising and business. It can provide us with an understanding of social issues and human trends. Then we can make informed decisions when we plan for the future, or make a new advertisement, or address social issues. We usually collect data from a fairly small group (called the sample). The sample must be big enough and it must be randomly chosen from the population. This is to make sure that it fairly represents the trends in the larger group of people (called the population). PAY SPECIAL ATTENTION NB NB population HINT! hint Sample: Some data randomly chosen from population. e.g. EG - worked examples exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 248 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 13.1 Bar graphs and frequency tables PAY SPECIAL ATTENTION NB NB Data can be represented with a frequency table or with a bar graph. Each bar represents a group of data and the bars can be compared to each other. Bar graphs HINT! must be labelled on one axis and show the numbering hint on the other axis. e.g. 1 EG - worked examples In a Geography class, 23 learners completed a test out of 10 marks. Here exams is a list of their results: 4; 1; 2; 2; 6; 9; 6; 10; 6; 8; 9; 6; 7; 7; 8; 4; 6; 6; 5; 7; 9; 10; 6. We can use a frequency table to record this data. activity Frequency table ACTIVITIES boy and girl-left and right of page Mark out Tally Number of learners who achieved this mark activity of 10 (frequency) 1 / 1 2 Step by step // comment comment 2 3 0 4 // 2 5 / 1 6 //// // 7 7 /// 3 8 // 2 9 /// 3 10 // 2 We can also make a bar graph to show this data. Use the marks from 1 to 10 on the horizontal axis. Use the number of learners who got that score on the vertical axis. The number of learners is the frequency. Geography test results out of 10 8 7 6 Number of learners 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 Marks out of 10 © Department of Basic Education 2015 UNIT 13 STATISTIC S 249 Mind the Gap Mathematics

13 Unit 13.2 Measures of central tendency 13.2.1 Ungrouped Data Measures of central tendency are different measures of finding the ‘middle’ or ‘average’ of a set of data. The three kinds of ‘middle’ of a set of data that we use are the mean, the median and the mode. It is advisable that we start by arranging the set of data in an ascending order before attempting questions. 1. Mean The mean of the data is the average if you add all the values and divide by the number of values. We use the symbol for the mean. PAY SPECIAL ATTENTION NB NB ) = ___ –     ∑fx mean (​ x​ ​  n ​ HINT! This hint formula will be on the information sheet provided in examinations. e.g. 2 EG - worked examples In a Mathematics class, 23 learners completed a test out of 25 marks. Here is a list of their results: exams 14; 10; 23; 21; 11; 19; 13; 11; 20; 21; 9; 11; 17; 17; 18; 14; 19; 11; 24; 21; 9; 16; 6. activity Calculate the mean of this data. ACTIVITIES boy and girl-left and right of page Solution activity    sum of values in set mean (​ x​) = _________________ – ​      numberof values in set ​ = 14 + 10 + 23 + 21 + 11 + 19 + 13 + 11 + 20 + 21 + 9 + 11 + 17 + 17 + 18 + 14 + 19 + 11 + 24 + 21 + 9 + 16 + 6 __________________________________________________________________________________ ​          Step by step      23comment comment ​3 = 15,4347… 3 (2) 250 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB Unit 13 2. MedianHINT! hint The median is the middle number in an ordered data set. e.g. 3 EG - worked examples In a Mathematics class, 23 learners completed a test out of 25 marks. Here is a list of their results: exams 14; 10; 23; 21; 11; 19; 13; 13; 20; 21; 9; 13; 17; 17; 18; 14; 19; 13; 24; 21; 9; 16; 6. Calculate the median of this data. Solution activity • First put the data ACTIVITIES in order, boy and girl-left and from lowest to highest. right of page 6; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24. activity There are 23 numbers, so the middle number is the12th number out of 23 numbers. So 16 is the median, the number in the middle of the data. • When there is an even number of values in the data set, the median lies halfway between the middle Step by step comment comment two values. • We can add these two values and divide by 2. For [Example], what if another learner wrote the test and her result was 7? We can add this to the ordered data set. 6; 7; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24. Now there are 24 numbers and the middle two numbers are the 12th and 13th numbers. The middle two numbers are 14 and 16. Add 14 and 16 to get 30 and divide by 2 to get a median of 15. 14+16 _____ NB PAY SPECIAL ATTENTION NB ​  2 ​= 15 3. Mode The mode HINT! is the number or value that appears most frequently in the hint data set. e.g. 4 EG - worked examples In a Mathematics class, 23 learners completed a test out of 25 marks. exams Here is a list of their results: 14; 10; 23; 21; 11; 19; 13; 13; 20; 21; 9; 13; 17; 17; 18; 14; 19; 13; 24; 21; 9; 16; 6. Find the mode of this data. activity ACTIVITIES boy Solution and girl-left and right of page The mode of the results of the test is 133 (13 appears 4 times). activity (1) Summary Step by step comment comment mean (​ x​) = ___     – ∑fx ​  n ​ median: middle score of an ordered list of data mode: the most frequent score © Department of Basic Education 2015 UNIT 13 STATISTIC S 251 Mind the Gap Mathematics

13 Unit Activity 1 The table below represents Mathematics test scores and frequency for each score. Scores (x) Frequency (f) 13 5 17 6 20 4 25 10 (a) Determine the median (2) (b) Determine the mean (2) [4] Solutions (a) ∑ f = 25 3 i.e. there are 25 scores. To determine the median, find the position of the median by adding the frequencies until you reach the position of the median. Median lies in position 13, hence median = 20 3 (2) 5(13) + 6(17) + 4(20) + (10)23 477 (b) mean = ​ ______________________     ___   ​= ​  25 ​= 19,08 25 33 (2) [4] 252 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

