MATHSCLINIC SMARTPREP GR12 ENG 2023 V2.1

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VIDEO LESSONS! We have free video lessons for every topic in this book! These videos can only be accessed via Science Clinic's website - follow these steps to access this amazing resource! DOWNLOAD A BOOK Visit our website to download a SmartPrep book. We'll send you an email with download links. VIDEOS WEBPAGE Go to our Videos webpage. Log in with the email address that you used to download a book. SELECT YOUR TOPIC From your videos page, select a topic. You'll be redirected to a playlist with video lessons. ABOUT OUR SMARTPREP BOOKS Our SmartPrep books provide crystal-clear summaries of the essential theory for Grade 8-12 Core Mathematics and Grade 10-12 Physical Science. Our books are available in English and Afrikaans. COPYRIGHT NOTICE AND LICENSE The content in this SmartPrep Book is the original work of Science Clinic (Pty) Ltd. You may distribute this material, provided that you adhere to the conditions in our copyright notice and license: ¢ If you copy and distribute this book electronically or in paper form, you must keep this copyright notice intact. ¢ If you use questions or syllabus summaries from this book, you must acknowledge the author. ¢ This Smart Prep Book is meant for the benefit of the community and you may not use or distribute it for commercial purposes. ¢ You may not broadcast, publicly perform, adapt, transform, remix, add on to this book and distribute it on your own terms. ° By exercising any of the rights to this Smart Prep Book, you accept and agree to the terms and conditions of the license provided on this link: CONTENT ACKNOWLEDGEMENT Thank you to all the amazing people involved in the production, translation and moderation of this book: S Bouwer, E Britz, G Kyle, D Kotze, Q Meades, S Sapsford, S Stevens, G Swanepoel, GM van Onselen, L Vosloo QD ww sciencectinie.coza © exciencectinic EB ew acodook.comisciencectiniesa

TABLE OF CONTENTS CORE THEORY SUMMARIES Paper 1 Nature of Roots 2 Quadratic Equations 5 Exponents and Surds 6 Sequences and Series 10 Finance, Growth and Decay 15 Functions, Graphs and Inverses 20 Differential Calculus 29 Probability 34 Paper 2 Euclidean Geometry 39 Trigonometry 49 Analytical Geometry 57 Statistics 62 www

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Mathematics 9 DBE/2021 Examination Guidelines 4. ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY In order to have some kind of uniformity, the use of the following shortened versions of the theorem statements is encouraged. 4.1 ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY (ENGLISH) THEOREM STATEMENT ACCEPTABLE REASON(S) LINES The adjacent angles on a straight line are supplementary. ‘s on a str line If the adjacent angles are supplementary, the outer arms of these angles adj ‘s supp form a straight line. The adjacent angles in a revolution add up to 360q. ‘s round a pt OR ‘s in a rev Vertically opposite angles are equal. vert opp ‘s = If AB || CD, then the alternate angles are equal. alt ‘s; AB || CD If AB || CD, then the corresponding angles are equal. corresp ‘s; AB || CD If AB || CD, then the co-interior angles are supplementary. co-int ‘s; AB || CD If the alternate angles between two lines are equal, then the lines are alt ‘s = parallel. If the corresponding angles between two lines are equal, then the lines corresp ‘s = are parallel. If the co-interior angles between two lines are supplementary, then the coint ‘s supp lines are parallel. TRIANGLES The interior angles of a triangle are supplementary. ‘ sum in ' OR sum of ‘s in ∆ OR Int ‘s ' The exterior angle of a triangle is equal to the sum of the interior ext ‘ of ' opposite angles. The angles opposite the equal sides in an isosceles triangle are equal. ‘s opp equal sides The sides opposite the equal angles in an isosceles triangle are equal. sides opp equal ‘s In a right-angled triangle, the square of the hypotenuse is equal to the Pythagoras OR sum of the squares of the other two sides. Theorem of Pythagoras If the square of the longest side in a triangle is equal to the sum of the Converse Pythagoras squares of the other two sides then the triangle is right-angled. OR Converse Theorem of Pythagoras If three sides of one triangle are respectively equal to three sides of SSS another triangle, the triangles are congruent. If two sides and an included angle of one triangle are respectively SAS OR S‘S equal to two sides and an included angle of another triangle, the triangles are congruent. If two angles and one side of one triangle are respectively equal to two AAS OR ‘‘S angles and the corresponding side in another triangle, the triangles are congruent. If in two right-angled triangles, the hypotenuse and one side of one RHS OR 90°HS triangle are respectively equal to the hypotenuse and one side of the other, the triangles are congruent Copyright reserved Please turn over

Mathematics 10 DBE/2021 Examination Guidelines THEOREM STATEMENT ACCEPTABLE REASON(S) The line segment joining the midpoints of two sides of a triangle is Midpt Theorem parallel to the third side and equal to half the length of the third side The line drawn from the midpoint of one side of a triangle, parallel to line through midpt || to 2nd side another side, bisects the third side. A line drawn parallel to one side of a triangle divides the other two line || one side of ' sides proportionally. OR prop theorem; name || lines If a line divides two sides of a triangle in the same proportion, then the line divides two sides of ∆ in prop line is parallel to the third side. If two triangles are equiangular, then the corresponding sides are in ||| 's OR equiangular ∆s proportion (and consequently the triangles are similar). If the corresponding sides of two triangles are proportional, then the Sides of ∆ in prop triangles are equiangular (and consequently the triangles are similar). If triangles (or parallelograms) are on the same base (or on bases of same base; same height OR equal length) and between the same parallel lines, then the triangles (or equal bases; equal height parallelograms) have equal areas. CIRCLES The tangent to a circle is perpendicular to the radius/diameter of the tan A radius circle at the point of contact. tan A diameter If a line is drawn perpendicular to a radius/diameter at the point where line A radius OR the radius/diameter meets the circle, then the line is a tangent to the converse tan A radius OR circle. converse tan A diameter The line drawn from the centre of a circle to the midpoint of a chord is line from centre to midpt of chord perpendicular to the chord. The line drawn from the centre of a circle perpendicular to a chord line from centre A to chord bisects the chord. The perpendicular bisector of a chord passes through the centre of the perp bisector of chord circle; The angle subtended by an arc at the centre of a circle is double the ‘ at centre = 2 ב at circumference size of the angle subtended by the same arc at the circle (on the same side of the chord as the centre) The angle subtended by the diameter at the circumference of the circle ‘s in semi-circle OR is 90q. diameter subtends right angle OR 1 ‘ in ~ 2 If the angle subtended by a chord at the circumference of the circle is chord subtends 90q OR 90q, then the chord is a diameter. converse ‘s in semi-circle Angles subtended by a chord of the circle, on the same side of the ‘s in the same seg chord, are equal If a line segment joining two points subtends equal angles at two points line subtends equal ‘s OR on the same side of the line segment, then the four points are concyclic. converse ‘s in the same seg Equal chords subtend equal angles at the circumference of the circle. equal chords; equal ‘s Equal chords subtend equal angles at the centre of the circle. equal chords; equal ‘s Equal chords in equal circles subtend equal angles at the circumference equal circles; equal chords; equal ‘s of the circles. Copyright reserved Please turn over

Mathematics 11 DBE/2021 Examination Guidelines THEOREM STATEMENT ACCEPTABLE REASON(S) Equal chords in equal circles subtend equal angles at the centre of the equal circles; equal chords; equal ‘s circles. The opposite angles of a cyclic quadrilateral are supplementary opp ‘s of cyclic quad If the opposite angles of a quadrilateral are supplementary then the opp ‘s quad supp OR quadrilateral is cyclic. converse opp ‘s of cyclic quad The exterior angle of a cyclic quadrilateral is equal to the interior ext ‘ of cyclic quad opposite angle. If the exterior angle of a quadrilateral is equal to the interior opposite ext ‘ = int opp ‘ OR angle of the quadrilateral, then the quadrilateral is cyclic. converse ext ‘ of cyclic quad Two tangents drawn to a circle from the same point outside the circle Tans from common pt OR are equal in length Tans from same pt The angle between the tangent to a circle and the chord drawn from the tan chord theorem point of contact is equal to the angle in the alternate segment. If a line is drawn through the end-point of a chord, making with the converse tan chord theorem OR chord an angle equal to an angle in the alternate segment, then the line ‘ between line and chord is a tangent to the circle. QUADRILATERALS The interior angles of a quadrilateral add up to 360q. sum of ‘s in quad The opposite sides of a parallelogram are parallel. opp sides of ||m If the opposite sides of a quadrilateral are parallel, then the opp sides of quad are || quadrilateral is a parallelogram. The opposite sides of a parallelogram are equal in length. opp sides of ||m If the opposite sides of a quadrilateral are equal , then the quadrilateral opp sides of quad are = is a parallelogram. OR converse opp sides of a parm The opposite angles of a parallelogram are equal. opp ‘s of ||m If the opposite angles of a quadrilateral are equal then the quadrilateral opp ‘s of quad are = OR is a parallelogram. converse opp angles of a parm The diagonals of a parallelogram bisect each other. diag of ||m If the diagonals of a quadrilateral bisect each other, then the diags of quad bisect each other quadrilateral is a parallelogram. OR converse diags of a parm If one pair of opposite sides of a quadrilateral are equal and parallel, pair of opp sides = and || then the quadrilateral is a parallelogram. The diagonals of a parallelogram bisect its area. diag bisect area of ||m The diagonals of a rhombus bisect at right angles. diags of rhombus The diagonals of a rhombus bisect the interior angles. diags of rhombus All four sides of a rhombus are equal in length. sides of rhombus All four sides of a square are equal in length. sides of square The diagonals of a rectangle are equal in length. diags of rect The diagonals of a kite intersect at right-angles. diags of kite A diagonal of a kite bisects the other diagonal. diag of kite A diagonal of a kite bisects the opposite angles diag of kite Copyright reserved Please turn over

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Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © NUMBERS NATURE: Refers to the type of numbers that the roots are. Non-real (R’) (Sometimes referred to as imaginary) ROOTS: The x-intercepts/solutions/zeros of a quadratic equation. Real (R) Two real roots One real root No (0) real roots Rational (Q) Irrational (Q’) y y y Integers and Fractions EXAMPLE Integers (Z) Positive and negative whole numbers π 2 Whole numbers (N0) x 0+N All decimals that cannot x x be written as a fraction. Natural numbers (N) 1, 2, 3... OR OR OR y y y x x x QUADRATIC FORMULA THE DISCRIMINANT −b ± b 2 − 4ac x= −b ± b 2 − 4ac 2a x= 2a a = the coefficient of x2 DETERMINING THE NATURE OF THE ROOTS b = the coefficient of x Indicated by Δ . c = the constant term The DISCRIMINANT is used to determine the nature of the roots. 2 ∴ Δ = b − 4ac Used to factorise quadratic equations. Δ EXAMPLE EXAMPLE Δ<0 Δ≥0 3x 2 + 2x − 4 = 0 NO Real Real 3x 2 +2x − 4 = 0 a = 3 roots roots b = 2 a = 3 c = −4 b = 2 c = −4 Δ>0 Δ=0 2 unequal 2 equal, −b ± b 2 − 4ac Δ = b 2 − 4a c roots real roots x = 2a Δ = (2)2 − 4(3)(−4) −(2) ± (2) 2 − 4(3)(− 4) Then solve for Δ x = Δ = perfect square Δ ≠ perfect square 2(3) irrational roots rational roots Then solve for x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 2

Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © DISCRIMINANT b2 - 4 a c EXAMPLES NATURE OF ROOTS NUMBER OF REAL ROOTS (∆ = b2 - 4ac) a>0 a<0 y y x2 + x + 1 = 0 Δ = b2 – 4ac x = (1)2 – 4(1)(1) 0 =1–4 Non real =–3 a b c Δ<0 x y y x2 – 6x + 9 = 0 Δ = b2 – 4ac x = (–6)2 – 4(1)(9) Real (Δ = +) 1 (2 of the same) = 36 – 36 Rational (Δ = perfect =0 square) a b c Δ=0 Equal (Δ = 0) x x2 – 5x – 6 = 0 Δ = b2 – 4ac = (–5)2 – 4(1)(–6) Real (Δ = +) 2 y y = 25 + 24 Rational (Δ = perfect = 49 square) a b c Δ > 0 (perfect square) Unequal (Δ ≠ 0) x x 2x2 + 3x – 7 = 0 Δ = b2 – 4ac = (3)2 – 4(2)(–7) Real (Δ = +) 2 = 9 + 56 Irrational (Δ ≠ = 65 perfect square) a b c Δ > 0 (not perfect square) Unequal (Δ ≠ 0) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 3

Grade 12 Maths Essentials Nature of Roots SCIENCE CLINIC 2022 © DETERMINING THE NATURE FOR WHICH VALUES OF k PROVE THE NATURE OF THE ROOTS OF ROOTS WITHOUT WILL THE EQUATION HAVE The nature of the roots will be supplied and the discriminant can be used to prove the nature, with either one, or no, SOLVING THE EQUATION EQUAL ROOTS? unknown value. The roots of an equation can be deter- The discriminant (Δ) can be used to calculate Steps to prove the nature of roots (NO unknown): Steps to prove the nature of roots (ONE unknown): mined by calculating the value of the the unknown value of k. (e.g. Ask yourself, for discriminant (Δ). which values of k will the discriminant be 0?) 1. Put the equation in its standard form 1. Put the equation in its standard form 2. Substitute the correct values in and calculate the dis- 2. Substitute the correct values in and calculate the dis- Steps to determine the roots using Steps to determine the values of k using criminant criminant the discriminant: the discriminant: 3. Determine the roots and confirm whether they are as 3. Determine the roots and confirm whether they are as 1. Put the equation in its standard form 1. Put the equation in its standard form supplied supplied 2. Substitute the correct values in and 2. Substitute the correct values in and calculate EXAMPLE EXAMPLE calculate the discriminant the discriminant Prove the equation has two, unequal, irrational roots: For the equation x(6x – 7m) = 5m2, prove that the roots 3. Determine the nature of the roots of 3. Equate the discriminant to 0 and solve for k x2 = 2x + 9 are real, rational and unequal if m > 0 the equation (quadratic equation) EXAMPLE 1. Standard form 1. Standard form EXAMPLE x2 – 2x – 9 = 0 6x2 – 7mx – 5m2 = 0 Determine the nature of the roots of For which values of k the equation will have x2 = 2x +1 without solving the equa- a b c a b c equal roots? tion REMEMBER: Δ = 0 for equal roots 2. Calculate the discriminant 2. Calculate the discriminant 1. Standard form Δ = b2 – 4ac Δ = b2 – 4ac x2 = 2x +1 Δ = (–2)2 – 4(1)(–9) Δ = (–7m)2 – 4(6)(– 5m2) x2 – 2x – 1 = 0 1. Standard form Δ = 4 + 36 Δ = 49m2 + 120m2 x2 + 2kx = –4x – 9k Δ = 40 Δ = 169m2 a b c x2 + 2kx + 4x + 9k = 0 3. Determine the roots 3. Determine the roots 2. Calculate the discriminant a b c The Roots are: The Roots are: Δ = b2 – 4ac Real (Δ > 0) Real (Δ > 0) Δ = (–2)2 – 4(1)(–1) 2. Calculate the discriminant Unequal (Δ ≠ 0) Unequal (Δ ≠ 0) Δ=4+4 Δ = b2 – 4ac Irrational (Δ ≠ perfect square) Rational (Δ = perfect square) Δ=8 Δ = (2k + 4)2 – 4(1)(9k) Δ = 4k2 +16k +16 – 36k 3. Determine the nature of the Δ = 4k2 – 20k + 16 roots The Roots are: 3. Equate to zero (0) and solve for k Real (Δ > 0) 0 = 4k2 – 20k + 16 (÷ 4) Unequal (Δ ≠ 0) 0 = k2 – 5k + 4 Irrational (Δ ≠ perfect square) 0 = (k – 1)(k – 4) Therefore k = 1 or k = 4 k needs to either be 1 or 4 to ensure that the discriminant of the equation is 0 (the discriminant must be 0 in order for equal roots) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 4

Grade 12 Maths Essentials Quadratic Equations SCIENCE CLINIC 2022 © Quadratic Equations are equations of the second degree (i.e. the highest exponent of the variable is 2). The degree of the equation determines the maximum number of real roots/solutions/x-intercepts/zeros. The standard form of a quadratic equation is: a x 2 + bx + c = 0 wh ere a ≠ 0 SOLVING QUADRATIC EQUATIONS DIFFERENCE OF TWO COMPLETE THE FRACTIONS AND FACTORING QUADRATIC FORMULA ONE ROOT TWO ROOTS SQUARES SQUARE RESTRICTIONS 1. Find the LCD and list restrictions 1. Put in standard Substitute into the quadratic 1. Substitute the known 2. Solve for x 1. Write in standard form 1. Substitute the roots into form formula: 2 root −b ± b − 4a c 2. Move C across the equation 3. Check your answers against your 2. Apply the zero x = 2. Solve for the variable 2a 3. Divide both sides by A restrictions factor law 3. Substitute the value 2. Use “FOIL” for the 4. Add (½ x b)2 to both sides REMEMBER: 3. List possible where a = coefficient of x2, b = of the variable and quadratic equation 5. Factorise and solve 0 x solutions coefficient of x, c = constant term solve for the root = 0 BUT = u nd e f i ne d x 0 x Eg. x2 = –2x + 63 Eg. –3x2 = –12 + 7x Either method may be used Eg. x2 + 2x = 1 Eg. x2 + mx – 15 = 0, Eg. –9 and 7 are the Eg. x−2 = x −1 3 − 2 −2 x x2 + 2x – 63 = 0 –3x2 – 7x + 12 = 0 (½ . 2)2 where 5 is a root. roots of a quadratic x a = –3; b = –7; c = 12 Eg. x2 = 25 (1) (5)2 + (5)m – 15 = 0 equation x−2 = x −1 3 + x −2 2 Find factors of 63 so x2 – 25 = 0 x2 = ± 25 x 2 + 2x + 1 = 1 + 1 25 + 5m – 15 = 0 x = –9 or x=7 LCD: (x − 2)(x − 3) −(−7) ± (−7) 2 − 4(−3)(12) (x – 5)(x + 5) = 0 5m = –10 x+9=0 x–7=0 that F1 x F2 = –63 x = (x + 1)2 = ± 2 Restrictions: x − 2 ≠ 0; x − 3 ≠ 0 2(−3) x = 5 or x = –5 x = ±5 m = –2 and F1 + F2 = 2 x(x − 2)(x − 3) 1(x − 2)(x − 3) 2(x − 2)(x − 3) 7± 49 + 144 x + 1= ± 2 (x + 9)(x – 7) = 0 = + (x − 2) (x − 3) (x − 2) x = Therefore x = ±5 x2 – 2x – 15 = 0 x2 – 7x + 9x – 63 = 0 (x + 9)(x – 7) = 0 −6 x + 1= − 2 or x + 1= 2 (x – 5)(x + 3) = 0 7± 193 x2 + 2x – 63 = 0 x(x – 3) = 1(x – 2) + 2(x – 3) x + 9 = 0 or x – 7 = 0 x = −1− 2 x = −1+ 2 x=5 or x = –3 x = x2 – 3x = x – 2 + 2x – 6 x = –9 x=7 −6 (given) 7+ 193 7− 193 x2 – 3x – x – 2x + 2 + 6 = 0 x = or x = x2 – 6x + 8 = 0 −6 −6 (x – 4)(x – 2) = 0 x–4=0 or x–2=0 Answer in surd form or can be x=4 x=2 calculated/rounded off to 2 decimals Check restrictions: x ≠ 2, x ≠ 3 x = –3,48 OR x = 1,15 Thus, x = 4 is the only solution. Remember: General Steps for solving Quadratic Equations Cannot Sum * LHS = Left Hand Side * RHS = Right Hand Side factorise * Fractions and their restrictions 2 perfect squares Finding factors: Binomial Difference * Factor 1 + Factor 2 = b * Factor 1 x Factor 2 = c 2. Factor out Highest Zero factor law: 1. Standard Form Common Factor If A x B = 0 then A, B or both = 0 Trinomial Factoring EXAMPLE Factorise Quadratic Formula x.y=0 x = 0 or y = 0 or (x + 1)(x – 3) = 0 Complete the x + 1 = 0 or x – 3 = 0 square Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 5