PAY SPECIAL ATTENTION NB NB Unit 13 HINT! 13.2.2 hint Grouped data e.g. 5 EG - worked examples Fifty shoppers were asked what percentage of their income they spend on exams groceries. Six answered that they spend between 10% and 19%, inclusive. The full set of responses is given in the table below. activity PERCENTAGE FREQUENCY (f) ACTIVITIES boy and girl-left and 10 < x < 19 right of page 6 activity 20 < x < 29 14 30 < x < 39 16 40 <Stepxby< step 49 comment comment 11 1. Determine 50 < x < 59 3 midpoints of each interval. Since we do not have the exact values in (a) Calculate the mean percentage of family income allocated to grouped data, we use these groceries. approximations (b) In which interval does the median lie? 2. Add al the frequencies to get the number of items (c) Determine the mode percentage of income spent on groceries. in adata set 3. Determine the total of all Solutions (a) PERCENTAGE Midpoint Interval Frequency f Total (fx) 10 < x < 19 14,5 6 14,5 × 6 = 87 20 < x < 29 24,5 14 24,5 × 14 = 343 30 < x < 39 34,5 16 34,5 × 16 = 552 40 < x < 49 44,5 11 44,5 × 11 = 489,5 PAY SPECIAL ATTENTION 50 < x < 59 54,5 3 54,5 × 3 = 163,5 NB Sum n=50 ∑(fx) = 1635 3 ∑fx Mean = ___ ​  50 ​= 32,7 3 (2) HINT! (b) 30 < x < 39 3 (median is in position 25,5 of the data. When we hint add frequencies from above, then position 25,5 lies in the interval There are 50 scores. median 30 < x < 39) (1) lies between positions 25 (c) 30 < x < 39 3 (the interval with the highest frequency) (1) e.g. and 26 6+14=20 EG - worked examples 20+16=36 exams activity ACTIVITIES boy and girl-left and right of page © Department of Basic Education 2015 UNIT 13 STATISTIC S 253 Mind the Gap Mathematics activity

13 Unit 13.3 Measures of dispersion (or spread) The measures of dispersion give us information about how spread out the data is around the median. The measures of central tendency give us information about the central point of the data, but we still need to know if the data is concentrated in one place, or evenly spread out. We first look at these measures of dispersion: range and interquartile range. 1. Range The range PAY isSPECIAL theATTENTION difference between the highest value (or maximum) NB and the lowest value (or minimum)NB in a data set. RangeHINT!= largest value in the data set – smallest value in the data set hint e.g. 6 EG - worked examples Find the range of the Mathematics test results: exams 6; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24. Solution activity ACTIVITIES boy 24 – 6 = 18. So the range ofpagethe test results is 18. and girl-left and right of activity 2. The interquartile range • The interquartile range depends on the median. So organise the data first and find the median. Step by step comment comment • The data is divided into four parts (quarters, which we call quartiles). First, the median (Q2) divides the data into two halves. • The lower quartile (Q1) divides the data below the median (Q2) into two equal sets of data. • The upper quartile (Q3) divides the data above the median into two equal sets of data. • The difference between the lower and the upper quartile (Q3 – Q1) is called the interquartile range. This tells us how spread out the middle half of the data is around the median. lower quartile median upper quartile Q1 Q2 Q3 data data data data interquartile range 254 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

HINT! hint Unit 13 e.g. 7 EG - worked examples We can also use the ff formulae Find the interquartile range of the Mathematics test results: to calculate the exams position of Q1, Q2 and Q3. 6; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; Position of Q2 20; 21; 21; 21; 23; 24. (n+1) (23 + 1) = ______ ​  4 ​= __________ ​  2 ​= 12 Q2 is the value in Solution activity position 12 which is 16 • We already know thatACTIVITIES the boy median is 16. and girl-left and Position of Q1 • Lower half of the data has right of page 11scores, so Q is the 6 th data item. 1 (n + 1) (23 + 1) ∴ Q1 = 13 activity = ________ ​  4 ​= __________ ​  4 ​= 6 • Upper half of the data set has 11 scores, so Q3 is the 6th score is the value in position 6 of the upper half data set ∴ Q3 = 20 which is 13 Step by step comment comment Position of Q3 lower quartile upper quartile 3(n + 1) 3(23 + 1) meridianQ2 ​ _________ 4 ​= ___________ ​    4 ​= 18 Q1 Q3 Q3 is the value in position 18 which 6; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24 is 20 interquartile range = Q3 – Q1 = 20 – 13 = 7 [8] IAL ATTENTION NB Median is not included in the lower half and upper half of the data when calculating Q1 and Q3 Activity 2 ed examples If the test scores in another class are represented by the data below, find the interquartile range of the test results: 6; 7; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24. [8] Solution We already know that the middle two numbers are 14 and 16. • The data set has an even number of points, so the median will lie between 14 and 16 (3). Use the lower value 14 in the lower half and the upper value 16 in the upper half. (3) 11 + 13 • Lower half: 12 numbers, so use the 6th and 7th numbers to find the lower quartile. ______ ​  2 ​= 12 (3) 19 + 20 • Upper half: 12 numbers, so use the 6th and 7th numbers to find the upper quartile. ______ ACTIVITIES boy and girl-left and ​  2 ​= 19,5 (3) right of page lower quartile meridian upper quartile Q1 3 Q2 3 Q3 3 6; 7; 9; 9; 10; 11;13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24 ep comment comment interquartile range = 19,5 – 12 = 7,5 3 [8] © Department of Basic Education 2015 UNIT 13 STATISTIC S 255 Mind the Gap Mathematics

13 Unit 13.4 Five number summary and box and whisker plot 1. Five number summary The five number summary is a ‘summary’ description of a data set. It is made of these five numbers: – thePAYminimum SPECIAL ATTENTION value NB NB – the lower quartile – the median – theHINT! upper quartile hint – the maximum value e.g. 8 EG - worked examples What is the five number summary for the set of data we have used so far? exams 6; 9; 9; 10; 11; 13; 13; 13; 13; 14; 14; 16; 17; 17; 18; 19; 19; 20; 21; 21; 21; 23; 24. – the minimum value: 6 – the lower quartile: activity 13 – the median: ACTIVITIES boy and girl-left and 16 – the upper quartile: right of page 20 activity – the maximum value: 24 2. Box and whisker plot comment We can represent the five number summary on a box and whisker Step by step comment plot (or diagram). The box represents the middle half of the data (the interquartile range) The line in the box shows the median. The ‘whiskers’ show the minimum and maximum values. minimum lower upper maximum value quartile median quartile quartile 0 2 4 6 8 10 12 14 16 18 20 22 256 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 Quartiles divide data into four equal sets of data. The longer whisker and box means that the lower 50% of the scores is more spread out than the upper 50%. Skewed to the right (Positively skewed) means that the upper half of the data is more spread out than the lower half. Skewed data A box and whisker plot can show whether a data set is symmetrical, positively skewed or negatively skewed. This box and whisker plot is not symmetrical because the whiskers are not the same length and the median is not in the centre of the box. The whisker on the left is a bit longer than the whisker on the right, which shows that the data on the left of the PAY SPECIAL ATTENTION NB NB box is more spread out. The box is also longer to the right of the median than to the left of the median. We say that the data is negatively skewed. (or skewed to the left). HINT! 3.hint Identification of outliers e.g. 9 EG - worked examples Determine whether the minimum in the [Example] above is an outlier or exams not. Solution Inter-quartile range = Q3 – Q1 activity =ACTIVITIES 20 –boy13 and girl-left and =right7of page activity Q1 – 1,5 × IQR = 13 – 1,5 × 7 PAY SPECIAL ATTENTION NB = 2,5 NB ∴ 6 > 2,5 6 is not an outlier Step by step comment comment HINT! hint To determine outliers: • Determine the interquartile range e.g. 1,5 ×examples • Determine Q1EG–- worked IQR • If the minimum < the value of Q1 – 1,5 × IQR, then it is an outlier. • Determine Q3 – 1,5 × IQR • If the maximum > Q3 – 1,5 × IQR, exams Then it is an outlier. activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 UNIT 13 STATISTIC S 257 Mind the Gap Mathematics Step by step comment comment

13 Unit Activity 3 1. These are the scores of ten students in a Science test: 90; 85; 10; 75; 70; 60; 78; 80; 82; 80; 55; 84 a) Draw a box and whisker diagram for the given data. (5) b) Determine the interquartile range. (2) c) State whether the data is skewed or not. (1) d) State whether 10 is an outlier or not. (2) [10] Solutions a) First write all the scores in ascending order. 10; 55; 60; 70; 75; 78; 80; 80; 82; 84; 85; 90 Work out the five number summary: Minimum number: 10 Maximum number: 90 78+80 Median: 12 numbers, so use the 6th and 7th numbers _____ ​  2 ​= 79 Lower quartile: Use the first 6 numbers. The 3rd and 4th numbers are 60 and 70 Upper quartile: Use the last 6 numbers. The 3rd and 4th numbers are 82 and 84. 653 793 833 10; 55; 60; 70; 75; 78; 80; 80; 82; 84; 85; 90 60 + 70 ______ 78 + 80 ______ 82 + 84 ______ ​  2 ​ ​  2 ​ ​  2 ​ 10 65 79 83 90 3 whiskers 3 box 10 20 30 40 50 60 70 80 90 100 (5) b) Interquartile range = upper quartile – lower quartile3 = 83 – 65 = 183 (2) c) The data is skewed to the left (negatively skewed). 3 (1) The whisker on the left is longer, i.e. the length on the left of the box is longer than the length on the right. d) Interquartile range (IQR) = Q3 – Q1 = 83 – 65 = 18 Q1 – 1,5 × IQR = 65 – 1,5 × 18 = 38 338 10 < 38 3 conclusion ∴ 10 is an outlier (2) [10] 258 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 Activity 4 The five number summary of heights of trees three months after they were planted is (23 ; 42 ; 50 ; 53 ; 75). This information is shown in the box and whisker diagram below. 23 42 53 20 30 40 50 60 70 80 a) Determine the interquartile range. (2) b) What percentage of plants has a height excess of 53 cm?  (2) c) Between which quartiles do the heights of the trees have the least variation? Explain. (2) [6] Solutions a) Interquartile range = 53 – 42 3 = 11 3 (2) b) 25% 33 (2) c) Between Q2 (50) and Q3 (53) 3 The distance between these two quartiles is the smallest 3 (2) [6] © Department of Basic Education 2015 UNIT 13 STATISTIC S 259 Mind the Gap Mathematics