Grade 12 Maths Essentials Exponents and Surds SCIENCE CLINIC 2022 © WHAT ARE: LAWS OF EXPONENTS Exponents: Exponents occur when multiplying or dividing expressions/ Laws of exponents only apply to multiplication, division, brackets and roots. NEVER adding or subtracting bases/variables numerous times by similar expressions/bases/variables Surds: A surd is the Mathematical terminology for irrational roots, when Algebraic Notation Exponential Notation Exponential Law in operation numbers are left in “root-form” as opposed to rounding them off to a deci- mal place. 1 16 = 2 × 2 × 2 × 2 16 = 24 When we MULTIPLY the SAME bases we ADD the exponents. HELPFUL HINTS FOR EQUATIONS\EXPRESSIONS F 1. Express larger numbers in exponential form by prime factorising 2 64 26 When we DIVIDE the SAME bases we MINUS the exponents (always top = 4 = 22 2. Remove a common factor if two unlike terms are separated by a +/- 16 24 minus bottom). 3. Ensure your surds are always expressed in their simplest form 3 When we have the exponents outside the BRACKETS we DISTRIBUTE 4. Express surds in exponential form for simplification 43 = 64 (22 )3 = 26 5. Take note of the following: them into the brackets. A common error, when solving for an unknown base with a fraction as 4 64 26 an exponent, is to multiply the exponents on both sides by the unknown = 1 = 20 = 1 Any base to the POWER OF ZERO is equal to one. (But 00 is undefined). 64 26 exponent’s inverse (so that the exponent will be 1). However, if you ex- press these fractions as surds, you will notice the following: 5 3 3 a. An even power will always produce a positive AND negative solution 64 = 4 26 = 22 The POWER inside the root is DIVIDED by the size of the root. 4 x3 = 3 6 4 × 9 = 36 22 × 32 = 62 3 x4 = 3 When we have non-identical bases, but identical exponents, we keep the exponents and multiply the bases. (This same rule will also apply for x4 = 27 1 1 1 7 2× 3= 6 division). x = ± 4 27 2 2 × 32 = 6 2 b. A negative number inside an even root cannot solve for a real solution 1 1 8 2× 2= 4= 2 1 2 2 × 2 2 = 21 Any square root multiplied by itself will equal the term inside the root. (−2) 2 = x −2 = x CONVERTING SURDS INTO OPERATIONS WITH SURDS No real solution EXPONENTIAL FORM Steps for working with surds: c. An unknown inside an even root cannot solve for a negative solution (AND VICE VERSA) 1. Express the surd in its simplest surd form 3 x4 = −2 The power inside the root becomes 2. Identify like terms (+ and −) or use Laws of Exponents ( × and ÷ ) 4 x3 = − 2 the NUMERATOR and the size of the Note: If you use your calculator, make sure to show the changes you made root becomes the DENOMINATOR. i.e. 50 = 25 × 2 = 25 × 2 = 5 2 No real solution D N EXAMPLE 1 EXAMPLE 2 EXAMPLE 3 ADDING AND SUBTRACTING LIKE-TERMS xN = x D Simplify Simplify A rectangle has a length of Like terms are terms in an equation/expression that have identical 31 7 + 1 and a width of variables and exponents. To add/subtract these, simply add/subtract EXAMPLE 50 + 3 18 − 98 5 81 × 4 27 3 20 = 7 − 1. Determine the their coefficients. Exponents never change when the operator is +/- 1. 5 x 2 = 5 2+ 9 2−7 2 5 9 length of the diagonal. 9× 3 3 10 2 2 2 = 7 2 ( 7 + 1) + ( 7 − 1) = r EXAMPLE 2 = x5 5 34 × 4 33 13 = = 3 20 5 32 × 3 7 + 2 7 + 1 + 7 − 2 7 + 1 = r2 1. 3x 2 y 4 − 5x 3 y + 2x 2 y 4 + x 3 y = 5x 2 y 4 − 4x 3 y 3 2. x 4 4 3 16 = r 2 35 × 34 20 = 2 1 = 313 4= r 2. 3 2 + 5 3 − 8 2 + 3 = −5 2+ 6 3 = 4 x3 35 × 32 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 6

Grade 12 Maths Essentials Exponents and Surds SCIENCE CLINIC 2022 © RATIONALISING THE DENOMINATOR The process of finding an equivalent fraction that can be expressed without a surd in the denominator Steps for rationalising monomial denominators: Steps for rationalising binomial denominators: 1. Multiply the numerator and denominator by the 1. Multiply numerator and denominator by the denominator’s surd binomial in the denominator with the opposite sign 2. Simplify (conjugate) 2. Simplify EXAMPLE 1 Why do we do this? Express the following with rational Multiplying the binomial by itself will give us a trino- denominators: mial with an irrational middle term. To avoid this, we multiply the binomial by its conjugate (same 3 6+ 3 2 binomial with the opposite sign) to create a differ- 1. 2. 7 2 3 ence of two squares. 3 7 6+ 3 2 3 = × = × EXAMPLE 1 7 7 2 3 3 Express the following fractions with rational 3 7 6 3+ 3 6 denominators: = = 7 2×3 3 7 1. 2. 6 3+ 3 6 5− 7 x− 1 = x 6 5+ 7 3 1 = × x+ 2 3+ 6 7 x = 5− 7 5+ 7 = × 2 x− 1 x+ 1 15 + 3 7 x x = EXAMPLE 2 25 − 7 7 7 x+ x2 + 2 15 + 3 7 x If x = 3 + 2, simplify: and express the an- = = x −2 18 x − 1x swer with a rational denominator 5+ 7 7x + 7 x2 + 2 = 1. 6 x x −2 = x2 − 1 ( 3 + 2)2 + 2 x = ( 3 + 2) − 2 7x + 7 x2 − 1 = ÷ x x 3+ 4 3+ 4+ 2 = 7(x + 1) x 3 = × x (x + 1)(x − 1) 9+ 4 3 3 = × 7x x 3 3 = × x (x − 1) x 9 3+ 4⋅3 = 7x x 3 = x (x − 1) = 3 3+ 4 7 x = (x − 1) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 7

Grade 12 Maths Essentials Exponents and Surds SCIENCE CLINIC 2022 © FACTORISING Factorising is the opposite of distribution, which means that you will subtract the exponents when “taking out” factors. There are 6 different types of factorisation. 1. Common Factor: 2. Difference of two squares: 3. Sum or difference of two cubes: Remove the highest common factor from the coefficients and Applied when there are two perfect squares separated by a ‘−’ sign. Applied when there are two perfect cubes separated by a + /−.The common variables. The square root of both terms will be in both pairs of brackets, one final answer will be a binomial in the one bracket and a trinomial in with a + and the other with a − the other. EXAMPLES A perfect square is a term whose number will not leave an irrational A perfect cube is a term whose number will not leave an irrational solution once square-rooted, and whose exponents are divisible by 2. solution once cube-rooted, and whose exponents are divisible by 3. Factorise the following: EXAMPLES EXAMPLES 4x 5 8x 3 16x 2 Factorise the following: 1. 3x 5 y 4 + 9x 3y 5 − 12x 2 y 4 2. 3 − + Factorise the following: 9y 27y 2 3y = 3x 2 y 4(x 3 + 3x y − 4) 3y ( 3y 2 9y ) 4x 2 x 3 2x 1. x 3 − 8 = + + 4 1. 9x 2 − 4y 6 2. x 4 − 16 = (x − 2)(x 2 + 2x + 4) 3 3 2 2 = (3x + 2y )(3x − 2y ) = (x + 4)(x − 4) 4. Exponential Factorising: = (x 2 + 4)(x + 2)(x − 2) x2 − 7 2. 27x 6 + 64y 9 Similar to common factorising (1). Remove the highest common 3. factor, in this case, a base with its exponent(s). Exponents are sub- x+ 7 = (3x 2 + 4y 3)(9x 4 − 12x 2 y 3 + 16y 6 ) tracted from the same bases. (x + 7)(x − 7) 4. a 2 + 2a b + b 2 − x 2 6. Trinomials: = EXAMPLES x+ 7 = (a + b)2 − x 2 Note: Ratio of exponents of term 1 to term 2 is 2:1. A combination of factors of term 1 and term 3 must give you term 2. Factorise the following: = (a + b + x)(a + b − x) = x− 7 EXAMPLES 9 x+ 2 − 32 x 5. Grouping: Factorise the following: (Q2 - Q6 are conceptually the same) 1. 2x+ 3 − 2x+ 1 2. x 3 3 ⋅ 2 × 3x ⋅ 5 Remove the common binomial factor from the expression = 2x (23 − 2) (32 ) x+ 2 − 32 x = EXAMPLES 1. 3x 2 − 5x − 2 2. x 2 + 3x − 10 = 2x ⋅ 6 32 x ⋅ 8 ⋅ 5 Factorise the following: = (3x + 1)(x − 2) = (x + 5)(x − 2) 32 x+ 4 − 32 x = 32 x ⋅ 40 1. x (y − 4) + 3(y − 4) 2. a 2 + 2a b + b 2 − 3a − 3b 5x − 5x−2 2 1 3. 2x 3 (3 − 1)4 3. x 4 + 3x 2 − 10 4. x 3 + 3x 3 − 10 2 ⋅ 5x − 5x = = (y − 4)(x + 3) 2 = (a + b) − 3(a + b) 32 x ⋅ 40 1 1 5x (1 − 5−2 ) = (a + b)(a + b − 3) = (x 2 + 5)(x 2 − 2) = (x 3 + 5)(x 3 − 2) = 80 5x (2 − 1) = 40 3. 5x − 15y + 9a y − 3a x 1 1 − 25 5. 52 x + 3 ⋅ 5x − 10 6. 32 x + 3x+ 1 − 10 = = 2 = 5(x − 3y) + 3a(3y − x) 1 = 5(x − 3y) − 3a(x − 3y) = (5x + 5)(5x − 2) = 32 x + 3 ⋅ 3x − 10 24 = 25 = (x − 3y)(5 − 3a) = (3x + 5)(3x − 2) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 8

Grade 12 Maths Essentials Exponents and Surds SCIENCE CLINIC 2022 © EQUATIONS 1. Linear Equations: 3. Simultaneous Equations: 5. Exponential Equations: Move all the variables to the one side, and the constants to the Solve for two unknowns in two different equations using the substitu- Make sure that you get a term on the one side of the equation that other to solve. Linear equations have only one solution. tion method. Remember to solve for both unknowns by substituting has a base that is equal to the base with the unknown exponent. them back into the original equation. Then, drop the bases, equate the exponents and solve. EXAMPLES EXAMPLES Solve: Hints: Solve: • NEVER drop the base if the terms are separated by a + or − 1. 3(x − 2) + 10 = 5 − (x + 9) 2. (x − 2)2 − 1 = (x + 3)(x − 3) - Remove common factors until the equation is in its simplest 1. Equation 1: 2x + 3y = 18 2. Equation 1: y + 3x = 2 3x − 6 + 10 = 5 − x − 9 x 2 − 4x + 4 − 1 = x 2 − 9 form and then solve Equation 2: − 3x + 5y = 11 Equation 2: y 2 − 9x 2 = 16 • Always convert decimals to fractions and then to bases with 3x + 4 = − x − 4 −4x + 3 = − 9 negative exponents From 1: 2x + 3y = 18 From 1: y + 3x = 2 4x = − 8 −4x = − 12 2x = − 3y + 18 y = − 3x + 2...1a x = −2 x = 3 −3y + 18 x = ....1a Sub 1a into 2: y 2 − 9x 2 = 16 EXAMPLES 2 (−3x + 2)2 − 9x 2 = 16 1. 4x = 8 2. 0,0625x = 64 2. Quadratic Equations: Sub 1a into 2: −3x + 5y = 11 1 x ( 16 ) 9x 2 − 12x + 4 − 9x 2 = 16 22 x = 23 ( ) Move everything to one side and equate to zero. By factorising the −3y + 18 = 26 −3 + 5y = 11 trinomial, you should find two solutions. 2 −12x = 12 2x = 3 1 x ( 24 ) 9y − 54 x = − 1...3 3 = 26 EXAMPLES + 5y = 11 x = 2 2 Sub 3 into 1: y + 3(−1) = 2 Solve: (Q3 - Q6 are the most likely exam-type questions) 9y − 54 + 10y = 22 2−4x = 26 y = 5 1. x 2 + 5 = 6x 2. (3x − 4)(5x + 2) = 0 19y = 76 3. 2 ⋅ 3x+ 1 + 5 ⋅ 3x = 33 −4x = 6 (−1; 5) −3 x 2 − 6x + 5 = 0 3x = 4 or 5x = − 2 y = 4....3 3x (2 ⋅ 3 + 5) = 33 x = 4 −2 2 (x − 5)(x − 1) = 0 x = or x = Sub 3 into 1: 2x + 3(4) = 18 3x (11) = 33 3 5 x = 5 or x = 1 2x = 6 3x = 31 9 x = 3 x = 1 5. 0,5x ⋅ 1+ = 10 2 1 16 3. x 4 + 3x 2 − 10 = 0 4. x 3 + 3x 3 − 10 = 0 (3; 4) 1 x (2) 25 1 1 4. 273x+ 1 = 812 x+ 5 ⋅ = 10 (x 2 + 5)(x 2 − 2) = 0 (x 3 + 5)(x 3 − 2) = 0 4. Surd Equations: 16 1 1 Isolate the surd on the one side of the equation. Power both sides of (33)3x+ 1 = (34 )2 x+ 5 5 x2 = − 5 or x 2 = 2 x3 = − 5 or x 3 = 2 2−x ⋅ = 10 the equation by the root. Ensure that you check your solutions by 39x+ 3 = 38x+ 20 4 N/A or x = ± 2 x = − 125 or x = 8 substituting your answers back into the original equation. 2−x = 8 9x + 3 = 8x + 20 EXAMPLES 2−x = 23 x = 17 5. x + 3 x − 10 = 0 6. 22 x − 6 ⋅ 2 x − 16 = 0 Solve: −x = 3 1 1. x −2 = 3 2. x + 5−x = 3 x+ 3x 2 − 10 = 0 (2 x + 2)(2 x − 8) = 0 x = −3 1 1 x −2 = 9 x+ 5= x+ 3 Check: (x 2 + 5)(x 2 − 2) = 0 2x = − 2 or 2 x = 8 L HS = (−1) + 5 − (−1) 1 1 x = 9+ 2 x+ 5= x2 + 6x + 9 L HS = 3 R HS = 3 x2 = − 5 or x 2 = 2 N/A or 2 x = 23 ∴ x = −1 x = 11 0= x2 + 5x + 4 L HS = (−4) + 5 − (−4) x = − 5 or x = 2 x = 3 L HS = 5 R HS = 3 0 = (x + 1)(x + 4) N/A or x = 4 ∴ x ≠ −4 x = − 1 or x ≠ − 4 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 9

Grade 12 Maths Essentials Sequences and Series SCIENCE CLINIC 2022 © Patterns/ Sequences: ordered set of numbers REMINDERS: Linear: Quadratic: F Constant first difference between consecutive terms. Constant second difference between consecutive terms. 1. Consecutive: directly follow one another Tn = general term Tn = general term Tn = d n+ c Tn = a n2 + bn+ c 2. Common/constant difference: difference d = constant difference n = number of the term between two consecutive terms in a pattern Notice how this is similar to a linear function y = m x + c Notice how this is similar to the quadratic equation and formula for the parabola d = T2 − T1 Steps to determine the nth term: Steps to determine the nth term: 3. General term Tn: 1. Find the constant difference 1. Find the constant second difference Also referred to as the nth term. 2. Substitute the constant difference (d) and the term value, along with the 2. Use the value of the second difference to find “a” term number • General term for linear patterns: 3. Use the “a” value and first difference to find “b” 3. Substitute the c- and d-values to define the nth term. Tn = d n + c 4. Use “a” and “b” to find “c” • General term for quadratic patterns: Tn = a n2 + b n + c EXAMPLE EXAMPLE 1. Determine the nth term of the following sequence: Determine the nth term of the following sequence: 4. T1; T2; . . . . T100: Terms indicated by T T1 T2 T3 T4 T1 T2 T3 T4 and the number of the term as a sub- Term 1 2; 7; 12; 17 6; 17; 34; 57 script. 7−2 12 − 7 17 − 12 (a + b + c) 17 − 6 34 − 17 57 − 34 5. Objective: First difference 5 5 5 11 17 23 (3a + b) a. Find the values of the variables. 17 − 11 23 − 17 b. Use the values to find the general Second difference term Using term 3 where T3 = 12 6 6 (2a) c. Use the general term to calculate Tn = 5n + c OR Tn = a + (n− 1)d specific term values 12 = 5(3) + c = 2 + (n− 1)(5) Second difference = 2a d. Use specific term values to find the 12 = 15 + c = 2 + 5n− 5 6 = 2a term number 12 − 15 = c Tn = 5n− 3 3= a −3 = c First difference = 3a + b 11 = 3a + b ∴ Tn = 5n− 3 11 = 3(3) + b 2= b 2. Determine the 100th term T100 = 5(100) − 3 Term 1 = a + b + c = 500 − 3 6 = (3) + (2) + c = 497 6−3−2 = c 1= c ∴ T100 = 497 ∴ Tn = 3n2 + 2n + 1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 10