13 Unit 13.5 Histograms and frequency polygons • Histograms and frequency polygons are graphs used to represent grouped and continuous data. They show the frequency and the distribution (spread) of the data. • Continuous data is data that is not just measured in whole numbers. For example, length, mass, volume or time are measured in continuous amounts. • The horizontal axis of a histogram and a frequency polygon have a continuous scale. • The vertical axis shows the frequency, or number of times the data is listed. PAY SPECIAL ATTENTION NB NB 1. Grouped data: Instead of recording every piece of data separately, we can group the data to HINT! make it easier to read. Grouped data can be represented on a hint histogram or a frequency polygon. e.g. 10 EG - worked examples A grocer wants to record the mass of each packet of chicken pieces he exams sells. He groups the masses into intervals of 0,2 kg. He makes a frequency table. Mass of chicken in kg activity Number 0,8 < mass of chicken ≤ 1,0 0 ACTIVITIES boy and girl-left and These 8 packets have any mass right of page 1,0 < mass of chicken ≤ 1,2 3 between a bit more than 1,2 kg activity and 1,4 kg. 1,2 < mass of chicken ≤ 1,4 8 So 1,2 < mass of chicken < 1,4 1,4 < mass of chicken ≤ 1,6 6 Step by step comment comment 1,6 < mass of chicken ≤ 1,8 2 1,8 < mass of chicken ≤ 2,0 1 2,0 < mass of chicken ≤ 2,2 0 260 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 2. Histograms PAY SPECIAL ATTENTION From the frequency table, he draws up a histogram. NB Histogram of mass of chicken pieces A histogram is a graphical 8 display of data using bars of different heights. It is similar HINT! to 7 hint a bar graph but for a histogram Use vertical axis Number of packets of 6 there are no gaps between the 5 bars. for frequency 4 x-coordinate: use upper limit of e.g. chicken pieces EG - worked examples the interval 3 y-coordinate: frequency 2 1 exams 0 0,2 0,4 0,6 0,8 1 1,2 1,4 1,6 1,8 2,0 2,2 Mass of chicken in kg Use horizontal axis for mass. Use intervals of 0,2 kg. 3. Frequency polygons activity AC We can also make a frequency polygon using this data. A frequency and polygon uses lines to join the mid-points of each interval. The righ polygon must begin and end on the horizontal axis. So we can add an activity interval at the beginning and the end of the data which both have a frequency of 0. Frequency polygon of mass of chicken pieces Step by step Number of packets of chicken pieces 8 PAY SPECIAL ATTENTION 7 NB NB Use vertical axis 6 Notice the polygon shape. This 5 one has 6 sides, including the for frequency 4 HINT! x-axis. 3 hint 2 1 0 0,2 0,4 0,6 e.g.1 0,8 1,2 EG - worked examples 1,4 1,6 1,8 2,0 2,2 Mass of chicken in kg Use horizontal axis for mass. exams Use intervals of 0,2 kg. activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 UNIT 13 STATISTIC S 261 Mind the Gap Mathematics

13 Unit The frequency polygon can also be drawn using the midpoints of the bars of the histogram, as shown below. 10 9 8 7 6 Frequency 5 4 3 2 1 0 0,8 1 1,2 1,4 1,6 1,8 2 2,2 Mass of chicken (kg) PAY SPECIAL ATTENTION NB NB Frequency polygons are useful to compare the distribution of two or more sets of data on the same set of axes. HINT! hint To plot a frequency polygon: • Plot midpoints of each interval • Join the midpoints by straight lines • AddEG -anworked interval at the beginning and end of the data, with both frequencies e.g. equals to 0. examples • Frequency polygon is a closed figure, hence must start and end at the x-axis. exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 262 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 13.6 Cumulative frequency tables and graphs (ogives) 1. Cumulative frequency tables PAY SPECIAL ATTENTION NB NB • Cumulative frequency gives us a running total of the frequency. So we keep adding onto the frequency from the first interval to the last interval.HINT! •hint We can show these results in a cumulative frequency table. e.g. 11 EG - worked examples In an English class, 30 learners completed a test out of 20 marks. Here is a exams list of their results: PAY SPECIAL ATTENTION 14; 10; 11; 19; 15; 11; 13; 11; 9; 11; 12; 17; 10; 14; 13; 17; 7; 14; 17; 13; 13; NB 9; 12; 16; 6; 9; 11; 11; 13; 20. Markactivity Tally Frequency Cumulative ACTIVITIES boy out of and girl-left and (number of learners) frequency HINT! hint Keep adding onto right of page 20 activity frequency from row 6 / 1 1 before. 7 / 1 1+1=2 For example, 8 Step by step comment comment 0 2+0=2 e.g. EG - worked examples 7 + 6 = 13 9 /// 3 2+3=5 10 // 2 5+2=7 exams 11 //// / 6 13 12 // 2 15 13 //// 5 20 14 /// 3 23 15 / 1 24 activity 16 / 1 25 PAY SPECIAL ATTENTIONACTIVITIES boy NB and girl-left and 17 /// 3 28 right of page 18 0 28 activity 19 / 1 29 HINT! 20 / 1 30 hint The last number is the same as total With this data set, it would be more useful to group the data. numberStepof learners by step comment e.g. EG - worked examples exams © Department of Basic Education 2015 UNIT 13 STATISTIC S 263 Mind the Gap Mathematics

13 Unit We can use intervals of 5 and make a cumulative frequency table for grouped data. Class interval Frequency Cumulative frequency 1<x<5 0 0 5 < x < 10 7 7 10 < x < 15 17 24 15 < x < 20 6 30 2. Cumulative frequency graph (ogive) • We can represent the cumulative results from a cumulative frequency table with a cumulative frequency graph or ogive. PAY SPECIAL ATTENTION NB NB • This graph always starts on the x-axis and usually forms an S-shaped curve, ending with the cumulative frequency (y-value). • TheHINT! endpoint of each interval is plotted against the cumulative hint frequency. e.g. 12 EG - worked examples Represent the data in the cumulative frequency table of grouped data with a exams cumulative frequency graph. • the x-axis needs the points 5; 10; 15; and 20 to mark the end of each interval. • y-axis represents the cumulative frequency from 0 to 30. theactivity • For plotting the ACTIVITIES points, and girl-left anduse the end of each class interval on the boy x-axis and the cumulative frequency on the y-axis. So you need to right of page activity plot these points: (5; 0); (10; 7); (15; 24); (20; 30) • Join the plotted points. Step by step comment comment PAY SPECIAL ATTENTION NB NB 35 30 Cummulative Frequency HINT! 25 hint To plot ogive: 20 • x-coordinate - use upper e.g. 15 limit of each EG - interval. worked examples • y-coordinate - cumulative 10 frequency • If the frequency of the 5 exams first interval is not 0, then include an interval before 0 the given one and make 0 5 10 15 20 25 use 0 as its frequency. Marks activity ACTIVITIES boy and girl-left and 264 UNIT 13 STATISTICS  right of page © Department of Basic Education 2015  Mind the Gap Mathematics activity

Unit 13 Activity 5 An ice cream vendor has kept a record of sales for October and November 2012. The daily sales in rands is shown in the histogram below. 24 Frequency (number of days) 21 18 15 12 9 6 3 0 10 20 30 40 50 60 70 80 90 100 110 120 Daily sales (rand) 1.1 Draw up a cumulative frequency table for the sales over October and November. (2) 1.2 Draw an ogive for the sales over October and November. (3) 1.3 Use your ogive to determine the median value for the daily sales. Explain how you obtain your answer. (1) 1.4 Estimate the interval of the upper 25% of the daily sales. (2) [8] Solutions 1.1 Cumulative frequency table: Daily sales (in rand) Frequency Cumulative frequency 60 ≤ rand < 70 5 5 70 ≤ rand < 80 11 16 80 ≤ rand < 90 22 38 3 1st three correct 90 ≤ rand < 100 13 51 100 ≤ rand < 110 7 58 110 ≤ rand < 120 3 61 3 last three correct (2) © Department of Basic Education 2015 UNIT 13 STATISTIC S 265 Mind the Gap Mathematics

13 Unit 1.2 70 60 cummulative frequency 3 1st three points plotted 50 correctly 40 3 last three points plotted correctly 30 3 grounding at 0 20 PAY SPECIAL ATTENTION NB 10 NB 0 0 20 40 60 80 100 120 140 Marks HINT! hint We can find the median, the range 1.3 There are 61 data points, so the median is the 31st data point. and the interquartile range from a cumulative frequency We can read the data point off the graph at 31. It gives a rand value e.g. graph. EG - worked examples of R87. 3 (1) We cannot find the mean 1.4 The upper 25% lies above 75% of 61 = 45,75. 3 from a cumulative frequency Read from the y-axis across to the graph and down to the x-axis. graph. exams The upper 25% of sales lies in the interval: 96 ≤ sales < 120 3(2) [8] activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 266 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 13.7 Variance and standard deviation Sometimes the mean is a more useful measure of central tendency than the median. The measures of dispersion (spread) around the mean are called the variance and the standard deviation. 1. Standard deviation The standard deviation is the square root of (the sum of the squared differences between each score and the mean divided by the number of scores). The formula for standard deviation is: _______    √ ∑ (x – ​ x​) –2    σ = ​ ​ _______ n ​ ​ where x is each individual value, ​ x​ – is the mean and n is the number of values. The symbol sigma ∑ ​means ‘the sum of’. This formula will be on the data sheet. Make sure you can use the formula properly. 1.1 Calculating the standard deviation using the formula: _______    –2 σ = ​ _______ ​  n ​ √ ∑(x – ​x ​)  ​ 1. Find the mean of all the numbers in the data set.    2. Find each value of x – ​ x​ – . In other words, work out by how much each of these values differs (or deviates) from the mean.    3. Square each deviation. Find each value of (x – ​ x​ –2 )    4. Add all the answers together. In other words, find ∑ (x – ​ x​ –2 ). 5. Divide this sum by the number of values, n.    ∑ ​(x – ​ x​ –2 )​ ​ 6. You havePAYnow ​  1 n .​ This value is called the variance. found _______ SPECIAL ATTENTION NB NB _______    √ ∑(x – ​ x​) –2 ​  n ​ ​ 7. Find the square root of the variance to find the standard deviation. ​ ______ By working HINT! through these steps, you have found the standard deviation using the formula. hint e.g. 13 Finding the variance and standard deviation EG - worked examples These exams are the results of a mathematics test for a Grade 11 class of 20 students. 52 44 62 66 60 57 95 78 71 62 100 69 62 72 73 55 32 83 78 80 activity 1. Calculate the mean mark for the class. ACTIVITIES boy (2) 2. and girl-left and Complete the table below and use it to calculate the standard right of page deviation activity of the marks. (3) 3. What percentage of the students scored within one standard deviation of the mean? (2) Step by step comment comment © Department of Basic Education 2015 UNIT 13 STATISTIC S 267 Mind the Gap Mathematics