Grade 12 Maths Essentials Sequences and Series - Arithmetic Sequence SCIENCE CLINIC 2022 © ARITHMETIC SEQUENCE EXAMPLE 2 EXAMPLE 3 A sequence formed by adding a common difference (d) to the previous term m ; 2m + 2 and 5m + 3 are three consecutive terms of an arithmetic Find the arithmetic sequence if the 6th term is 10 sequence. and the 14th term is 58. ∴ d = T2 − T1 = T3 − T2 Tn = general term a) Determine the value of m. For n = 6; Tn = a + (n− 1)d n = number of the term b) Determine the sequence. Tn = a + (n− 1)d 10 = a + 5d . . . . . . . . . . . eq1 a = T1 Steps to determine the nth term: a) Arithmetic sequence, thus: d = first difference 1. Find the constant difference (d) T2 − T1 = T3 − T2 For n = 14; 2. Use T1 as “a” (2m + 2) − m = (5m + 3) − (2m + 2) Tn = a + (n− 1)d m + 2 = 3m + 1 58 = a + 13d . . . . . . . . . . . eq 2 −2m = − 1 1 EXAMPLE 1 m = 2 eq2 − eq1; Given: –6; –2; 2; 6; 10; ...; 110. Determine: a + 13d = 58 b) d = T2 − T1 −a + 5d = 10 a) the general term. 8d = 48 = (2m + 2) − m b) which term in the sequence exceeds 84. ∴d = 6 = 2( 1 ) + 2 − 1 c) the value of the 14th term. 2 2 5 d) How many terms there are in the sequence. = b into eq1; 2 10 = a + 5d Tn = a + (n− 1)d 10 = a + 5(6) –6; –2; 2; 6; 10; ...; 110 28 = a + (12 − 1)( 5 ) a = −20 –2–(–6) 2–(–2) 6–2 10–6 2 a+ ( 2 ) 55 28 = 4 4 4 4 First difference (d) NOTE: If asked for sequence, it means the 1 first three terms a = 2 a) Tn = a + (n− 1)d (2) 1 5 Tn = − 6 + (n− 1)4 ∴ Tn = + (n− 1) Tn = a + (n− 1)d 2 Tn = 4n− 10 T1 = −20 + (1 − 1)(6) NOTE: If asked for sequence, it means the first three terms T1 = −20 b) 4n− 10 > 84 4n > 94 T1 = m n > 23,5 1 Tn = a + (n− 1)d T1 = ∴ 24th term 2 T2 = −20 + (2 − 1)(6) T1 = −14 c) Tn = 4n− 10 T2 = 2m + 2 T14 = 4(14) − 10 T2 = 2( 1 ) + 2 Tn = a + (n− 1)d 2 T14 = 46 T2 = 3 T1 = −20 + (3 − 1)(6) T1 = −8 T3 = 5m + 3 d) Tn = 4n− 10 5( ) + 3 1 110 = 4n− 10 = 2 4n = 120 ∴ − 20; − 14; − 8 11 n = 30 = 2 1 11 ∴ ; 3; 2 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 11

Grade 12 Maths Essentials Sequences and Series - Geometric Sequence SCIENCE CLINIC 2022 © GEOMETRIC SEQUENCE EXAMPLE 2 EXAMPLE 3 A sequence formed by multiplying the previous term 4g–2 ; g+1 ; g–3 are the first three terms of a The 6th term of a geometric sequence is 3, and the 11th term is 27. by a common ratio (r). geometric sequence. Determine the sequence. a) Determine the value of g if g is an integer and hence show Tn = general term that the first three terms are 18; 6; 2. T2 T For n = 6; ∴r = = 3 n = number of the term T1 T2 b) Write down the nth term of the sequence Tn = a ⋅ r n−1 a = T1 3 = a ⋅ r 5 . . . . . . . . . . . eq1 Tn = a ⋅ r n−1 r = common ratio T2 T3 a) r = = For n = 11; T1 T2 Steps to determine the nth term: g + 1 = g −3 Tn = a ⋅ r n−1 4g − 2 g + 1 1. Find the common ratio 27 = a ⋅ r 10 . . . . . . . . . . . eq 2 (g + 1)(g + 1) = (g − 3)(4g − 2) 2. Use T1 as “a” g 2 + 2g + 1 = 4g 2 − 14g + 6 3g 2 − 16g + 5 = 0 eq2 ÷ eq1; EXAMPLE 1 (3g − 1)(g − 5) = 0 ar 10 27 = Given sequence 6; 18; 54; ....; 118 098. Determine: ar 5 3 1 a) the next 2 terms ∴g ≠ (not an integer) or g = 5 33 3 r5 = 1 b) the nth term 32 T1 = 4g − 2 3− 1 c) how many terms there are in the sequence. = 4(5) − 2 r5 = 3 2 5 = 18 r5 = 32 1 6; 18 18; 54 54; ....; 118 098 T2 = g + 1 r = 32 = 3 6 18 = (5) + 1 = 6 3 3 r into eq1; a) r = 3 T3 = g −3 3 = a ⋅ r5 = (5) − 3 5 Tn = a ⋅ r n−1 Tn = a ⋅ r n−1 = 2 3 = a ⋅ ( 3) T4 = 6 ⋅ 34−1 T5 = 6 ⋅ 35−1 1 ∴ 18; 6; 2 a = ( 3) T4 = 162 T5 = 486 4 T2 1 b) a = b) r = 9 Tn = a ⋅ r n−1 T1 6 = 6 ⋅ 3n−1 = 18 Tn = a ⋅ r n−1 Tn = a ⋅ r n−1 Tn = a ⋅ r n−1 = 2 × 3 ⋅ 3n−1 1 = 1−1 2−1 3−1 ( 9 )( 3 ) ( 9 )( 3 ) ( 9 )( 3 ) 1 1 1 = 2 ⋅ 3n 3 T1 = T2 = T3 = 1 1 c) Tn = 2 ⋅ 3n Tn = a ⋅ r n−1 T1 = 9 T2 = 3 T3 = 3 9 118 098 = 2 ⋅ 3n 1 n−1 = 18 ⋅ ( ) 59 049 = 3n 3 310 = 3n or n = log3 59 049 = 2 × 32 × 3−n+ 1 1 3 1 ∴n = 10 ∴n = 10 ∴ ; ; = 2 ⋅ 33−n 9 9 3 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 12

Grade 12 Maths Essentials Sequences and Series - Series SCIENCE CLINIC 2022 © A series is the sum of a number of terms in a sequence and is represented by Sn. General Term for an arithmetic series: ARITHMETIC SERIES Tn = a + (n− 1)d PROOF: Sum of the arithmetic series: n Sn = a + [a + d ] + [a + 2d ] + ... + [a + (n− 3)d ] + [a + (n− 2)d ] + [a + (n− 1)d ] Sn = [2a + (n− 1)d ] 2 + Sn = [a + (n− 1)d ] + [a + (n− 2)d ] + [a + (n− 3)d ] + ... + [a + 2d ] + [a + d ] + a Last term: 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 ] + [2a + (n− 1)d ] l = a + (n− 1)d ∴ 2Sn = n[2a + (n− 1)d ] n n ∴ Sn = [2a + (n− 1)d ] ∴ Sn = [a + l ] 2 2 EXAMPLE 1 EXAMPLE 2 Determine the sum of the following: The sixth term of an arithmetic sequence is 23 and the sum of the first six terms is 78. Determine the sum of the first twenty-one terms. a) The first 25 terms of 18 + 13 + 8 + ... b) The series 5 + 8 + 11 + ... + 35 n 1. Tn = a + (n− 1)d 3. q2 − q1 : 5. Sn = 2 [2a + (n− 1)d ] d = −5 a = 5; d = 3; l = 35 23 = a + (6 − 1)d 23 = a + 5d 21 − 26 = 2a + 5d S21 = [2(3) + (21 − 1)(4)] 23 = a + 5d . . . . . . . . . . . . . . . . . eq1 2 n 3 = a Sn = [2a + (n− 1)d ] Find n : = 903 2 25 n S25 = [2(18) + (25 − 1)(−5)] 35 = 5 + (n− 1)(3) Sn = n [a + l ] 2. Sn = 2 [2a + (n− 1)d ] 4. a into eq1 : 2 2 6 23 = a + 5d = −1050 30 = 3(n− 1) 11 78 = [2a + (6 − 1)d ] S11 = [5 + 35] 2 23 − (3) = 5d 10 = n− 1 2 = 220 26 = 2a + 5d . . . . . . . . . . . . . . . . . eq 2 d = 4 ∴n = 11 GEOMETRIC SERIES PROOF: General Term for a geometric series: Sn = a + ar + ar2 + a r3 + ... + a r n−2 + a r n−1 Tn = a ⋅ r n−1 r Sn = ar + ar2 + a r3 + ... + a r n−2 + a r n−1 + arn Sum of the geometric series: Sn − r Sn = a − arn a (1 − r n) a (r n − 1) Sn(1 − r) = a (1 − r n) Sn = if r < 1 OR Sn = if r > 1 (1 − r) (r − 1) a(1 − r n) ∴ Sn = (1 − r) EXAMPLE 1 EXAMPLE 2 Given the series –2 +6 –18 +54 +.... Determine: 3 1533 The sum of the geometric series 6 + ⋯ + is . a) the sum to nine terms b) the value of n if the sum of the series is –797 162. 128 128 Determine the common ratio and the number of terms in the sequence. T2 6 a(1 − r n) a(1 − r n) r r = = = −3 Sn = Tn = a r n−1 Sn = rn = (1 − r) 256 T1 −2 3 (1 − r) −2(1 − (−3) n) = 6r n−1 1 ( 256 ) 1 n (2) 128 2 a(1 − r n) −797 162 = 6 1− r = Sn = (1 − (−3)) 1 256 (1 − r) = r n−1 1533 = ( ) 1 n (2) 256 −3 188 648 = −2 1 − (−3) n 128 1−r = 1 −2(1 − (−3)9 ) 1 768(1 − 256 ) S9 = 1 594 324 = 1 − (−3) n r n ⋅ r −1 = 15339(1 − r) = r 512 (1 − (−3)) 256 2−n = 2−9 = −9 842 (−3) n = −1 594 323 r n⋅ r 1 = 1 1533 − 1533r = 768 − 3r 256 ∴n = 9 (−3) n = (−3)13 r 1530r = 765 rn = 1 ∴n = 13 256 r = 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 13

Grade 12 Maths Essentials Sequences and Series - Series SCIENCE CLINIC 2022 © SUM TO INFINITY (S∞) SIGMA NOTATION Consider the following values of Sn as n →∞, in the following series: Sigma notation is denoted by Σ, which means ‘the sum of’. 1. 2+ 4+ 6+ 8+ ⋯= ∞ (Arithmetic and d = 2) end term EXAMPLE 7 Start term : 3 2. 2+ 0−2−4−6+ ⋯= ∞ (Arithmetic and d = − 2) Σ Formula for the series (T ) n ∑ 2n End term : 7 3. 2 + 4 + 8 + 16 + ⋯ = ∞ (Geometric and r = 2) start term n= 3 Series formula : Tn=2n 4. 2 − 4 + 8 − 16 + 32 + ⋯ = ∞ (Geometric and r = − 2) r ∑7n= 3 2n = 2(3) + 2(4) + 2(5) + 2(6) + 2(7) ∑ (Tn) 1 1 5. 8+ 4+ 2+ 1+ 2 + ⋯= ∞ (Geometric and r = 2 ) n= p = 6 + 8 + 10 + 12 + 14 = 50 Note for 5: T1 = a n Sn = [a + l ] a=6 n= r −p + 1 2 8(1 − ( 1 ) 20 ) 8(1 − ( 1 )30 ) 8(1 − ( 1 )40 ) 8(1 − ( 1 )80 ) = 5 [6 + 14] d=2 2 2 2 2 S20 = S30 = S40 = S80 = 3 1− 1 1− 1 1− 1 1− 1 Find the first three terms to check if arithmetic = 50 n=r–p+1 2 2 2 2 = 15,9999 = 15,9999 = 16 = 16 or geometric. Then use formula for Sn. =7–3+1 ∴ S∞ = 16 EXAMPLE 1 EXAMPLE 2 Thus only a geometric series with −1 < r < 1, r ≠ 0, , will have a sum to infinity, as S∞ Write the following in sigma Determine the following: will approach a set value. This is known as a convergent series, or a series that notation: ∞ 18 converges. 10 2−n ∑ ∑ a) b) 2 − 5n 2 + 10 + 50 + ⋯ + 781 250 1 1 n= 1 n= 4 If r = , p ∈ N, then as lim = 0, therefore 1 − r ∞ = 1 − 0 = 1, and therefore; p p→∞ p a First find the first three terms to check if arithmetic or geometric. r = 5; a = 2 S∞ = 1−r 10 + 1 + 0,1 + ⋯ −18 − 23 − 28 ∴ Tn = 2 ⋅ 5n−1 EXAMPLE 1 EXAMPLE 2 ∴ n = 0,1 ∴ d = −5 · For which value(s) of k will the series Use the sum to infinity to write 0,4 as 4(k − 2) + 8(k − 2)2 + 16(k − 2)3 + ⋯ converge? a proper fraction. Tn = 2 ⋅ 5n−1 S∞ = a n = r −p + 1 1−r 781 250 = 2 ⋅ 5n−1 10 = 18 − 4 + 1 S∞ = = 15 T2 · 0,4 = 0,4 + 0,04 + 0,004 + 0,0004 + ⋯ 390 625 = 5n−1 1 − 0,1 r = S∞ = 11,11 T1 58 = 5n−1 n 8(k − 2) 2 ∴n = 9 Sn = [2a + (n− 1)d ] 2 r = T2 4(k − 2) r = 15 T1 = [2(−18) + (15 − 1)(−5)] r = 2(k − 2) 2 9 r = 0,04 2 ⋅ 5n−1 = −795 ∑ 0,4 ∴ n= 1 −1 < r < 1 , r ≠ 0 r = 0,1 −1 < 2(k − 2) < 1 , 2(k − 2) ≠ 0 1 1 − < k −2 < , k −2 ≠ 0 S∞ = a 2 2 1−r 1 1 < k < 21 , k ≠ 2 = 0,4 2 2 1 − 0,1 4 = 9 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 14

Grade 12 Maths Essentials Finance - Simple and Compound Interest SCIENCE CLINIC 2022 © SIMPLE INTEREST A = accumulated amount COMPOUND INTEREST EXAMPLE REMINDERS: A = P(1 + in) P = original amount A = P(1 + i )n Calculate the future value of your investment after n = number of periods three years at an interest rate of 15% per annum 1.Inflation: OR OR compounded: r = interest rate as a % The rate at which prices increase r r n A = P(1 + 100 ) A = P(1 + 100 ) r over time n i = interest rate a) Annually 100 A = P (1 + i ) n 2.Consumer Price Index (CPI): = 15 000(1 + (0,15))3 the average prices of a basket of EXAMPLE = R 22 813,13 goods Determine the difference in the accumulated amounts when investing your savings of R 15 000 for 4 years b) Semi-annually 3.Exchange Rates: at two different banks, both offer 6,5% however one offers simple interest and the other compound. A = P (1 + i ) n The value of one currency for the 6 15 000(1 + ( 2 )) 0,15 purpose of conversion to another SIMPLE INTEREST = A = P (1 + i n) = R 23 149,52 4.Population Growth: Change of population size over A = 15 000(1 + (0,065)(4)) c) Quarterly MONEY time A = R18 900 ere st A = P (1 + i ) n int 12 15 000(1 + ( 4 )) nd 0,15 5.Hire Purchase: ou = COMPOUND INTEREST omp C t Short term loan, deposit payable. interes = R 23 331,81 A = P (1 + i ) n Simple Calculated using simple interest. A = 15 000(1 + (0,065))4 TIME d) Monthly 6.Reducing balance loan: A = R19 296,99 A = P (1 + i ) n Interest is paid on the reducing 36 15 000(1 + ( 12 )) 0,15 balance, the lower the balance, = the less you have to pay. HIRE PURCHASE POPULATION GROWTH = R 23 495,16 You buy a washing machine of R4 000 by 7.Nominal interest rate: signing a 2 year hire purchase agreement, you Notice: As compounding periods increase during the Pf u tu re = Ppresent (1 + i ) n Quoted period and compounded pay an R800 deposit. Calculate the year, so the period is different eg 15% per a) total amount you will repay if the interest accumulated amount increases. Pfuture = future population size annum compounded monthly. rate is 12% Ppresent = present population size 8.Effective interest rate: b) your monthly installment. i = average population (%) EXAMPLE Quoted period and compound n= number of years If R13 865 is received after 6 years of being invested period is equal eg 0,75% per and the interest rate was 16% compounded month compounded monthly. a) Deposit : R800 The population of lions is 2 567 in 2015. annually, what was the original amount invested? P = 4 000 − 800 A = P (1 + i )n = 3 200 If the growth rate is 1,34%, calculate the number of COMPOUND PERIODS 13 865 = P (1 + 0,16)6 lions in 2020. A = P (1 + i n) Annually: 1 per year 13 865 = 2,44P = 3 200(1 + (0,12)(2)) 13 865 = R 3 968 = P Semi-annually: 2 per year 2,44 2020 − 2015 = 5 P = 5 690,78 Quarterly: 4 per year b) A = R 3 968 Pfu tu re = Ppresent(1 + i)n ∴R 5 690,78 was the principal amount invested. Monthly: 12 per year 2 years = 24 equal payments = 2567(1 + 0,0134)5 OR use the following formulae: Daily: 365 per year* = 2743 3 968 (note that the number of lions will be an integer) A = P (1 + i )n To find A *(excl leap years) = R165,33 24 P = A(1 + i )−n To find P Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 15