13 Unit PAY SPECIAL ATTENTION Solutions NB     – ​ x​= 52+44+62+66+60+57+95+78+71+62+100+69+62+72+73+55+32+83+78+80 1. ____________________________________________________________ ​          ​ = 67,55 3 20 2.         HINT! (x – ​ x​) (x – ​ x​)2 Mark obtained – – (%) hint     52 52 – 67,55 (–15,55)2 The squaring of (x – ​ x– )​ deals = –15,55 = 241,8 with the effect of the negative 44 –23,55 554,6 e.g. signs. EG - worked examples At the end, we find the 62 –5,55 30,8 square root of the whole 66 –1,55 2,4 answer to ‘reverse’ the effect ofexams the square. 60 –7,55 57,0 57 –10,55 111,3 95 27,45 753,5 78 10,45 109,2 71 3,45 11,9 activity ACTIVITIES boy 62 –5,55 30,8 and girl-left and right of page 100 32,45 1053,0 69 1,45 2,1 activity 62 –5,55 30,8 72 4,45 19,8 73 5,45 29,7 Step by step comment 55 –12,55 157,5 32 –35,55 1263,8 83 15,45 238,7 78 10,45 109,2 80 12,45 155,0    ∑ (x – ​–x ​)2 4 962,9 ______ √ 4 962,9 σ = ​ ​ ______ 20 ​ ​= 15, 7526... = 15,75 333 answer (correct to 2 decimal places) 3. One standard deviation from the mean lies between PAY SPECIAL ATTENTION       (​ x​– σ; ​ x​+ σ) = (67,55 – 15,75; 67,55 + 15,75) – – NB = (51,8; 83,3) 3 We can say this is a representative 16 scores lie in the interval (51,8; 83,3) set of data, because more than 16 out of 20 of the marks lie within one standard deviation of the 66,6% lie within one standard HINT! mean, __ 16 ​  20 ​× 100 = 80% 3 hint deviation of the mean. Answer: 80% of the students’ marks lie within one standard deviation from the mean. e.g. EG - worked examples 268 UNIT 13 STATISTICS  exams © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 1.2 Steps for calculating the standard deviation with a scientific calculator: Using a Casio fx–82 ES PLUS calculator: press Mode then STAT then 1 – VAR – enter all data one at a time pressing = after each entry. – press the orange AC button – press shift STAT then VAR    – in order to calculate the mean press 2: ​ x​ – . – once all these steps have been completed, simply press AC shift STAT then VAR – now press 3: σ to calculate the standard deviation. If you understand the calculator steps and use them properly, you will come to the same answer of 15,75 as we found before. Practise these steps so that you can do exam Examples using a calculator. CIAL ATTENTION NB • the interquartile range measures a spread around the median, so it has to do with the positions of data and not their actual values. • the standard deviation measures a spread around the mean, using the actual values of the data and not just their positions. ked examples ACTIVITIES boy and girl-left and right of page ep comment comment © Department of Basic Education 2015 UNIT 13 STATISTIC S 269 Mind the Gap Mathematics

13 Unit Activity 6 The data below shows the energy levels, in kilocalories per 100 g, of 10 different snack foods. 440 520 480 560 615 550 620 680 545 490 (a) Calculate the mean energy level of these snack foods. (2) (b) Calculate the standard deviation. (2) (c) The energy levels, in kilocalories per 100 g, of 10 different breakfast cereals had a mean of 545,7 kilocalories and a standard deviation of 28 kilocalories. Which of the two types of food show greater variation in energy levels? What do you conclude? (2) [6] Solutions 5500 (a) Mean = ____ ​  10 ​= 550 kilocalories 33 (2) (b) σ = 69,03 kilocalories 33 (2) (c) Snack foods have a greater variation. 3 The standard deviation for snack foods is 69,03 kilocalories whilst the standard deviation for breakfast cereals is 28 kilocalories. i.e energy levels of breakfast cereals is spread closer to the mean than in those of the snack food. 3 (2) [6] 2. The normal distributive curve The data can be plotted on a graph that shows the standard deviations. If the data is distributed symmetrically around the mean, the values form a normal distribution curve: mean Frequency    –             ​x ​– 2σ ​–x ​ – σ ​–x ​ ​–x ​ + σ ​–x ​ + 2σ 270 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 13.8 Bivariate data and the scatter plot (scatter graph) • A scatter plot is a graph using the x- and y-axes to represent bivariate data. • Bivariate data means that each point on the graph represents two variables that are independent of each other. • In a scatter plot, we plot a point for each pair of coordinates and look at the overall pattern or trend in the data. • The points in the data are compared to see if there is a correlation or some kind of pattern (or trend) in the data. • When a point does not fit the trend of the other points, it is called an outlier. • OutliersPAYare SPECIALeasy ATTENTIONto identify on a scatter diagram or a box and NB NB whisker diagram. • We can sometimes represent the trend in the data with a line or curveHINT!of best fit. The line or curve can be represented by an hint equation that could be linear, quadratic, exponential, hyperbolic etc. e.g. 14 EG - worked examples A science teacher compares the marks for the mid-year examination with exams the marks for final examinations achieved by 11 learners. mid-year 80 68 94 72 74 83 56 68 65 75 88 marks final marks activity 72 71 96 77 82 72 58 83 78 80 92 ACTIVITIES boy 1. Draw a scatter graph of this data. and girl-left and right of page (3) 2. Describe activitythe curve of best fit. (2) 3. Use the scatter plot to estimate the final mark of a learner who had a midyear mark of 75%. (1) Step by step comment comment [6] © Department of Basic Education 2015 UNIT 13 STATISTIC S 271 Mind the Gap Mathematics