Grade 12 Maths Essentials Finance SCIENCE CLINIC 2022 © NOMINAL VS EFFECTIVE INTEREST RATES CHANGING INTEREST RATES DEPRECIATION (DECAY) (COMPOUND INTEREST) If the interest rate changes after a set period of time: Depreciation is the loss or decrease of value at a specified rate over time. Annual effective rate is equivalent to the nominal rate per 1. Determine the accumulated amount after the first period Depreciation: Loss of value over time 2. Set the accumulated amount as the initial amount for the A = Book or scrap value annum compounded monthly, because it produces the Book value: Value of equipment at a given time after second period P = Present value same accumulated amount. depreciation 3. Determine the accumulated amount after the second period. i = Depreciation rate iNom n Scrap value: Book value of equipment at the end of its 1 + ieff = (1 + n ) n = time period EXAMPLE useful life R100 000 is invested for 6 years at an interest rate of LINEAR DEPRECIATION COMPOUND DEPRECIATION ieff = effective rate (annual) 16% per annum compounded quarterly. Thereafter the Also known as simple decay or Also known as iNom = nominal rate accumulated amount is reinvested for 5 years at an straight line depreciation depreciation on a reducing balance n = number of compoundings per year interest rate of 14% compounded semi-annually. Calculate the value of the investment at the end of this period. A = P(1 − in) A = P(1 − i)n EXAMPLE Straight Line Depreciation Reducing Balance Depreciation A = P (1 + i ) n Convert a nominal rate of 18% per annum 24 periods : 0,16 24 100 000(1 + 4 ) compounded monthly to an annual effective rate. A = Value of Asset (R) Value of Asset (R) 4 period (quarterly) p.a. over 6 years A = R 256 330,4165 0,18 12 1 + ieff = (1 + 12 ) A = P (1 + i ) n 0,18 12 (1 + 12 ) ieff = −1 10 periods : 0,14 10 256 330,4165(1 + 2 ) A = 2 period (semi-annual) ieff = 0,196 p.a. over 6 years A = R 504 240,73 ∴ reff = 19,6 % Number of periods Number of periods EXAMPLE EXAMPLE My new car , to the value of R 200 000 depreciates at a rate of 9% per annum. EXAMPLE R30 000 was left to you in a savings account. The interest What would the value of my car be after 6 years? Compare a linear depreciation You invest R25 000 at 14% per annum compounded to a reducing balance depreciation. monthly for a period of 12 months. Use the annual rate for the first 4 years is 12% per annum compounded effective rate to show that the same accumulated semi-annually. Thereafter the rates change to 18% per LINEAR DEPRECIATION REDUCING BALANCE DEPRECIATION amount will be obtained as when using the nominal annum compounded monthly and you leave the money for A = P (1 − i n) A = P (1 − i ) n another 3 years. What is the future value of the = 200 000(1 − (0,09)(6)) = 200 000(1 − 0,09)6 rate. investment after the savings period. = R 92 000 = R113 573,85 0,14 12 1 + ieff = (1 + 12 ) The exponent A = P (1 + i ) n 8 periods : EXAMPLE (12) is calculated 0,12 8 30 000(1 + 2 ) 0,14 12 (1 + 12 ) ieff = −1 by noting there A = 2 period (semi-annual) The value of a piece of equipment depreciates from R15 000 to R5 000 in four will be 12 p.a. over 4 years years. What is the rate of depreciation calculated on the: ieff = 0,1493 A = R47 815,44 compounding ∴ reff = 14,93 % a) Straight line method b) Reducing balance depreciation periods: once a A = P (1 + i ) n A = P (1 − i n) A = P (1 − i ) n month for 12 A = P (1 + i ) n 0,18 36 36 periods : 47 815,44(1 + 12 ) months. A = 12 period (monthly) 5 000 = 15 000((1 − (x)4) 5 000 = 15 000(1 − i )4 12 25 000(1 + ( 12 )) 0,14 5 000 = 15 000 − 60 000x 1 = p.a. over 3 years = (1 − i )4 A = R 81 723,25 −10 000 = −60 000x 3 = R 28 733,55 x = 0,1667 4 1 −1 = −x Alternatively: A = P (1 + i ) n × (1 + i ) n 3 A = P (1 + i )n 0,12 8 0,18 36 −0,2401... = −i 30 000(1 + 2 ) × (1 + 12 ) = Depreciation rate = 16,67% = 25 000(1 + 0,1493)1 i = 0,2401... × 100 = R 28 733,55 = R 81 723,26 r = 24 % Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 16

Grade 12 Maths Essentials Finance SCIENCE CLINIC 2022 © CALCULATING TIME PERIOD ADDITIONAL PAYMENTS OR WITHDRAWALS OF A LOAN OR INVESTMENT Timelines assist in visualising and keeping track of different rates and payments. Set up each When calculating the time of your in- section with information about the number of terms, compound periods and interest rates. vestment (n) in a compound formula EXAMPLE EXAMPLE you need to make use of Logs: You deposit R5 000 into a savings account and 4 years later you add R4 500. The You take out a loan to buy a new iPad. You make an additional payment of R 5 000 interest for the first 3 years was 11% compounded quarterly, then it changed to four years after taking out the loan. Two years later you repay the final amount of 12,5% compounded monthly. Calculate the savings at the end of the 6th year. y = b x (Exponential form) R 6 000. During the first four years of the loan the interest rate is 14% per annum compounded semi annually. For the last two years the rate changed to 11% per annum compounded quarterly. How much did you initially borrow? x = log b y (Logarithmic form) n = 3 year × 4 n = 1 year × 12 n = 2 year × 12 = 12 = 12 = 24 n = 4 year × 2 n = 2 year × 4 EXAMPLE =8 =8 Your car, costing R 250 000, needs 0,11 0,125 0,125 to be traded in when its value has != != != 0,14 0,11 Y0 4 Y3 12 Y4 12 Y6 != != depreciated to R 90 000. How Y0 2 Y4 4 Y6 long will you drive the car if the reducing balance depreciation rate is 7,5% per annum. +4500 +5000 +6000 A = P (1 − i ) n Working backwards, we need the value of the the loan at Y4: 90 000 = 250 000(1 − 0,075) n Y3 = P (1 + i ) n 0,36 = 0,925n 0,11 12 5000(1 + 4 ) = P = A(1 + i )−n n = log0,925 0,36 A = P (1 + i )n To find A 0,11 −8 6000(1 + 4 ) = R 6923,92 P = n = 13,1 P = A(1 + i )−n To find P P = R4 829,44 You can drive the car for 13 years before it depreciates to R90 000. Y4 = P (1 + i ) n + 4500 0,125 12 That means that the loan amount at the end of Y4, before the 6923,92(1 + 12 ) = + 4500 payment of R5000 was made, was R9 829,44. = R12 340,76 Work backing backwards, we determine the initial amount bor- rowed / loan amount at Y0. Y6 = P (1 + i ) n 12 340,76(1 + 0,125 24 A(1 + i )−n 12 ) = P = If withdrawals are made 0,14 −8 9829,44(1 + 2 ) = R15 825,37 from the savings account, P = use subtraction instead P = R 5 720,82 OR combine the steps into a single calculation 6000[(1 + 5000][(1 + 2 ) ] 0,11 −8 0,14 −8 4 ) OR P = + = R 5 720,82 [5000[(1 + 4 ) ][( 12 ) ] 4500][(1 + 12 ) ] 0,11 12 0,125 12 0,125 24 Y6 = 1+ + = R15 825,37 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 17

Grade 12 Maths Essentials Finance SCIENCE CLINIC 2022 © FUTURE VALUE ANNUITIES PRESENT VALUE ANNUITIES Equal investment payments made at regular intervals, subject to a rate of interest over a period of time. Reducing balance loan, a sum of money borrowed and paid back with interest in regular payments at Eg annuity fund, a retirement fund, a savings account. equal intervals. F = total accumulated at the end of the time period P = present value x = monthly installment x[1 − (1 + i )−n] x = monthly installment x[(1 + i )n − 1] P= i = interest rate per annum i i = interest rate p.a. F= i n = number of installments/payments n = number of time periods to repay the loan EXAMPLE EXAMPLE R2000 is deposited into a bank account. One month later a further R2 000 is deposited, and another You borrow money for a year at an interest rate of 18% per annum compounded quarterly and R2 000 a month after that. The interest rate is 7% per annum compounded monthly. How much will monthly repayments will be R 1 367,20. The payments will start one month after receiving the loan. be saved after 2 months? What was the initial amount borrowed? Compound interest formula: A = P (1 + i ) n Compound interest formula: P = A(1 + i )−n At the end of each month, a) interest is determined on the accumulated amount, and The payment at the end of each month can be used to determine the b) an additional deposit is made: present value of that payment at M0: 0,18 −1 M1 : P = 1367,20(1 + 4 ) M0 : 2000 The money borrowed 0,07 1 in the beginning is M1 : 2000(1 + 12 ) + 2000 The money saved 0,18 −2 the present value of M2 : P = 1367,20(1 + 4 ) at the end is the the annuity. 0,07 2 0,07 1 Future value of M2 : 2000(1 + 12 ) + 2000(1 + 12 ) + 2000 = R 6 035,07 the investment Present value of all payments (loan) at M0: 0,18 −1 0,18 −2 0,18 −12 P = 1367,20(1 + ) + 1367,20(1 + ) + ⋯1367,20(1 + 4 ) = R12 466,92 4 4 The values van be expressed as a series: 0,07 2 0,07 1 0,07 0 The values van be expressed as a series: M2 : 2000(1 + ) + 2000(1 + ) + 2000(1 + 12 ) M12: 1367,20(1 + 0,18 −1 0,18 −2 0,18 −12 4 ) + 1367,20(1 + 4 ) + ⋯1367,20(1 + 4 ) 12 12 0,07 This is a geometric series where a = 2000 , r = (1 + 12 ) , n = 3 (3 payments). 0,18 −1 0,18 −1 This is a geometric series where a = 1367,20(1 + 4 ) , r = (1 + 4 ) , n = 12 (12 payments) a (r n − 1) Sn = a (r n − 1) r −1 Sn = 0,07 3 r −1 2000[(1 + ) − 1] 12 0,18 −1 0,18 −12 F3 = 1367,20(1 + ) [(1 + ) − 1] 0,07 4 4 (1 + )−1 P = 12 0,18 −1 (1 + ) −1 4 F3 = 6035,07 P = 12 466,92 Future value annuity formula: Present value annuity formula: x [(1 + i ) n − 1] x [1 − (1 + i )−n] F = Formula only holds when P = Formula only holds when there is a i payment commences one i period break between the loan and first 0,07 3 period from the present 0,18 −12 payment. Eg first monthly payment 2000[(1 + ) − 1] 1367,20[1 − (1 + ) ] 12 and ends after n periods. starts 1 month after the loan is granted. = = 4 0,07 0,18 12 4 = R 6 035,07 = R12 466,92 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 18

Grade 12 Maths Essentials Finance SCIENCE CLINIC 2022 © CALCULATE MONTHLY INSTALLMENTS ANALYSES OF INVESTMENT AND LOAN OPTIONS Monthly installments are calculated using the present value annuity formula, and solving for x, When analysing investment and loan options, consider both the payment amounts, as well as the total the repayment amount. payment made at completion. NB: If repayments commence one month after the initiation of the loan; Investments with higher future values are preferable. Loans with smaller present values are preferable. a) Calculate the growth of the loan during the first month to determine the new present value b) Subtract one month from the total repayment terms. EXAMPLE CALCULATE OUTSTANDING BALANCE Outstanding balance is calculated using the present value annuity formula. The present value You have to take a home loan of R800 000 and there are 2 options to consider. after the nth installments is the outstanding balance, also known as the settlement amount. A- a 20 year loan at 17% interest per annum compounded monthly. B- a 30 year loan at 17% interest compounded monthly. EXAMPLE In order to buy a car John takes a loan of R 25 000. The bank charges an annual interest Monthly repayments: rate of 11% compounded monthly. The installments start a month after he has received the money from the bank. x [1 − (1 + i )−n] A. P = Calculate i x[1 − (1 + ] 0,17 −240 a) his monthly installments if he has to pay back the loan over a period of 5 years. 12 ) b) the outstanding balance of his loan after two years immediately after the 24th installment. 800 000 = 0,17 12 0,17 0,17 −240 800 000 × = x[1 − (1 + ) ] a) Total number of terms: (5 × 12) = 60 12 12 11 333,33 = 0,9658x x = R11 734,40 x [1 − (1 + i )−n] P = i x [1 − (1 + i )−n] x[1 − (1 + ] 0,11 −60 12 ) B. P = 25 000 = i x[1 − (1 + ] 0,17 −360 12 ) 0,11 12 800 000 = 0,17 12 0,17 0,17 −360 = x[1 − (1 + ] x = R 543,56 800 000 × ) 12 12 11 333,33 = 0,99368x b) n = 60 − 24 = 36 x = R11 405,40 x [1 − (1 + i )−n] Total repayments: Pafter 2 years = i A : R11 734,40 × 240 = R 2 816 256 543,56[1 − (1 + ] 0,11 −36 B : R11 405,40 × 360 = R4 105 944 12 ) = 0,11 12 Option B has a lower monthly repayment, but a total amount of almost double that of OPTION A, therefore A is the better option. = R16 602,97 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 19

Grade 12 Maths Essentials Functions and Graphs Grade 11 Recap SCIENCE CLINIC 2022 © Grade 11 Functions: Quadratic Functions Steps for sketching y = a x 2 + b x + c Finding the equation in the form y = a (x − p)2 + q Finding the equation in the form y = a x 2 + b x + c 2 y = a(x − p) + q Given the x-intercepts and another point 1. Determine the shape (‘a’) Given the turning point and another point 2. Find the x- and y-intercepts 1. Substitute the turning point into y = a (x − p)2 + q 1. Substitute the x-intercepts into Shape: Horizontal shift: Vertical shift: −b y = a (x − x1)(x − x 2 ) 3. Find the turning point ( ) 2. Substitute the other point into the equation to • a>0 • x-value of 2a find ‘a’ 2. Substitute the other point in to find ‘a’ • y-value of turning pt turning pt 4. Plot points and sketch graph 3. Determine the equation of the graph 3. Write/simplify your final equation • a<0 • axis of EXAMPLE 2 EXAMPLE 1 EXAMPLE 2 symmetry Sketch f (x) = x 2 − 4x + 3 Find the equation of the following graph: Find the equation of the following graph: Steps for sketching y = a (x − p)2 + q y 1. Determine the shape (‘a’) • Shape: a > 0 ∴ y 2. Find the x- and y-intercepts • x-intercept (y = 0) 3. Find the turning point Option 1: 0 = x 2 − 4x + 3 (0;3) 4. Plot points and sketch graph 0 = (x − 3)(x − 1) (-1;0) x EXAMPLE 1 x = 3 OR x = 1 (1;0) (3;0) x Sketch f (x) = (x + 1)2 − 9 −b ± b 2 − 4a c (1;0) Option 2: x = 2a (1;-9) • Shape: a > 0 ∴ y −(−4) ± (−4)2 − 4(1)(3) • x-intercept (y = 0) x = 2(1) • x-intercepts 0 = (x + 1)2 − 9 4± 2 x = 1 OR x = 3 (-4;0) (2;0) x = • Turning point (p;q) 9 = (x + 1)2 2 Formula: y = a (x − x1)(x − x 2 ) x p = 1 and q = − 9 ± 9= x+ 1 x = 3 OR x = 1 y = a (x − 1)2 − 9 y = a (x − 1)(x − 3) + 3 = x + 1 OR − 3 = x + 1 • y-intercept (x = 0) y = 3 • Other point • Other point 2 = x OR − 4 = x (−1; 0) (0; 3) (0;8) (0;–8) • Turning point (p;q) • y-intercept (x = 0) −b −(−4) 1) x-value of TP = = 0= a (−1 − 1)2 − 9 3 = a (−1)(−3) (-1;-9) 2a 2(1) y = (0 + 1)2 − 9 9 = 4a 1= a x = 2 y = −8 2) Subst. into original eq: 9 a = • Equation • Turning point (p;q) y = (2)2 − 4(2) + 3 4 Remember: y = 1(x − 1)(x − 3) y = −1 (−1; − 9) (x − (−1))2 − 9 y • Equation TP (2; − 1) y = x 2 − 4x + 3 (x − p)2 + q 9 • Axis of symmetry • Axis of symmetry y = (x − 1)2 − 9 4 NOTE: x = −1 x = 2 (0;3) If you need to write this equation in the NOTE: • Domain form y = a (x − p)2 + q complete the square • Domain f (x) = x 2 x ∈R (3;0) x ∈R → moved 1 unit to the left y = x 2 − 4x + 4 + 3 − 4 → moved 9 units down • Range (1;0) • Range y ≥ −1 y = (x − 2)2 + 3 − 4 (2;-1) y ≥ −9 y = (x − 2)2 − 1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 20

Grade 12 Maths Essentials Functions and Graphs Grade 11 Recap SCIENCE CLINIC 2022 © Exponential Graphs Hyperbola a Finding the equation in the form y = a b x−p + q Finding the equation in the form y = + q x−p a x −p y = ab + q y= + q Given the asymptote and another point x −p Given the asymptotes and another point 1. Substitute the asymptote into the equation Vertical shift: 1. Substitute the asymptotes into the equation Shape: Shape: Horizontal shift: Vertical shift: y y 2. Substitute the other point in x • asymptote x 2. Substitute the other point into the equation • b>1 3. Write/simplify your final equation • a>0 • asymptote • asymptote [y = q] to find ‘a’ [x = p] [y = q] y y • 0<b<1 • a<0 x x 3. Write/simplify your final equation EXAMPLE 2 Find the equation of the following graph given a EXAMPLE 2 Steps for sketching y = a b x−p + q Steps for sketching y = + q y = b x+ 1 + q : x −p Find the equation of the following graph: y 1. Determine the asymptote (y = q) 1. Determine the asymptotes (y = ‘q’ and x = ‘p’) y 1. Determine the shape (‘a’) 1. Determine the shape (‘a’) 2. Find the x- and y-intercepts 2. Find the x- and y-intercepts y=1 3. Plot points (at least 2 others) and sketch graph 3. Plot points (at least 2 others) and sketch graph (2;0) x (-3;2) EXAMPLE 1 EXAMPLE 1 Sketch f (x) = 2x+ 1 + 1 −1 x Sketch f (x) = −1 y = -2 x −2 x = -1 • Asymptote • Asymptote • Asymptotes y = 1 y y = 1 and x = − 1 x = 2 a • Shape: a > 0 ∴ x f (x) = + 1 • Asymptote y = −1 x − (−1) • x-intercept (y = 0) y q = −2 a • Shape: a < 0 ∴ f (x) = + 1 0= 2 x+ 1 + 1 x x+ 1 y = b x+ 1 − 2 • x-intercept (y = 0) −1 = 2 x+ 1 • Other point • Other point −1 Not possible to solve for x 0= −1 (2; 0) x −2 (−3; 2) a ∴ No x-intercept −1 0= + 1 2 = b −3+ 1 − 2 1= 2+ 1 x −2 • y-intercept (x = 0) a 4 = b −2 x −2 = −1 −1 = y = 20+ 1 + 1 3 1 x = 1 y = 3 4= −3 = a y b2 • y-intercept (x = 0) y • Domain 1 • Equation b2 = −1 4 y = −1 −3 x ∈R f (x) = + 1 (1;5) 1 1 −2 x+ 1 • Range b = ± b ≠− 1 (0;3) 2 2 y = − (1;0) (-1;2) 2 x y> 1 y=1 1 Lines of Symmetry: b = + y = -1 2 • Domain (0;-½) Use point of intersection of asymptotes. (-1;1) x (3;-2) • Equation x ∈R; x ≠ 2 y = x+ c (−1; 1) y = −x + c (−1; 1) 1 1= −1+ c 1= 1+ c y = ( ) x+ 1 − 2 • Range 2 x=2 2= c 0= c y ∈R; y ≠ − 1 y = x+ 2 y = −x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 21

Grade 12 Maths Essentials Functions and Graphs Grade 11 Recap SCIENCE CLINIC 2022 © Deductions from Graphs DISTANCE NOTATION EXAMPLE 1 A f (x) = a x 2 + bx + c and g (x) = k x are sketched. D is the turning point of f(x) with the axis of symmetry at x=2. Steps for determining VERTICAL DISTANCE • f (x) > 0 (above the line y = 0) AB is 6 units. 1. Determine the vertical distance B y y y Vertical distance = top graph − (bottom graph) 2. Substitute the given x-value to derive your answer (-2;9) x x D F Steps for determining HORIZONTAL DISTANCE (i.e. where y is positive) A E B x 1. Find the applicable x-values • f (x) < 0 A B A B = xB − xA (largest − smallest) (below the line y = 0) y y Steps for determining MAXIMUM DISTANCE x x 1. Determine the vertical distance Vertical distance = top graph − bottom graph Questions: Solutions: (i.e. where y is negative) 2. Complete the square a. Determine the value of k. a. (−2; 9) 1 2 4 d. y = − (2) + (2) + 1 y = a (x − p)2 + q A 9 = k −2 5 5 • f (x) ⋅ g (x) ≤ 0 b. Determine the x-values of A and B. 1 9 9 3. State the maximum distance 9= y = ∴ D = (2; ) B −1 4 k2 5 5 y = a (x − p)2 + (q) → q is the max distance c. Show that a = and b = . 1 (one graph lies above y = 0 and 5 5 k = ± NOTE: one graph lies below y = 0) 3 d. Determine the coordinates of D. 1 k = 9 • Distance is always positive 3 e. units (y-value of co- • f (x) ≥ g (x) e. Determine the maximum distance of DE. 5 • Distance on a graph is measured in units top bottom ordinate D is also TP) (i.e. f(x) lies above g(x)) f. Determine the values of p for which: INTERSECTION OF GRAPHS −1 2 4 b. E = (2; 0) and A B = 6 u ni t s • f (x) = g (x) x + x+ p< 0 Steps for determining POINTS OF INTERSECTION 5 5 A = (−1; 0) x = − 1 (point of intersection) 1 4 g. Determine for which values of x: B = (5; 0) x = 5 f. − x 2 + x+ p< 0 1. Equate the two functions ROOTS & PARABOLAS 5 5 f (x) = g (x) i. f (x) ≥ 0 4 p< − • Equal, real roots f (x) c. y = a (x − x1)(x − x 2 ) 5 2. Solve for x (look for the applicable x-value: A or B) y y ii. > 0 g (x) (−1; 0) and (5; 0) NOTE: 3. Substitute the applicable x-value into any of the y = a (x + 1)(x − 5) • Interpret question as: two equations to find ‘y’ x x iii. f (x) is increasing f(x) Use F (0; 1) How many units must the 1 = a (+ 1)(−5) graph move for the max. g(x) 1 value to be < 0 A • Non-real/ No real roots − = a y y 5 B 1 y = − (x + 1)(x − 5) 5 g. x x 1 INCREASING/DECREASING y = − (x 2 − 4x − 5) i. x ∈[−1; 5] 5 1 4 ii. x ∈(−1; 5) • Real, unequal roots y = − x2 + x+ 1 Dec y y 5 5 asin iii. x ∈(−∞; 2) Dec asin g 4 g s b = s rea 5 rea Inc re Inc x x re ing ing Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 22

Grade 12 Maths Essentials Functions and Graphs Grade 11 Recap SCIENCE CLINIC 2022 © EXAMPLE 2 Solutions: f (x) = m x + c and g (x) = a(x − p)2 + q are sketched below. T is the turning point of g(x). a. y = a (x + 2)2 + 8 (0; 0) g. −2x 2 − 8x = − 2x − 8 −8 = 4a 0 = 2x 2 + 6x − 8 −2 = a and p = − 2 and q = 8 0 = 2(x − 1)(x + 4) y ∴ g (x) = − 2(x + 2)2 + 8 x = 1 or x = − 4 (NA) D (−4; 0) y = − 2(1) − 8 8 (-2;8) T ∴ m = −4 = − 2 and c = − 8 y = − 10 ∴ FP = 10 units B b. OD = 4 units G A P h. Max length is given by TP of parabola (L (x)) J O x c. TR = 8 units D R given by L (x) = g (x) − f (x). Find the TP by completing the square. M -8 F ∴ Max BM = g (x) − f (x) d. TR: x = − 2 = − 2x 2 − 8x − (−2x − 8) H f = − 2x 2 − 6x + 8 g e. g (x) = − 2(x + 2)2 + 8 = − 2(x 2 + 3x − 4) 3 9 = − 2(x 2 + 4x + 4) + 8 = − 2[(x + )2 − 4 − ] 2 4 = − 2x − 8x − 8 + 8 3 25 = − 2[(x + )2 − ] 2 4 = − 2x 2 − 8x 3 25 = − 2(x + )2 + Questions: BM = g (x) − f (x) 2 2 25 Determine: BM = − 2x 2 − 8x − (−2x − 8) ∴ Max of BM = units 2 a. The value of a , p, q, m and c. = − 2x 2 − 6x + 8 BM = − 2(−1)2 − 6(−1) + 8 b. The length of OD. BM = 12 units i. k = -8 c. The length of TR. d. The equation of TR. f. 28 = − 2x − 8 − (−2x 2 − 8x) j. x ∈(−∞; 0); x ≠ − 4 e. BM if OA = 1 unit. 28 = − 2x − 8 + 2x 2 + 8x f. OJ if GH = 28 units. 0 = 2x 2 + 6x − 36 0 = 2(x 2 + 3x − 18) g. The length of FP. 0 = 2(x + 6)(x − 3) h. The maximum length of BM. x = −6 or x = 3 (NA) i. The value of k for which −2x 2 − 8x + k has two equal roots. ∴ OJ = 6 units f (x) j. For which value(s) of x will < 0? g (x) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 23

Grade 12 Maths Essentials Functions and Inverses SCIENCE CLINIC 2022 © Relation: set of ordered pairs INVERSE OF A FUNCTION SKETCHING INVERSE FUNCTIONS Function: relation where each of the values in the domain (x-values) is associated with only ONE value in The symmetry if a graph is a mirror image of the range (y-value) the curve around a specific line. The inverse Method 1: Table Method of a function is the symmetry of a graph 1. Using the original graph, create a table How to determine whether a graph is a function: 2. Sketch the graph of the inverse by switching the about the line y = x. Use the Vertical Ruler Test. If the ruler crosses the graph: values of x and y for each point ! Once, it IS a function ! More than once, it IS NOT a function EXAMPLE How to determine the inverse of a function: EXAMPLE Original function: 1. Write in standard form Determine whether each of the following graphs is a function or not x -1 0 1 2. Switch x and y 1. 2. 3 y -2 0 -2 3. Make y the subject of the formula Inverse: NOTE: x -2 0 -2 For Many-to-One functions (Parabolas): The domain needs to be restricted in y -1 0 1 order for the inverse to be a function EXAMPLE Method 2: Using the original function Is not a function Is a function Is not a function 1. Find important properties on the original function Find the inverse equation of each of the (i.e. x- and y-intercepts, turning points, ONE-TO-ONE FUNCTION MANY-TO-ONE FUNCTION NON-FUNCTION following functions. Write them in the asymptotes, domain and range) form f −1(x) = . . . 2. Sketch the graph of the inverse by switching the A function where every element A function where two or more elements One element of the domain values of x and y for each important point of the domain has a different of the domain may be associated with the associated with two or more element in the range. If you do same element of the range. If you do a elements of the range. 1. f (x) = 4x − 5 EXAMPLE a horizontal line test, the graph horizontal line test, the graph will be cut x = 4y − 5 Original function: will only be cut by your line by your line MORE THAN ONCE. x-intercept - 0 = 2x − 4 ONCE. x + 5 = 4y –1 –2 –3 9 –3 9 x = 2 0 0 0 2 0 2 1 5 y-intercept - y = 2(0) − 4 y = x+ 2 4 3 10 3 10 4 4 y = −4 3 6 4 1 5 −1 ∴ f (x) = x+ y y 4 4 Inverse: y x-intercept - x = − 4 (−4; 0) (0;2) y-intercept - y = 2 (0; 2) (0;1) (2;0) Method 3: Using the inverse equation 2. f (x) = (x − 8)2 + 1 x x x 1. Determine the equation of the inverse x = (y − 8)2 + 1 2. Calculate the important properties (i.e. x- and y- x − 1 = (y − 8)2 intercepts, turning points, asymptotes, intervals that are ascending/descending, domain and range) ± x −1 = y −8 3. Sketch the graph of the inverse y y y 8± x −1 = y EXAMPLE ∴ f −1(x) = 8± x −1 Original function: y = − 2x 2 (2;0) x = − 2y 2 x x x −x = y2 2 −x (0; –2) Inverse: y = ± (not a function) 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 24

Grade 12 Maths Essentials Functions and Inverses SCIENCE CLINIC 2022 © INVERSE OF A LINEAR FUNCTION INVERSE OF A PARABOLA EXAMPLE Given h (x) = − x 2 with domain x ≤ 0, sketch the Straight line: f (x) = a x + q Parabola: f (x) = a x 2 x −q graph of h (x) and its inverse on the same set of Inverse: f −1(x) = x axes. Indicate the line of symmetry. a Inverse: f −1(x) = ± a 1 2 Given f (x) = − 3x − 2, the inverse is f −1(x) = − x− Given f (x) = x 2, the inverse is f −1(x) = ± x 3 3 h (x) = − x 2 y General properties of the inverse: General properties of the inverse: For h −1(x), switch x and y f (x) y=x Domain: x ≥ 0 if a > 0 x = − y2 Domain: x ∈R y x ≤ 0 if a < 0 y2 = − x (–2;0) Range: y ∈R f (x) y=x Range: y ∈R y = ± −x x Shape: ascending if a > 0 Shape: if a > 0 f –1(x) ∴ h −1(x) = ± −x descending if a < 0 if a < 0 (0;–2) f –1(x) 1 Interval ascending (thus average BUT domain for h (x) given as x ≤ 0 Average Gradient: m = a x gradient is positive): thus, h −1(x) = − −x if a > 0; x ∈[0; ∞) y if a < 0; x ∈(−∞; 0] NOTE: y=x Interval descending (thus average The inverse of y = k (where k is a constant) will be perpendicular to the x -axis gradient is negative): The inverse of x = k (where k is a constant) will be perpendicular to the y -axis if a > 0; x ∈[0; ∞) EXAMPLE if a < 0; x ∈(−∞; 0] x Given g (x) = x − 3, sketch the graph of g (x) and its inverse on the same set h –1(x) of axes. The inverse of a parabola is not a function, because there are two elements of the range for every element of the domain, BUT if we restrict the domain of the y g –1(x) original function, we get an inverse that is a function and will look like this: h (x) g (x) = x − 3 y=x x For g −1(x), switch x and y (0;3) g(x) Given g (x) = 3x 2 with domain x ≥ 0, the inverse is g −1(x) = + . Both the Determine the following relating to the inverse: 3 x = y −3 restricted parabola and its inverse appear on the graph below. 1. Domain: x ≤ 0 (–3;0) y = x+ 3 (3;0) x 2. Range: y ≤ 0 y g (x) y=x ∴ g −1(x) = x+ 3 3. x-intercept: x = 0 or (0; 0) (0;–3) 4. y-intercept: y = 0 or (0; 0) g –1(x) 5. Interval ascending: x ∈(−∞; 0] Determine the following relating to the inverse: x 1. Domain: x ∈R 2. Range: y ∈R 3. x-intercept: x = 3 or (3; 0) 4. y-intercept: y = − 3 or (0; − 3) Therefore, the domain of a parabolic function should be limited, either as x ≥ 0 0 − (−3) or x ≤ 0, in order to create an inverse that is a function. 5. Average gradient: m = = 1 3−0 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 25

Grade 12 Maths Essentials Functions and Inverses SCIENCE CLINIC 2022 © WORKING WITH LOGARITHMS LOG LAWS EXPONENTIAL AND LOG FUNCTIONS A logarithm (or ‘log’) is a mathematical notation that has been defined to allow us 1. Sum to product law: For exponential functions: to make an exponent the subject of a formula. A logarithm function is the inverse loga x + loga y = loga x y if a > 1 if 0 < a < 1 of an exponential function, therefore if y = a x (a > 0, a ≠ 1) then the inverse is y y x = a y. If written in standard function form then x = a y is y = loga x. 2. Difference to quotient law: x NOTE: loga x − loga y = loga y loga a = 1 x = ay then y = loga x loga1 = 0 3. Power law: loga x = log10 a x x Index/(log) : y Log/(Index) : y loga x m = m loga x Number : x Number : x 4. Change of base law: Base : a Base : a logb x loga x = EXAMPLES logb a increasing function decreasing function 1. Write the following in log form: For logarithmic functions: 1 x (5 ) a. y = b. 31 = 3 c. x = 4 y EXAMPLES if a > 1 if 0 < a < 1 Simplify the following: x = log 1 y 1 = log33 y = log 4 x 5 y 81 1. log3 3. (logx16 − logx 4) ÷ logx 8 2. Write the following in exponential form: 729 16 a. y = log 1 x b. log61 = 0 c. x = logy 5 = log381 − log3729 = logx ÷ logx 8 4 x x 2 y (2 ) 1 = 4log33 − 6log33 = logx 4 ÷ logx 8 x = 1 = 60 5 = yx = 4(1) − 6(1) 2logx 2 = = −2 3logx 2 2 increasing function decreasing function 3. Solve the following equations: = 3 (4) m x a. log 4 = 3 b. log3(5x − 3) = 3 c. log = −3 3 1 m x 2. log3 + log51 + log 48 = 43 5x − 3 = 33 = 10−3 9 3 4 27 + 3 4 log28 m = 64 × 3 = 192 x = = 6 x = = − 2log33 + log51 + 5 1000 log2 4 3log2 2 = − 2(1) + 0 + 2log2 2 4. Determine the value of the following: 3 3 ( 729 ) 1 = −2+ a. log2 256 b. log 1 c. log 10 2 1 = − ( ) 1 2 x = log2 256 x = log 1 x = log 10 3 729 1 x (3) 1 256 = 2 x = 10 x = 10 729 1 x = 8 x = 6 x = 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 26

Grade 12 Maths Essentials Functions and Inverses SCIENCE CLINIC 2022 © INVERSE OF AN EXPONENTIAL FUNCTION MIXED EXAMPLE 1 Questions: Exponential: f (x) = a x; a > 0 and a ≠ 1 Questions: 2. The growth of a virus cell is given by g (t ) = 1,95t, where t is in minutes Inverse: f −1(x) = loga x ; a > 0 and a ≠ 1 1. Given f (x) = log 1 x 2 a. Determine the number of virus cells after 8,5 minutes Given f (x) = 3x, the inverse is f −1(x) = log3 x a. If f (x) = − 3, determine x. b. After how many minutes will there be 2164 virus cells in the b. Draw the graph of f (x). body? General properties of the inverse: c. For which value of x is f (x) < − 3? y f (x) Domain: x > 0 y=x Range: y ∈R d. Determine f −1(x) and then draw the graph on the same set of axes as f. Shape: if a > 0 f –1(x) if a < 0 Asymptote: x = 0 Solutions: x Solutions: a. g (8,5) − 1,958,5 If there are no vertical or horizon- a. log 1 x = − 3 Number of virus cells ≈ 292 tal shifts (i.e. no p - or q -values) 2 −3 (2 ) 1 the log graph will cut the x -axis x = b. 1,95t = 2164 at 1. x = 8 log1,95t = log2164 EXAMPLE t log1,95 = log2164 1 x log2164 (3 ) b. Given h (x) = sketch the graph of h (x) and its inverse on the same t = log1,95 set of axes. Indicate the line of symmetry. f –1(x) y t ≈ 11,5 minutes y=x 1 x (3 ) h (x) = Alternative solution (b): 1,95t = 2164 For h −1(x), switch x and y (0;1) t = log1,952164 1 y (3 ) x = t ≈ 11,5 minutes (1;0) x y = log 1 x 3 ∴ h −1(x) = log 1 x f (x) 3 h (x) y y=x (–1;3) x c. x ∈(8; ∞) OR x > 8; x ∈R h –1(x) (3;–1) 1 x (2 ) d. f −1(x) = = 2−x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 27

Grade 12 Maths Essentials Functions and Inverses SCIENCE CLINIC 2022 © MIXED EXAMPLE 2 MIXED EXAMPLE 3 MIXED EXAMPLE 4 MIXED EXAMPLE 5 Questions: Questions: Questions: Questions: Given g −1(x) = − x Given f (x) = 3x a. If h (x) = 2x 2, determine the equation of the The sketch represents the graph of g (x) = a x and inverse of h (x) h (x), the reflection of g (x) in the y-axis. a. Write down the domain and range of a. Draw a sketch graph of f and give the domain g −1(x) = − x and range. b. Is the inverse a function? If it isn’t a function, h (x) y g (x) how can we restrict the original function so b. Determine g (x). b. Determine f −1(x) = .... that it is a function? A c. Draw sketch graphs of g and g −1 on the same c. Sketch the graph of f −1 on the same set of c. Sketch the graph h (x) and it’s inverse as deter- set of axes axes and give the domain and range mined in (b). d. Give the equation of the line symmetrical to f −1 about the x-axis x Solutions: 1 Solutions: a. Calculate the value of a if A(2; 2 ) is a point on a. Domain: x ∈[0; ∞) 4 a. x = y 2 g (x). Range: y ∈(−∞; 0] Solutions: y = ± x a. b. Write down the equation of h (x). y f (x) b. x = − y c. Write down the equation of g −1(x) in the form b. The inverse is not a function as one x has two y = ... (x)2 = (− y)2 different y values. Thus, we must restrict the domain of h (x) . There are two ways: i) x ≤ 0 d. Give the domain and range of g (x), h (x) and x2 = y (0;1) f –1(x) g −1(x) or ii) x ≥ 0 ∴ g (x) = x 2; x ∈(−∞; 0] (1;0) x h (x) y c. i) Solutions: c. g –1(x) y 1 a. 2 = a 2 4 9 Domain: x ∈R = a2 Range: y ∈(0; ∞) x 4 3 h –1(x) ∴a = x 2 b. y = log3 x g (x) 3 −x (2 ) ii) b. h (x) = c. Domain: x ∈(0; ∞) y h (x) 2 x (3 ) Range: y ∈R h (x) = d. y = − log3 x h –1(x) c. g −1(x) = log 3 x 2 x d. g : x ∈R and y ∈(0; ∞) h : x ∈R and y ∈(0; ∞) g −1 : x ∈(0; ∞) and y ∈R Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 28