13 Unit Solutions 1. 110 105 100 95 90 85 Final examination 80 75 70 65 60 55 50 45 40 40 45 50 55 60 65 70 75 80 85 90 95 100 Mid-year examination marks 333 all points plotted correctly 2. The ‘curve’ or line of best fit is a straight line. 3 There should be about five dots above the line and five dots below the line. 3 3. A line from 75 on the x-axis to the trendline takes us to about 78 on the y-axis. So we can predict that a learner with a midyear mark of 75% can expect to get about 78% 3 in the final exam. [6] 272 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 Activity 7 The outdoor temperature (in °C) at noon is measured. It is compared with the number of units of electricity used to heat a house each day. Temp in °C 7 11 9 2 4 7 0 10 5 3 units of electricity used 32 20 27 37 32 28 41 23 33 36 (a) Draw a scatter graph to represent this data. (3) (b) Draw in a line of best fit. (1) (c) Use your line of best fit to estimate the noon temperature when 30 units of electricity are used. (1) [5] Solutions (a) Graph 333 (3) (b) Line of best fit 3 (1) 45 40 35 Units of electricity 30 25 20 15 10 0 1 2 3 4 5 6 7 8 9 10 11 12 Noon temperature in °C (c) If the noon temperature is 6,25°C, about 30 units 3 of electricity will probably be used in the house. (1) [5] © Department of Basic Education 2015 UNIT 13 STATISTIC S 273 Mind the Gap Mathematics

13 Unit 13.9 The linear regression line (or the least squares regression line) The line of best fit for a set of bivariate numerical data is the linear regression line. So far, we have seen this trend line on a scatter graph. Now we use a scientific calculator to determine the equation for this line. We know the straight line equation: y = mx + c Statistics (as used on the CASIO x-82ES PLUS calculator) uses y = A + Bx, where B is the gradient and A is the cut on the y-axis of the straight line of best fit. So the gradient is B instead of m and the y-intercept is A instead of c. The Regression Coefficient ‘r’ This is a statistical number that measures the strength of the correlation (relationship) between two sets of data. • This number is calculated from two sets of data using a calculator. • r always lies between –1 and +1. • The closer r is to –1, the stronger the negative correlation. • The closer r is to +1, the stronger the positive correlation. • If r = 0, there is no correlation between the two sets of data. The number line shows the r values and the strength of the correlation between bivariate data. –1 –0,8 –0,6 –0,4 –0,2 0 0,2 0,4 0,6 0,8 1 strong medium weak weak medium strong negative negative negative positive positive positive correlation correlation correlation correlation correlation correlation perfect no linear perfect negative correlation positive correlation correlation PAY SPECIAL ATTENTION NB NB We only study the r value of bivariate data when the line of best fit is a straight line. HINT! hint e.g. EG - worked examples 274 UNIT 13 STATISTICS  © Department of Basic Education 2015  exams Mind the Gap Mathematics

Unit 13 A negative correlation means that as x increases, y decreases. The more closely the points are clustered together around the line, the stronger the correlation. x x x x x x x x x x x x x x x x x x x x x x x x x x A strong negative A weaker negative A strong positive A weaker positive correlation correlation correlation correlation A positive correlation means that as x increases, y also increases A correlation of zero means that there is no relationship between x and y. PAY SPECIAL ATTENTION x x NB NB x x x x HINT! x hint e.g. 15 EG - worked examples A diesel engine turns at a rate of x revolutions per minute. The corresponding exams horse power of the engine is measured by y in the table below: x (revolutions per minute) 400 500 600 700 750 y (horse power) 580 1 030 1 420 1 880 2 100 activity and least squares regression line: y = A + B x ACTIVITIES boy 1. Find the equationand of the girl-left right of page (correct to two decimal places). activity 2. Determine the regression coefficient r. Discuss the correlation between x and y. 3. Use this regression line to comment estimate the power output when the Step by step comment engine runs at 800 rev per min. 4. About how fast is the engine running when it has an output of 1 200 horse power? © Department of Basic Education 2015 UNIT 13 STATISTIC S 275 Mind the Gap Mathematics

13 Unit Solutions 1. Use a calculator 2. Keep all the information in the calculator – Mode 2: STAT from 1. – 2: A + B x – Press AC – Enter the x values first: – SHIFT STAT – 400 =; 500 =; 600 =; 700 =; 750 = – 5:REG – Use arrows to move right to y column and up – Then 3: r = 0,9996821357 to start next to 400. There is a strong positive correlation between x – Enter y values: and y (r is very close to +1) – 580 =; 1030 =; 1420 =; 1880 =; 2100= – Press (orange) AC button 3. Substitute x = 800 into line of best fit equation: – Press SHIFT STAT (at 1) y = –1 145,8 + 4,32(800) – Press 5: Reg y = 2 310,2 – Press 1: A = and get –1145,792683 800 revolutions will generate 2 310,2 output This is the y-intercept of the regression line of horse power. – Press orange AC button – Press SHIFT STAT 4. Let y = 1 200 – Press 5: Reg 1 200 = –1 145,8 + 4,32 x – Now press 2: B = and get 4,318292683 1 200 + 1 145,8 = 4,32 x This is the gradient of the regression line 2 345,8 = 4,32 x Remember to Answer: 2 345,8 change the MODE The least squares regression line: ​  4,32 ​ = x ______ back to 1:COMP y = –1 145,8 + 4,32 x (correct to 2 decimal places) 543,0092593 = x when doing normal There are about calculations 543 revolutions per min for an output of 1 200 horse power. The scatter plot and the line of best fit show the trend in the relationship between the revolutions and the horse power. 2250 2000 1750 1500 Horsepower 1250 1000 750 500 250 0 300 350 400 450 500 550 600 650 700 750 800 Revolutions per minute 276 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 Activity 8 1. Pick ’n Pay wants to survey how long in seconds (y) it takes a teller to scan (x) items at the till. The table shows the results from 9 shoppers. Shoppers A B C D E F G H I x (no of items) 5 8 12 15 15 17 20 21 25 y (time in seconds) 3 11 9 6 15 13 25 15 13 a) Use your calculator to determine the equation of the line of best fit (the regression line or the least squares regression line) correct to two decimal places. (3) b) Calculate the value of r, the correlation coefficient for the data. What can you say about the correlation between x and y?(3) c) How long would the teller take to scan 21 items at the till? (2) d) How many items could a teller scan in 21,28 seconds? (2) 2. A restaurant wants to know the relationship between the number of customers and the number of chicken pies that are ordered. number of customers (x) 5 10 15 20 25 30 35 40 number of chicken pies (y) 3 5 10 10 15 20 20 24 a) Determine the equation of the regression line correct to two decimal places. (3) b) Determine the value of r, the correlation coefficient. Describe the type and strength of the correlation between the number of people and the number of chicken pies ordered.(3) c) Determine how many chicken pies 100 people would order. (2) d) If they only have 12 pies left, how many people can they serve? (2) [20] © Department of Basic Education 2015 UNIT 13 STATISTIC S 277 Mind the Gap Mathematics

13 Unit Solutions 1. a) A = 2,68 3 B = 0,62 3 y = 2,68 + 0,62x 3 (3) b) r = 0,62847…. = 0,63 33 This is a weak positive correlation 3 (3) c) y = 2,68 + 0,62(21) 3 = 15,7 (about 16 seconds) 3 (2) d) 21,28 = 2,68 + 0,62 x 3 21,28 – 2,68 = 0,62 x 18,6 0,62 ​= x ​ ____ 30 = x 30 items can be scanned in 21,28 seconds. 3 (2) 2. a) A = –0,39285… 3 B = 0,61190 3 y = –0,4 + 0,6 x 3 (3) b) r = 0,9866… 33 This is a very strong positive correlation 3 (r is close to +1) (3) c) y = –0,4 + 0,6 x y = –0,4 + 0,6(100) 3 y = 59,6 About 60 chicken pies are ordered by 100 3 people. (2) d) 12 = –0,4 + 0,6 x 3 12 + 0,4 = 0,6 x 12,4 ​  0,6 ​= x ____ 20,6… = x About 21 people will order 12 pies. 3 (2) [20] 278 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 Activity 8 (continued) 3. A recording company investigates the relationship between the number of times a CD is played by a national radio station and the national sales of the same CD in the following week. The data below was collected for random sample of CDs. The sales figures are rounded to the nearest 50 Number of times 47 34 40 34 33 50 28 53 25 45 CD is played Weekly sales of 3 950 2 500 3 700 2 800 2 900 3 750 2 300 4 400 2 200 3 400 the CD a) Identify the independent variable. (1) b) Draw a scatter plot of this data.  (3) c) Determine the equation of the least squares regression line. (3) d) Calculate the correlation coefficient. (2) e) Predict, correct to the nearest 50, the weekly sales for a CD that was played 45 times by the station in the previous week. (2) f) Comment on the strength of the relationship between the variables. (1) [12] Solutions 3. a) the number of times the CD is played 3 (1) b) Scatter plot showing the number of times a CD was played vs the CD sales in the following week 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 0 10 20 30 40 50 60 333 (3) c) a = 264,326 3 d) r = 0,95 3 3 (2) b = 75,21 3 y = 264,33 + 75,21x 3 (3) © Department of Basic Education 2015 UNIT 13 STATISTIC S 279 Mind the Gap Mathematics

13 Unit e) y = 264,33 + 75,21x(45) 3 (substitution) ≈ 3 648,78 ≈ 3 648 ≈ 3 650 (to the nearest 50) 3 (2) f) There is a very strong positive relationship between the number of times that a CD was played and the sales of that CD in the following week. 3 (1) [12] 280 UNIT 13 STATISTICS  © Department of Basic Education 2015  Mind the Gap Mathematics

Unit 13 What you need to be able to do: 1. Determine mean, median and mode in grouped or ungrouped data 2. Draw and analyse the following methods for representing data: • box-and-whisker plots • histograms • frequency polygons • cumulative frequency curves (ogives) • Calculate the variance and the standard deviation of a set of ungrouped data. • Comment on whether a data set is symmetric or skewed, by analysing the representation of the data. • Identify outliers in a set of data by looking at the box-and-whisker plot or scatterplot. • Determine the equation of the line of best fit of bivariate data using a calculator. (This line could be called the “least squares regression line”.) • Determine the regression correlation coefficient “r”. • Use the line of best fit to draw conclusions. Well done! © Department of Basic Education 2015 UNIT 13 STATISTIC S 281 Mind the Gap Mathematics

The Mind the Gap study guide series assists you to make the leap by studying hard to achieve success in the Grade 12 exam. This publication is not for sale. © Copyright Department of Basic Education www.education.gov.za This publication has a Creative Commons Attribution NonCommercial Sharealike license. Call Centre 0800 202 933