Grade 12 Maths Essentials Differential Calculus SCIENCE CLINIC 2022 © AVERAGE GRADIENT: FINDING THE DERIVATIVE USING THE DIFFERENT NOTATIONS DIFFERENTIATION RULES Gradient between two points on a curve: DEFINITION (FIRST PRINCIPLES) USED FOR DERIVATIVES NB: Use these rules unless you are specifically asked to use The derivative of a curve is the gradient of the curve. The derivative of y = f (x) can be “FIRST PRINCIPLES” y2 − y1 written as: Ave m = f ′(x) is the derivative of f (x). x2 − x1 1. The derivative of a constant is ZERO f (x + h ) − f (x) • f ′(x) f ′(x) = lim dy a. (−4) = 0 EXAMPLE h →0 h dx Determine the average gradient of d f (x) b. If f (x) = 3 then f ′(x) = 0 EXAMPLE • dx f (x) = x 2 + 2 between x = 2 and x = 4. Find the derivative of f (x) using first principles. d 2. The derivative of x n is nx n−1 (when n is a constant) y − y1 1. f (x) = x 2 + 4 Ave m = 2 • d x [ f (x)] dy x 2 − x1 a. If y = x 2 then = 2x 2−1 = 2x ∴ f (x + h ) = (x + h )2 + 4 dx f (4) − f (2) = = x 2 + 2x h + h 2 + 4 −3 4−2 b. Dx[x −3] = − 3x −3−1 = − 3x −4 = • y′ x4 f (x + h ) − f (x) [(4)2 + 2] − [(2)2 + 2] f ′(x) = lim = h →0 h 4−2 [x 2 + 2x h + h 2 + 4] − [x 2 + 4] dy 3. The derivative of a constant k multiplied by f (x) is k . f ′(x) = 6 = lim • dx h →0 h d ( − 2x ) = − 2(3x ) = − 6x a. 3 3−1 2 2x h + h 2 Note: First change all GRADIENT AT A POINT dx = lim monomial denominators • Dx[ f (x)] [ x6 ] h →0 h 3 −18 The gradient of f (x) at the point where b. Dx = Dx[3x −6 ] = − 18x −7 = to negative exponents h (2x + h ) x7 x = a is given by: = lim h →0 h 1 x −2 1 1 dy c. y = = then y ′ = ( − 2x −2−1) = − x −3 = − f (a + h ) − f (a) = 2x + 0 • d t means the derivative of the 2x 2 2 2 x3 lim h →0 h = 2x function of y with respect to the variable t 4. The derivative of a sum, is the sum if the derivatives EXAMPLE d x(2 ) d x(2 ) 1 d 1 3 1 d 1 3 2. f (x) = a. x + −π = x + x −1 − π Determine the gradient of f (x) = 2x 2 − 1 x • Dk[ f (k)] means the derivative x at the point where x = 1. 1 of f (k) with respect to the 3 2 1 ∴ f (x + h ) = variable k = x − x+ h 2 x2 f (x + h ) − f (x) f (a + h ) − f (a) f ′(x) = lim b. If s = (t + 1)(t − 4) = t 2 − 3t − 4 lim h →0 h h →0 h ds 1 f (1 + h ) − f (1) − 1x then = 2t − 3 = lim = lim x+ h dt h →0 h h →0 h x 2 − 5x − 6 (x − 6)(x + 1) c. If f (x) = = = x+ 1 h →0 [ (x + h )x ] [2(1 + h )2 − 1] − [2(1)2 − 1] x − (x + h ) x −6 (x − 6) = lim = lim ÷h h →0 h then f ′(x) = 1 4h + 2h 2 −h 1 = lim = lim × x3 − 3 x3 3 h →0 h Note: DO NOT h →0 x (x + h ) h d. If y = = − = x − 3x −2 x 2 x2 x2 h (4 + 2h ) substitute zero −1 = lim until h is no longer = dy 6 h →0 h x (x + 0) then = 1 + 6x −3 = 1 + in the denominator dx x3 = 4 + 2(0) −1 (0 ) A = is undefined x2 = 4 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 29

Grade 12 Maths Essentials Differential Calculus SCIENCE CLINIC 2022 © TANGENTS TO A CURVE AT A POINT CUBIC CURVES SKETCHING CUBIC GRAPHS The gradient of a tangent to y = f (x) at the point where x = c is given by m tan = f ′(c). Standard form: Steps for sketching a cubic curve: Steps: EXAMPLE 2 y = a x3 + b x2 + c x + d y-intercept 1. Determine the stationary points (SPs) and their natures 1. Find the point of contact (POC) Find the equation of the tangent(s) to 2. Determine the x- and y-intercepts a>0 or OR a<0 or 2. Find f ′(x) at the POC (i.e. f ′(c)) f (x) = x 3 − 1 that is perpendicular to 3. Sketch the graph 1 3. m tan = f ′(c) y = − x+ 8 THE FIRST DERIVATIVE TEST 3 4. Find y = m x + c and sub in m tan Stationary points (SP) are found when f ′(x) = 0 and the POC 1. ⊥ lines: m1 × m 2 = − 1 • f (c) is a local max if : • f (c) is a local min if : • If f (c) has f ′(x) with the 1 − × m⊥ = − 1 f ′(x) > 0 to the left of c f ′(x) < 0 to the left of c same sign on either side EXAMPLE 1 3 f ′(x) < 0 to the right of c f ′(x) > 0 to the right of c of c. Then f (c) is neither a Find the equation of the tangent to ∴ m⊥ = 3 f l (x) = 0 local to the right of c the curve y = 2x 2 + 1 at the point f l (x) SP f l (x) BUT m tan = f ′(x) f l (x) SP f l (x) where x = 2 f l (x) f l (x) = 0 f l (x) 3= 3x 2 • POC: x = 2 f l (x) = 0 EXAMPLE 1 y = 2(2)2 + 1 x2 = 1 = 9 Sketch f (x) = (x − 2)3 + 8 (2; 9) ∴ x = ± 1 f l (x) So there are 2 tangents to f (x) that have • Standard form: f l (x) • f ′(x) f (x) = (x − 2)(x 2 − 4x + 4) + 8 f l (x) = 0 a gradient of 3 Let y = f (x) = x 3 − 6x 2 + 12x ∴ f ′(x) = 4x Tangent 1: Tangent 2: EXAMPLE 2 At x = 2: y = (1)3 − 1 = 0 y = (−1)3 − 1 = −2 • Stationary points (SPs at f ′(x) = 0): Sketch f (x) = 6x − 2x 3 f ′(2) = 4(2) = 8 3x 2 − 12x + 12 = 0 POC (1; 0) POC (−1; − 2) • Stationary points (SPs at f ′(x) = 0): • ∴ m tan = 8 x 2 − 4x + 4 = 0 y = 3x + c y = 3x + c (x − 2)2 = 0 6 − 6x 2 = 0 • y = mx + c x2 = 1 0 = 3(1) + c −2 = 3(−1) + c ∴ x = 2 y = 8x + c (Sub in (2; 9)) x = ± 1 9 = 8(2) + c c = −3 c = 1 f (2) = 8 ∴ (2; 8) ∴ (1; 4) or (−1; − 4) c = −7 ∴ y = 3x − 3 ∴ y = 3x + 1 ∴ y = 8x − 7 is the tangent to Nature of SPs → first derivative test Nature of SPs → first derivative test y = 2x 2 + 1 at x = 2 x SP (-1) SP (1) x SP 2 EXAMPLE 3 f ′(x) 0 0 f ′(x) 0 Find the normal to the tangent of the curve f (x) = 2x 2 + 4 at the point where x = 0 f (x) y NOTE: THE NORMAL IS THE LINE ⊥ TO THE TANGENT AT THE POC • y-intercept (x = 0) y • y-intercept (x = 0) • POC: x = 0 • m tan = 0 f (0) = 0 f (0) = 0 y = 2(0)2 + 4 f (x) (1;4) thus, tangent is a horizontal line y = 4 • x-intercept (y = 0) = 4 • x-intercept (y = 0) (2;8) (0; 4) • tan ⊥ normal f (x) = 0 f (x) = 0 x (0;–√3) ∴ the normal will be a vertical line (x − 2)3 + 8 = 0 6x − 2x 3 = 0 x • f ′(x) (0;√3) through the POC 2x (3 − x 2 ) = 0 f ′(x) = 4x (x − 2)3 = − 8 At x = 0: ∴ x = 0 x = 0 or x 2 = 3 x − 2 = 3 −8 (–1; –4) f ′(0) = 4(0) = 0 x = ± 3 x −2 = −2 ∴ x = 0 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 30

Grade 12 Maths Essentials Differential Calculus SCIENCE CLINIC 2022 © CONCAVITY FACTORISING MORE COMPLEX CUBIC POLYNOMIALS Concavity explains how the gradients of the tangents to a curve The factor theorem states that if f (c) = 0 then (x − c) is a factor change from left to right EXAMPLE 1 EXAMPLE 2 • A section of a curve is concave up if the gradient of the tangent Factorise x 3 − 5x 2 − 2x + 24 by using inspection Factorise x 3 − 7x 2 + 14x − 8 by using synthetic division increases from left to right f (−2) = 0 ∴ (x + 2) is a factor f (1) = 0 ∴ (x − 1) is a factor → 0 → m m x 3 − 5x 2 − 2x + 24 1 1 − 7 + 14 − 8 ← coefficients of the polynomial increasing gradients x3 24 ↓+ 1−6+ 8 ∴ concave up = (x + 2)(x 2 − 7x + 12) 1−6+ 8 ⋅ m=0 2x 2 = (x − 1)(x 2 − 6x + 8) • A section of a curve is concave down if the gradient of the tangent −7x 2 decreases from left to right = (x + 2)(x − 4)(x − 3) = (x − 1)(x − 2)(x − 4) m=0 → 0 → MORE SKETCHING (USING THE SECOND DERIVATIVE) EXAMPLE 2 m m decreasing gradients EXAMPLE 1 ∴ concave down Sketch f (x) = x 3 − x 2 − 5x − 3 Sketch f (x) = x 3 − 5x 2 + 8x − 4 • Stationary points (SPs at f ′(x) = 0): TEST FOR CONCAVITY • Stationary points (SPs at f ′(x) = 0): 3x 2 − 2x − 5 = 0 Remove the highest common factor from the coefficients and 3x 2 − 10x + 8 = 0 (3x − 5)(x + 1) = 0 (3x − 4)(x − 2) = 0 5 common variables. x = or x = − 1 4 3 x = or x = 2 5 −256 f ′′(x) → second derivative 3 ( ; ) (−1; 0) 4 4 3 27 ( ; ) (2; 0) 3 27 Nature of SPs → 2nd derivative test OR 1st derivative test If: * f ′′(x) > 0 concave up * f ′′(x) < 0 concave down Nature of SPs → second derivative test f ′′(x) = 6x − 2 x SP (-1) SP (5/3) (3 ) f ′′(x) = 6x − 10 5 * f ′′(x) = 0 no conclusion about concavity f ′′ = 8 and 8 > 0 f ′(x) (3 ) 4 0 0 f ′′ = − 2 and −2 < 0 f ′′(2) = 2 and 2 > 0 POINT OF INFLECTION (POI) ∴ local min local local min ∴ local max ∴ local min f ′′(−1) = − 8 and −8 < 0 Occurs where concavity changes (i.e. where f ′′(x) = 0) max f (x) ∴ local max concave up • y-intercept (x = 0) y f ′′(x) > 0 f (0) = − 4 • y-intercept (x = 0) concave down y f ′′(x) < 0 POI f (0) = − 3 • x-intercept (y = 0) (⅔; 4⅔27) 4 f (x) concave up POI f (x) = 0 (1; 0) • x-intercept (y = 0) concave down (–1;0) f ′′(x) > 0 x 3 − 5x 2 + 8x − 4 = 0 x f (x) = − 3 f ′′(x) < 0 (2;0) (3;0) x f (1) = 0 ∴ (x − 1) is a factor x 3 − 5x 2 + 8x − 4 = 0 Second derivative test: 1 1−5+ 8−4 f (−1) = 0 ∴ (x + 1) is a factor Another way to find the nature of the stationary points: ↓+ 1−4+ 4 x 3 − x 2 − 5x − 3 = 0 (0;–3) 1−4+ 4 ⋅ (0;–4) x3 −3 If: * f ′′(x) > 0 at the stationary point, is LOCAL MIN = (x + 1)(x 2 − 2x − 3) (x − 1)(x 2 − 4x +4) = 0 * f ′′(x) < 0 at the stationary point, is LOCAL MAX x2 (⅔; 4 ⅔27) 5 -256 (x − 1)(x − 2)2 = 0 −2x 2 * f ′′(x) = 0 at the stationary point, is POINT OF INFLECTION x = 1 or x = 2 = (x + 1)(x + 1)(x − 3) = 0 ((x = 2) is double x-cut, ∴ also a SP) x = − 1 or x = 3 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 31

Grade 12 Maths Essentials Differential Calculus SCIENCE CLINIC 2022 © EXAMPLE 2: APPLICATION QUESTIONS FINDING THE EQUATION OF A CUBIC CURVE Questions: 1. Find the x-value of the POI and add it onto your sketch of f (x). x-intercepts given Stationary Points given Derivative graph given 2. State the values of x for which: a. f (x) ≥ 0 (above/on the x-axis) y = a (x − x1)(x − x 2 )(x − x 3) Remember: f ′(x) = 0 This is a sketch of f ′(x), the derivative of y f (x) = a x 3 + b x 2 + c x + 8 b. f ′(x) > 0 (gradient positive and increasing) [x1; x 2; x 3 : x-intercepts] (2;2) y c. f ′′(x) < 0 (concave down) f (x) f (x) f ′(x) graph is a parabola with x-intercepts given d. f (x) ⋅ f ′(x) > 0 (positive product) (0;2) y = p (x − x )(x − x ) y x 1 2 e. f ′′(x) ⋅ f ′(x) ≤ 0 (negative product) f l (x) (–1;0) y = p (x + 2)(x − 6) (2;0) (0;2) x (0;–2) Sub in (0; − 4) x (–2;0) (6;0) Solutions: −4 = p (0 + 2)(0 − 6) 1. y If f (x) = − x 3 + b x 2 + c x + d f (x) 1 (0;–4) y = a (x − x1)(x − x 2 )(x − x 3) then f ′(x) = − 3x 2 + 2b x + c p= 3 (–1;0) (⅔;0) 1 at x = 0 and x = 2 y = a (x + 1)(x − 2)(x − 2) (3;0) x 1 (0; 2) 2 = a (0 + 1)(0 − 2)2 ∴ y = (x + 2)(x − 6) 3 2 = a (0 + 1)(0 − 2)2 −3(0)2 + 2b (0) + c = 0 1 2 4 (0;–3) y = x − x −4 c = 0 3 3 4a = 2 1 2 4 1 −3(2)2 + 2b (2) = 0 ∴ f ′(x) = x − x −4 a = 3 3 2 b = 3 (⅔; 4 ⅔27) 5 -256 But f ′(x) = 3a x 2 + 2b x + c 2. 1 1 4 ∴ y = (x + 1)(x − 2)2 ∴ y = − x 3 + 3x 2 + d (Sub in (0; − 2) or (2; 2) ∴ 3a = 2b = − c = −4 2 3 3 a. x = − 1 or x ∈[3; ∞) 1 −2 = − (0)3 + 3(0)2 + d 1 2 (3 ) 5 = (x + 1)(x 2 − 4x + 4) a = b = − b. x ∈(−∞; − 1) ∪ ;∞ 2 ∴ d = −2 9 3 1 3 3 2 1 3 2 2 f (x) = x − x − 4x + 8 ( 3) 1 = x − x + 2 ∴ y = − x 3 + 3x 2 − 2 c. x ∈ − ∞; 2 2 9 3 d. f (x) f ′(x) ANALYSING DERIVATIVE GRAPHS below x-axis For which value of x does f (x) have the following characteristics? above x-axis f ′(x) graph → f (x) graph • Decreasing (i.e. f ′(x) < 0) • Concave up (grad incr L→R) ( 3) ( 2) 5 y x ∈(−∞; − 4) ∪ (1; ∞) 3 x ∈ − 1; ∪ (3; ∞) Negative (i.e. f ′(x) < 0 ) Slope is decreasing x ∈ − ∞; − • Increasing (i.e. f ′(x) > 0) e. f ′′(x) f ′(x) Positive (i.e. f ′(x) > 0 ) Slope if increasing • Concave down (grad decr L→R) x ∈(−4; 1) ( 2 ) (1;0) 3 conc down Zero/x-intercepts Stationary points (–4;0) x • SPs (i.e. f ′(x) = 0) x ∈ − ;∞ (i.e. f ′(x) = 0 ) (SPs) conc up f l (x) x = − 4 or x = 1 • Hence, the general shape of f (x) Turning point Point of inflection • POI (i.e. f ′(x) − TP) would be: x=1 [3 3 ] 1 5 −4 + 1 −3 x ∈(−∞; − 1] ∪ ; (i.e. f ′′(x) = 0 ) (POI) POI x = = x = –½ 3 2 2 x = –4 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 32

Grade 12 Maths Essentials Applications of Differential Calculus SCIENCE CLINIC 2022 © MINIMA AND MAXIMA RATE OF CHANGE CALCULUS OF MOTION dy Minima/ maxima are found when the derivative is zero (i.e. f ′(x) = 0) If y is a function of x, then the instantaneous rate of change is . dx Displacement s m EXAMPLE EXAMPLE ds The volume of a tank of water at a given time (t in minutes) is Velocity v m/s v= dt given by v = 10 + 8t − 2t 2 m3 dv d 2s Acceleration a m/s 2 a= or a = dt d t2 Questions: x 60 – 2x 1. What is the rate at which the volume is changing after EXAMPLE 1 minute? The displacement of a projectile is given by s = 5t 2 − 20t with s 2x 2. After how long will the water’s volume be a maximum? in metres and t in seconds. 3. When will the tank be empty? A rectangular box has the dimensions x × 2x × (60 − 2x) cm. Questions: Solutions: 1. What is the time when the displacement is a maximum? Questions: 1. v = 10 + 8t − 2t 2 m3 2. What is the velocity of the projectile after 5 seconds? 1. Determine an expression for the volume of the box in terms dv 3. What is the acceleration of the projectile? of x. Rate of change: = 8 − 4t dt 2. Determine the dimensions of the box that would give the maximum volume. At t = 1: 8 − 4(1) Solutions: 3. What is the maximum volume? = 4m3/minute ds dv 1. = v= 0 2. Max when = 0 dt dt Solutions: 10t − 20 = 0 ∴ 8 − 4t = 0 1. v = ℓ bh = x (2x)(60 − 2x) ∴ t = 2 seconds t = 2 minutes = 120x 2 − 4x 3 cm3 2. v = 10t − 20 3. v = 0 2. v′ = 0 at maximum volume = 10(5) − 20 10 + 8t − 2t 2 = 0 240x − 12x 2 = 0 30 m/s t 2 − 4t − 5 = 0 12x (20 − x) = 0 3. v = 10t − 20 (t − 5)(t + 1) = 0 dv x ≠ 0 or x = 20 t = 5 or t ≠ − 1 a= = 10 m/s 2 dt Dimensions: x × 2x × (60 − 2x) cm ∴ t = 5 minutes = 20 × 40 × 20 cm 3. Method 1: Method 2: v= 120x 2 − 4x 3 v = ℓ bh = 120(20)2 − 4(20)3 = (20)(40)(20) = 16 000 cm3 = 16 000 cm3 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 33

Grade 12 Maths Essentials Probability SCIENCE CLINIC 2022 © Theoretical Probability of an event happening: Relative frequency or Experimental probability: Theoretical Probability of an event happening: number of possible times event can occur number of times the event occured S = {sample set} P(E ) = P(E ) = number of possible outcomes number of trials done A = {event A} n(E ) B = {event B} = n(S) A ∪ B = {A union B} = in sets A or B A ∩ B = {A intersection B} = in sets A and B E = event S = sample space Addition Rule: Mutually Exclusive: Exhaustive Events: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) A ∩ B = {} Events are exhaustive when they cover all elements in the sample set. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) S A and B are mutually exclusive events as they have no elements in common. A and B are inclusive events as they have elements in common. A B S S A ∩ B = {2; 6} 1 2 5 NOTE: 3 4 6 A B A ∪ B = {1; 2; 3; 4; 5; 6; 7; 8} A B n(A ∪ B ) = n(A) + n(B ) n(A) = 5 1 5 1 2 5 6 n(A ∩ B ) = 0 4 2 7 n(B ) = 5 3 4 6 3 8 P (A ∪ B ) = P (A) + P (B ) n(A ∩ B ) = 2 9 9 n(A ∪ B ) = 8 Complimentary Events: Independent Events: Dependent Events: Events A and B are complimentary events if they are mutually Independent events are two events that do not affect each Dependent events are when the first event (A) affects the exclusive and exhaustive. other’s outcomes. E.g. choosing two coloured marbles from a bag, outcome of the second event (B). E.g. choosing two coins from a with replacement, thus, the first choice doesn’t affect the outcome wallet without replacing the first coin. The first choice affects the of the second choice. second choice as the coin in hand in no longer available for the second choice. n(not in A) = n(A′) = 1 − n(A) Thus, the multiplication rule holds P(A′) = 1 − P(A) = P(B) P(A and B) = P(A ∩ B) = P(A) × P(B) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 34

Grade 12 Maths Essentials Probability SCIENCE CLINIC 2022 © TREE DIAGRAMS CONTINGENCY TABLE (OR TWO-WAY TABLE) EXAMPLE 1 EXAMPLE 2 EXAMPLE 1 Questions: Questions: Questions: In a group of 26 learners, 10 wear glasses, 8 are left handed and 6 of the students The girls’ hockey team has a match on Saturday. A bag containing 4 red metal balls, 5 blue metal balls and 3 green are both left-handed and wear glasses. There is an 85% chance Makayla will play goalie. metal balls. A ball is chosen at random and not replaced and then 1. Complete the missing information in the table below. If she does there is a 70% chance the team will another ball is chosen at random again. Draw a tree diagram to list win but if she doesn’t play there is a 45% all possible events Left-handed Right-handed TOTAL chance they will win the game. Use the tree diagram to determine the probabilities of the following: a b c Glasses Draw a tree diagram to help answer the follow- 1. a blue metal ball in the first drawn. ing questions: No glasses d e f 2. a red metal ball then a green metal ball is drawn. 1. What is the probability of Makayla playing 3. two metal balls of the same colour are drawn. TOTAL g h i goalie and the team winning? 2. Calculate that if you pick any one of the 26 learners at random that they will be: 2. What is the probability of the team losing the Solutions: match on Saturday? a. Right handed 0,27 b. Right handed and not wear glasses Red (R;R) Solutions: 0,33 0,45 Solutions: Red Blue (R;B) 0,7 1. a = 6; b = 4; c = 10; d = 2; e = 14; f = 16; g = 8; h = 18; i = 26 Win 0,85 Mikayla 0,27 goalie Green (R;G) 18 2. P(RH) = = 0,69 (MG) 0,3 26 Loose 0,36 14 0,45 Red (B;R) 3. P(RH and NG) = = 0,54 Win 26 0,15 Mikayla 0,42 0,36 not goalie Blue Blue (B;B) EXAMPLE 2 (MNG) 0,55 Loose Questions: 0,27 A survey was conducted asking learners which hand they use to write with and Green (B;G) 1. P(MG and win) = 0,85 × 0,7 what colour ink they prefer. The results are summarised below. = 0,6 0,36 Hand used to write with Red (G;R) Left (L) Right (R) Total 2. P(MG and loose) + P(MNG and loose) 0,25 0,45 Ink Blue (Bu) a b 24 Green Blue (G;B) = 0,85 × 0,3 + 0,15 × 0,55 colour Black (Ba) c d 56 = 0,34 0,18 Total 50 30 80 Green (G;G) The survey concluded that “the hand used to write with” and “Ink colour” are 1. P(B and any) = 0,42 × 0,36 + 0,42 × 0,36 + 0,42 × 0,27 independent. Calculate the values of a, b and c. = 0,42 Solutions: P(L and BU) = P(L) × P(Bu) 2. P(R then G) = 0,33 × 0,27 a 50 80 = 80 × 24 80 = 0,09 a = 15 3. P(RR or BB or GG) = 0,33 × 0,27 + 0,42 × 0,36 + 0,25 × 0,18 ∴ b = 24 − 15 = 9 = 0,29 c = 50 − 15 = 35 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 35

Grade 12 Maths Essentials Probability - Venn Diagram SCIENCE CLINIC 2022 © EXAMPLE 1 EXAMPLE 2 V ∪M (V ∪ M )′ or V′ ∩ M′ Questions: Questions: V M V M Calculate from the Venn diagram for a grade 6 group 120 Gr 12 girls at Girls High where asked about their participation in which the number of equally likely ways the in the school’s cultural activities: events (Reading(R); Sports(S) and Art(A)) can occur • 61 girls did drama (D) has been filled in: • 29 girls did public speaking (P) • 48 girls did choir (C) • 8 girls did all three S S Gr 6 • 11 girls did drama and public speaking • 13 girls did public speaking and choir R S • 13 girls did no cultural activities M V 10 97 1. Draw a Venn diagram to represent this information. V M V M 20 2. Determine the number of Girls who participate in drama only 5 and choir only, respectively. 25 8 3. Determine the probability that a grade 12 pupil chose at ran- dom will: 3 S S a. only do choir. 2 b. not do public speaking. A c. participate in at least two of these activities. V∩M (V ∩ M )′ or V′ ∪ M′ V M V M 1. P(A⋂R⋂S) Solutions: 2. P(R and A and not S) 1. n(S) = 120 3. P(A or R) n(D) = 61 n(P) = 29 4. P((S or R) and not A) S S 50–x 3 13 Solutions: 8 5 1 x 5 V′ ∩ M V ∪ M′ 1. P(A ∩ R ∩ S) = = 170 34 35–x V M V M 25 5 13 2. P(R and A and not S) = = n(C) = 48 170 34 70 7 2. (50 − x) + 3 + 13 + x + 8 + 5 + (35 − x) + 13 = 120 3. P(A or R) = = 170 17 127 − x = 120 S S ∴x = 7 127 4. P(S or R and not A) = ∴ 50 − x = 43 and 35 − x = 28 170 M′ 3. 35 − 7 28 V M a. P(C only) = = = 0,23 120 120 120 − 29 91 b. P(P′) = = = 0,76 120 120 3+ 7+ 8+ 5 23 S c. P(at least 2) = = = 0,19 120 120 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 36

Grade 12 Maths Essentials Probability - Fundamental Counting Principles SCIENCE CLINIC 2022 © Given different choices c, d and e Order of arrangement is important (Permutation): Arrangements and Set Positions: n! n(s) = number of positions × number of arrangements in each position n(s) = c × d × e n(s) = or n(s) = nPr (n− r)! EXAMPLE: EXAMPLE: How many different outfits could you put How many ways can 5 Maths books, 2 Afrikaans books and 3 English books be arranged if they are together with 4 shirts, 6 skirts and 2 pairs of where; grouped in their subjects? shoes? r = number of specific choices Number of positions = 3 n(s) = 4 × 6 × 2 EXAMPLE: Number of arrangements for Maths books = 5! There are 7 players in a netball team who hope to Number of arrangements for Afrikaans books = 2! = 48 outfits Number of arrangements for English books = 3! be shooter or goal attack. How many different Arrangements with repetition: options are there? 7! n(s) = 3! × 5! × 2! × 3! n(s) = k x n(s) = (7 − 2)! Where; = 8 640 k = number of choices = 42 x = number of times you can choose EXAMPLE: EXAMPLE: EXAMPLE: or Questions: Questions: A four-digit code can be made from numbers 1 A password consists of 8 characters. The first How many ways van the letters in ‘ERIN’ be to 9 and 4 vowels. two characters must be any consonant and may arranged with repetition? n(s) = 7P 2 1. How many possible codes can be made not be repeated. The third letter is a vowel. n(s) = 44 with repetition? The next four characters form a four-digit = 42 2. How many codes can be formed if the number which must not start with 0 but digits = 256 vowels cannot be repeated? can repeat. The last character is a vowel which Use [nPr] key on calculator: 3. What is the probability of a code been must be different from the first vowel. EXAMPLE: [7][ nPr][2] [=] created with no numbers been repeated? For example: HG E 2558 A How many three letter codes can be made 1. How many different passwords are Identical items (repetition) in an arrangement: Solutions: from the letters d, g, h, m, r, and t, if the possible? 1. n(S) = n(no . codes) + n(vowel codes) letters can be repeated? n! = 9 4 + 54 2. What is the Probability the code will have 3 n(s) = n(s) = 6 m! × p! = 7 186 an even number between the letters and end with an A? = 216 2. n(E) = n(no . codes) + n(vowel codes) Solutions: where; = 9 4 + 5P4 = 6 681 1. 26 letters - 5 vowels = 21 consonants Arrangements without repetition: m and p: number of times different items are repeated 0;1;2;...9 is 10 numbers n(s) = p! (factorial notation) 3. n(E 2 ) = n(no . codes) + n(vowel codes) = p × ( p − 1) × ( p − 2) × ( p − 3) × . . . EXAMPLE: = 9P4 + 54 n(S) = 21 × 20 × 5 × 9 × 10 3 × 4 How many times can the letters in the name = 3 649 = 75 600 000 EXAMPLE: ‘VANESSA’ be arranged? n(E2 ) 2. n(E ) = 21 × 20 × 4 × 9 × 10 2 × 5 × 1 How many ways can the letters in Erin be ∴ P (E2 ) = = 7 560 000 arranged without repetition? n(S) There are 2 A’s and 2 S’s: 3649 n(s) = 4! = ∴ P (E ) = n(S) 7! 7186 n(E ) = 4×3×2×1 n(s) = 2! × 2! = 0,51 7 560 000 = = 24 = 1260 75 600 000 = 0,1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 37

Grade 12 Maths Essentials Probability SCIENCE CLINIC 2022 © EXAMPLE: EXAMPLE: EXAMPLE: Questions: Questions: Questions: Consider the word Matric-Vacation. 1. Events A and B are mutually exclusive. It is further given that: James has four R10, six R20, two R100 and three R200 notes in 1. How many different ways can the letters be arranged? Treat • 3P(B) = P(A), and his wallet. all letters as different. • P(A or B) = 0,64; 1. In how many different ways can the notes be arranged? 2. What is the probability that the M and V will always be next to 2. What is the probability that all the R200 notes are next to each each other? Treat all letters as different. Calculate P(A). other? 3. What is the probability that the ‘word’ will never start with cc? Treat all letters as different. 2. A and B are independent events such that P (A ∩ B ) = 0,27 and Solutions: 4. If the letters are not different, how many different ways can P (B ) = 0,36. 15! 1. n(S) = the letters be arranged? 4! × 6! × 2! × 3! Find P(A). = 6 306 300 Solutions: 1. n(s) = 14! Solutions: = 87 178 291 200 1. For mutually exclusive events: P(A) + P(B) = P(A or B) 13! 2. n(R200 next to each other) = 4! × 6! × 2! 3P (B ) = P (A) 2. MV or VM give 2!; MV can be treated as one letter group thus P (A) = 180 180 13 letters ∴ P (B ) = n(E1) = 2! × 13! 3 = 12 454 041 600 180 180 P (A) P (R200 next to each other) = P (A) + = 0,64 6 306 300 12 454 041 600 3 ∴ P (E1) = 1 87 178 291 200 3P (A) + P (A) = 1,92 = = 0,03 1 35 = = 0,143 ∴ P (A) = 0,48 7 3. P (c c) = 2! 2. For independent events: (A) × P(B) = P(A ∩ B) cc has to be a group and first, therefore 12 options remain following cc P (A) × P (B ) = P (A ∩ B ) P(not starting with cc) = 1 − P(starting with cc) P (A) × 0,36 = 0,27 2! × 12! P (A) = 0,75 = 1− 14! 90 = = 0,989 91 4. Note that there are 3 A’s, 2 T’s, I’s and C’s 14! n(S) = 3! × 2! × 2! × 2! = 1 816 214 400 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 38

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Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © FLASHBACK: Theory CIRCLE GEOMETRY from previous grades Theorem 1: Converse of Theorem 1: Converse two of Theorem 1: (line from centre ⟂ chord) (line from centre mid-pt. chord) (perp bisector of chord) A The line segment joining the centre of a circle to the The perpendicular bisector of a chord passes through A line drawn from the centre of a circle perpendicular midpoint of a chord is perpendicular to the chord. the centre of the circle. to a chord bisects the chord. K L J 1 2 N M P B C If JK = K L, then OK ⊥ J L O B̂ = C1̂ (∠ 's opp. = sides) Â + B̂ + C1̂ = 180∘ (sum ∠ 's of Δ) O C2̂ = Â + B̂ (ext. ∠ 's of Δ) EXAMPLE GIVEN: R T = R P and M R ⊥ T P Given circle centre M with a diameter of 20 cm and GIVEN: Circle centre O with chord N P ⊥ MO . chord DF of 12 cm. D RTP: M R goes through the centre of the circle. F RTP: N M = M P x PROOF: 2 1 2 1 K Choose any point, say M, on A D. M 3 PROOF: Join M T and M P Join ON and OP E x–3 In ΔM R P and ΔM R T G In ΔMON and ΔMOP PR = R T (given) K2̂ = M̂1 (corres. ∠ 's DE//GF) N ̂ MO = P ̂ MO (OM⟂PN, given) M R = M R (common) ON = OP (radii) M R̂ P = M R̂ T = 90∘ (∠‘s on a str. line) K2̂ = M̂3 (alt. ∠ 's DE//GF) OM = OM (common) Determine the length of of chord AC. K2̂ + M̂2 = 180∘ (co-int. ∠ 's DE//GF) ΔM R T ≡ ΔM R P (SAS) ∴ ΔMON = ΔMOP (RHS) ∴ M T = MP M̂1 = M̂3 (vert. opp. ∠ 's) NM = MP Join MF ∴All points on A D are equidistant from P and T and K2̂ + K1̂ = 180∘ (∠ 's on a str. line) DE = E F = 6 cm (line from centre ⟂ chord) the centre is equidistant from P and T. MF = 10 cm (radius) ∴The centre lies on A D. x 2 = 10 2 − 62 (Pythag. Th.) x 2 = 64 P x = 8 cm ∴ M B = 8 − 3 = 5 cm (given) T Join M A R M A ⊥ AC (line from centre mid-pt. chord0 P T 2 = PR 2 + R T 2 (Pythag. Th.) M A = 10 cm (radius) A B 2 = 10 2 − 52 (Pythag. Th.) A B 2 = 75 A B = 8,66 cm ∴ AC = 17,32 cm Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 39

Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © CIRCLE GEOMETRY Theorem 2: EXAMPLE 1 Theorem 3: Converse Theorem 3: (∠ at centre = 2 x ∠ at circum.) (∠ in semi-circle) (chord subtends 90°) Determine the value of x: The angle subtended by an arc at the cen- The angle subtended by the diameter at the If a chord subtends an angle of 90° at the circumference tre of the circle is twice the angle the arc circumference of a circle is a right angle. of a circle, then that chord is a diameter of the circle. subtends at any point on the circumference of the circle. If A MC is the diameter then B̂ = 90∘. If B̂ = 90∘ then A MC is the diameter. x = 54∘ ÷ 2 (∠ at centre = 2 x ∠ at circum.) ∴ x = 27∘ ALTERNATIVE DIAGRAMS: EXAMPLE In circle O with diameter AC, D C = A D EXAMPLE 2 and B2̂ = 56∘. Determine the size of D Â B Determine the value(s) of x and y: D B GIVEN: Circle centre M with arc A B x subtending A M̂ B at the centre and A Ĉ B at the circumference. 2 E y O1 3 RTP: A M̂ B = 2 × A Ĉ B 88° PROOF: CO = OB (radii) A M = B M = C M (radii) A C C2̂ = B2̂ = 56∘ (∠ ‘s opp. = sides) Â = C2̂ (∠ ‘s opp. = sides) O1̂ = 68∘ (sum ∠ ‘s of Δ) B̂ = C1̂ (∠ ‘s opp. = sides) x = 44∘ (∠ at centre = 2 x ∠ at circum.) A2̂ = 34∘ (∠ at centre = 2 x ∠ at circum.) OB = OC (radii) D̂ = 90∘ (∠ in semi-circle) M̂1 = Â + C2̂ (ext. ∠ of Δ) Ĉ = 44∘ (∠ ‘s opp. = sides) A1̂ = C1̂ (∠‘s opp. = sides, DC = AD) ∴ M̂1 = 2C2̂ O3̂ = 92∘ (sum ∠ ‘s of Δ) A1̂ = 45∘ (sum ∠ ‘s of Δ) O2̂ = 88∘ (vert. opp. ∠ ‘s) ∴ D Â B = 34∘ + 45∘ = 79∘ M̂2 = B̂ + C1̂ (ext. ∠ of Δ) ∴ M̂2 = 2C1̂ 88∘ + 92∘ + 88∘ y = 2 ∴ M̂1 + M̂2 = 2(C1̂ + C2̂ ) y = 134∘ (∠ at centre = 2 x ∠ at circum.) ∴ A M̂ B = 2 × A Ĉ B Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 40

Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © CIRCLE GEOMETRY Theorem 4: COROLLARIES: Converse Theorem 4: (∠ in same seg.) (line subt. = ∠’s) a) Equal chords (or arcs) subtend equal Angles subtended by a chord (or arc) at the If a line segment joining two points subtends equal angles at two other points on the same side of the line angles at the circumference. circumference of a circle on the same side of segment, then these four points are concyclic (that is, they lie on the circumference of a circle.) the chord are equal. K L = ST then P̂ = M̂ (= chords, = ∠’s) GIVEN: Circle centre N with arc R T subtending If Ŵ = U,̂ then W U Z Y is a cyclic quadrilateral. R P̂ T and R M̂ T in the same segment. b) Equal chords subtend equal angles at EXAMPLE 1 EXAMPLE 2 centre of the circle. RTP: R P̂ T = R M̂ T Given circle centre O with Ĉ = 36∘ Given circle A BC D with A B | | E F. PROOF: Join N R and N T to form N1̂ . 1 M̂ = × N1̂ (∠ at centre = 2 x ∠ at circum.) 2 1 P̂ = × N1̂ (∠ at centre = 2 x ∠ at circum.) 2 If A B = C D then O1̂ = O2̂ (= chords, = ∠’s) ∴ R M̂ T = R P̂ T c) Equal chords in equal circles subtend Questions: equal angles at their circumference. Calculate the values of angles: a) Prove C DE F is a cylindrical quad. O1̂ , Â and B.̂ b) If D2̂ = 38∘, calculate E2̂ Solutions: a) B1̂ = C1̂ (∠‘s same seg.) O1̂ = 2 × 36∘ = 72∘ (∠ at centre = 2 x ∠ at circum.) B1̂ = F1̂ (corres. ∠ ‘s, AB||EF) ∴ C1̂ = F1̂ Â = B̂ = Ĉ = 36∘ (∠‘s same seg.) ∴ C DE F cyc . quad (line subt = ∠‘s) b) D2̂ = E2̂ = 38∘ (∠‘s same seg quad CDEF) If HF = PQ then Ĝ = R̂ (= chords, = ∠’s) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 41

Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © CIRCLE GEOMETRY Theorem 5: Converse Theorem 5: Theorem 6: (opp. ∠‘s quad supp) EXAMPLE 1 (opp. ∠‘s cyc. quad) (ext. ∠ cyc quad) If the opposite angles of a quadrilateral are GFE is a double chord and H1̂ = 75∘ The opposite angles of a cyclic quadrilateral are The exterior angle of a cyclic quadrilateral is supplementary. supplementary, then the quadrilateral is cyclic. equal to the interior opposite angle. If Q̂ + Y ̂ = 180∘ or Ĉ + L̂ = 180∘ GIVEN: Circle centre C with quad QUA D. L Q̂ D = Â (ext. ∠ cyc quad) Then QC Y L Determine the value of D.̂ RTP: Q̂ + Â = 180∘ is cyclic Converse Theorem 6: H1̂ = F1̂ = 75∘ (ext. ∠ cyc quad) (ext. ∠ = int. opp. ∠) F1̂ = D̂ = 75∘ (ext. ∠ cyc quad) PROOF: If the exterior angle of a quadrilateral is equal to Join UC and D C the interior opposite angle, then the quadrilateral C1̂ = 2 Â (∠ at centre = 2 x ∠ at circum.) is cyclic. EXAMPLE 2 C2̂ = 2Q̂ (∠ at centre = 2 x ∠ at circum.) Given circle GH JK with GM ⊥ H J and EXAMPLE 2 GL ⊥ L J. G3̂ = 24∘ A BC D is a parallelogram and B Â D = F1̂ . C1̂ + C2̂ = 360∘ (∠‘s around a pt.) Prove that CE F G is a cyclic quad. ∴ 2 Â + 2Q̂ = 360∘ ∴ Â + Q̂ = 180∘ EXAMPLE 1 Calculate the value of α. If L Q̂ D = Â then QUA D is cyclic a) Is quadrilateral GL J M a cyclic quad? b) Is quadrilateral GL J H a cyclic quad? a) M̂2 = 90∘ (Given GM ⊥ H J) L̂ = 90∘ (Given GL ⊥ L J) B Â D = C1̂ (opp. ∠‘s parm) ∴ GL J M cyc quad (opp ∠ ’s quad suppl) B Â D = F1̂ (given) 55∘ − α + 41∘ + 3α = 180∘ (opp. ∠‘s cyc. quad) ∴ C1̂ = F1̂ 2α = 180∘ − 96∘ b) Ĥ = 180∘ − 24∘ − 90∘ (sum ∠‘s of Δ) ∴ CE F G is a cyc quad (ext. ∠ = int. opp. ∠) 2α = 84∘ Ĥ = 66∘ ∴ α = 42∘ GL J H not cyclic (opp ∠‘s = 156° not 180°) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 42

Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © CIRCLE GEOMETRY Theorem 7: EXAMPLE 1 Theorem 8: Theorem 9: Converse Theorem 9: (tan ⟂ radius) (tan from same pt.) (tan-chord th.) (∠ betw. line and chord) Given circle centre O with tangent Z Y U and M N = F G. If Ĥ = 18∘ determine the Two tangents drawn to a circle from the same The angle between a tangent to a circle and If a line is drawn through the end A tangent to a circle is size of Y2̂ . point outside the circle are equal in length. a chord drawn from the point of contact is point of a chord, making with the perpendicular to the radius equal to an angle in the alternate segment. chord an angle equal to an angle at its point of contact. in the alternate segment, then the line is a tangent to the circle. GIVEN: Tangents T PK and SR K to circle centre O. RTP: PK = R K GIVEN: Tangent TA N to circle O, and Y1̂ = Ĥ = 18∘ (equal chords, = ∠‘s) PROOF: chord AC subtending B.̂ If TA N is a tangent to Construct radii OR and OP and join OK. RTP: A1̂ = C2̂ If  = Ĉ or A2̂ = B,̂ circle P, then PA ⊥ TA N Y1̂ + Y2̂ = 90∘ (tan ⟂ radius) In ΔOPK and ΔOR K PROOF: TA N a tangent ∴ Y2̂ = 90∘ − 18∘ = 72∘ OP = OR (radii) Draw in diameter AOD an join D C. Converse Theorem 7: OK = OK (common) A1̂ + A2̂ = 90∘ (tan ⟂ radius) (line seg ⟂ radius) ̂ = O R̂ K = 90∘ (tan ⟂ radius) O PK EXAMPLE 2 C1̂ + C2̂ = 90∘ (∠ in semi-circle) A line drawn perpendicular Prove that T PK is a tangent to circle centre ∴ ΔOPK ≡ ΔOR K (RHS) A2̂ = C1̂ (∠’s in same seg) to the radius at the point O and radius of 8 cm, if OK = 17 cm and ∴ PK = R K ∴ A1̂ = C2̂ where the radius meets the PK = 15 cm. circumference is a tangent to EXAMPLE the circle. PK and K N are tangents to circle centre EXAMPLE 1 M. If N1̂ = 24∘, determine the size of P K̂ N. T R N is a tangent at R and SR = RQ. If R1̂ = x, find five angles equal to x. R1̂ = P1̂ = x (tan-chord th.) Q2̂ = x (tan-chord or ∠’s in same seg) Q2̂ = S2̂ = x (∠‘s opp. = sides) OK 2 = 172 = 289 S ̂2 = P ̂2 = x (∠‘s same seg) OP 2 + PK 2 = 82 + 152 P2̂ = R 4̂ = x (tan-chord th.) = 289 If PA ⊥ TA N, then TA N is a M N̂ K = 90∘ (tan ⟂ radius) tangent to circle P. ∴ OK 2 = OP 2 + PK 2 ∴ N2̂ = 66∘ ∴ OP ⊥ T PK (conv. Pythag. Th.) PK = N K (tan from same pt.) ∴ T PK is a tan to circle O (line seg ⟂ radius) N2̂ = N PK ̂ = 66∘ (∠ ‘s opp. = sides) ∴ P K̂ N = 48∘ (sum ∠‘s of Δ) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 43

Grade 12 Maths Essentials Euclidean Geometry Grade 11 Recap SCIENCE CLINIC 2022 © CIRCLE GEOMETRY EXAMPLE 2 In the figure, A D and A E are tangents to the circle DE F. The straight line drawn through A, parallel to F D meets E D produced at C and E F produced at B . The tangent A D cuts E B at G. Hints when answering Geometry Questions • Read the given information and mark on to the diagram if not already done. • Never assume anything. If not given or marked on diagram is not true unless proved. • As you prove angles equal or calculate angles mark them on to the dia- a) Prove that A BDE is a cyclic quadrilateral given E2̂ = x. gram and write down statement and reason there and then. b) If it is further given that E F = DF, prove that A BC is a tangent to the • Make sure that by the end of the question you have used all the given circle passing through the points B, F and D. information. • If asked to prove something, it is true. a) E2̂ = D2̂ = x (tan-chord th.) For EXAMPLE if asked to prove ABCD a cyclic quad, then it is; but if you D2̂ = A2̂ = x (alt ∠’s AB||FD) can’t then you can use it as one in the next part of the question. ∴ A BDE a cyc quad (line seg subt. = ∠’s) b) E2̂ = D3̂ = x (∠‘s opp. = sides) Note for Matric: F1̂ = E2̂ + D3̂ = 2x (ext. ∠ of Δ) All Grade 11 Theorems and their required A E = A D (tan from same pt.) proofs are also examinable in Grade 12 E1̂ + E2̂ = D2̂ + D3̂ = 2x (∠‘s opp. = sides) ∴ B3̂ = 2x (ext. ∠ cyc quad) B3̂ = F1̂ ∴ A BC tan to circle (∠ betw. line and chord) ALTERNATIVE F1̂ = B1̂ (alt ∠’s AB||FD) B ̂1 = D ̂2 + D ̂3 (∠‘s same seg) D1̂ = E1̂ (∠‘s same seg) E1̂ = D3̂ (tan-chord th.) ∴ B1̂ = D2̂ + D1̂ ∴ A BC tan to circle (∠ betw. line and chord) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 44

Grade 12 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © Theorem 1: Converse of Theorem 1: EXAMPLE 2 (Proportion Theorem) MR In circle O, RQ ⊥ PM, OQ = Q N and x = . A line drawn parallel to one side of a triangle cuts the other If a line divides two sides of a triangle in proportion, then MP two sides in proportion. (Proportion Th. or side 1 ∥ side 2 or the line is parallel to the third side of the triangle. Calculate x. P But MO = ON = r (radii) line ∥ one side △ ) (line divides 2 sides △ in prop. or conv. Prop. Th.) N 1 D R2 OQ = Q N = r (given) 1 2 D Q r + 1r MQ 2 O ∴ = H MN r+ r ⊥h 3 K ⊥k r 2 M = 2r Q 3 1 P = × P Q P̂ = 90∘ (∠ in semi-circle) 2 2 ∴ P̂ = R̂1 = 90∘ (given RQ ⊥ PM) 3 = ∴ RQ ∥ PN (corresp. ∠ ’s equal) 4 E F 3 E F MR MQ ∴ x = ∴ = (line ∥ one side △) MP MN 4 GIVEN: △ DE F, with P on DE and Q on DF and PQ ∥ E F DP DQ Given = , therefore PQ ∥ E F. DP DQ PE QF EXAMPLE 3 RTP: = PE QF In the diagram below A D ∥ E F ∥ BC. Prove that A F ⋅ BD = AC ⋅ DF. A PROOF: EXAMPLE 1 E AB BD B In △ A BD : = (Proportion Th. A D ∥ E F) Construct PF and EQ and draw in altitudes (perpendicular Calculate x AE DF I AB AC heights) In △ A BC : = (Proportion Th. ST ∥ RV) AE AF Q K ( ⊥ k) and PH ( ⊥ h) 5cm BD AC AB 1 K F ∴ = (both = ) Area of triangle: = × base × ⊥ h D DF AF AE 2 1 ∴ BD ⋅ A F = AC ⋅ DF x+1 J Area △ DPQ = × DQ × ⊥ h 2 1 x Area △ FPQ = × FQ × ⊥ h H 6cm L 2 1 C Area △ DPQ × DQ × ⊥ h DQ ∴ = 2 = Area △ FPQ 1 FQ EXAMPLE 4 × FQ × ⊥ h K L ∥ I J (given) 2 1 HK HL ST U R is a parallelogram, with SU X, T U W and RU V straight lines. Prove R T ∥ V W. Area △ DPQ × DP × ⊥ k DP ∴ = (proportion Th., K L ∥ I J or line ∥ one side Δ) Similarly: = 2 = KI LJ Area △ EQ P 1 PE T × PE × ⊥ k ST ∥ RU and SR ∥ T U (opp. sides parm) 2 S x+ 1 6 SU RW But Area △ FPQ = Area △ EQ P ∴ = In △ SR X : = (Proportion Th. SR ∥ T W) 5 x UX WX (Same base PQ and between same ∥ lines) SU TV x (x + 1) = 30 V In △ ST X : = (Proportion Th. ST ∥ RV) x 2 + x − 30 = 0 UX VX DQ DP U RW TV SU ∴ = (x + 6)(x − 5) = 0 ∴ = (both = ) FQ PE R X WX VX UX ∴ x ≠ − 6 or x = 5 W ∴ R T ∥ V W (line divides 2 sides △ in prop.) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 45

Grade 12 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © Mid-Point Theorem: Converse: Theorem 2: Similarity Theorem Converse Theorem EXAMPLE 1 (mid-pt. Th.) (conv. mid-pt. Th.) (AAA) (sides in prop.) Given QU = 24 cm, QC = 16 cm, D A = 6 cm The line segment joining the mid- The line passing through the mid- If the corresponding angles in two triangles If the corresponding sides of two triangles are and C D = x. points of two sides of a triangle, is point of one side of a triangle and are equal, then the corresponding sides are in proportion, then the corresponding angles parallel to the third side and half parallel to another side, bisects the in proportion (AAA) are in equal (sides in prop.) Questions: the length of the third side. third side. The line is also equal to G J 1. Prove △ QUC | | | △ D AC G J half the length of the side it is 2. Calculate x S parallel to. M NK M NK S L L U 1 2 Q P H I H I Q P Q 2 1 O R GIVEN: △ GH I and △ JK L, Ĝ = J;̂ GH GI HI GIVEN: ∴ = = C Ĥ = K̂ and I ̂ = L̂ JK JL KL 2 3 O R 1 4 Therefore if SQ = QO and SP = PR then PQ ∥ OR and Therefore if SQ = QO and GH GI HI RTP: Ĝ = J;̂ Ĥ = K̂ and I ̂ = L̂ RTP: = = 1 2 1 PQ ∥ OR then SP = PR and JK JL KL 2 1 QP = OR (mid-pt. Th.) D A 2 1 PROOF: QP = OR (conv. mid-pt. Th.) 2 PROOF: Construct M on GH such that GM = J K and Construct M on GH such that GM = J K N on GI such that GN = J L Solutions: EXAMPLE and N on GI such that GN = J L Join M N GH GI 1. In △ QUC and △ D AC In △ ACE, A B = BC, GE = 15 cm and A F = FE = E D. Join M N = (given) JK JL Ĉ2 = Ĉ4 (vert. opp. ∠ ’s) Determine the length of CE. In △ GM N and △ JK L GH GI Q̂ 2 = D̂ 2 (∠ ’s in same seg.) Ĝ = J ̂ (given) ∴ = (construction GM = J K and GM GN Û 1 = A1̂ (sum ∠ ’s of △ or ∠ ’s in same seg.) GM = J K (construction) GN = J L (construction) GN = J L) ∴ △ CQU | | | △ C D A (AAA) CQ QU CU In △ ACE ∴ △ GM N ≡ △ J K L (SAS) ∴ M N ∥ H I (line divides 2 sides △ in prop.) ∴ = = ( | | | △ ’s) CD DA CA A B = BC and A F = FE (given) ∴ G M̂ N = K̂ and G N̂ M = L̂ ( ≡ △ ’s) ∴ Ĥ = G M̂ N and I ̂ = G N̂ M 1 BUT Ĥ = K̂ and I ̂ = L̂ ∴ BF ∥ CE and BF = CE (mid-pt. Th.) (corresp. ∠ ’s M N ∥ H I) 2 CQ QU In △ DF B 2. = ∴ Ĥ = G M̂ N and I ̂ = G N̂ M CD DA FE = E D (given) ∴ △ GH I | | | △ GM N (AAA) 16 24 ∴ M N ∥ H I (corresp. ∠ ’s equal) BF ∥ GE (proven) HI GH = GH GI ∴ = x 6 1 ∴ = (proportion Th., M N ∥ H I) MN GM ∴ BG = GD and GE = BF GM GN GH 96 = 24x 2 = (construction) (conv. mid-pt. Th.) GH GI JK ∴ x = 4 cm ∴ = JK JL HI ∴ BF = 2GE = (given) KL ∴ BF = 2(15) = 30 cm GH HI ∴ MN = K L CE = 2BF (proven) Similarly, ∴ = (by constructing P JK KL ∴ △ GM N ≡ △ J K L (SSS) ∴ CE = 2(30) = 60 cm on H I such that H P = K L) ∴ Ĝ = J;̂ M̂ = K̂ and N̂ = L̂ GH GI HI ∴ = = ∴ Ĝ = J;̂ Ĥ = K̂ and I ̂ = L̂ JK JL KL Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 46

Grade 12 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © EXAMPLE 2 EXAMPLE 3 Theorem 3: Similar Right Angled Triangles DM is a diameter of circle A. R T D ⊥ RC, A, B, C and D are points on the circumference of a circle such that A B = BC. (⊥ from rt ∠ vert. to hyp.) R T = 3 cm, RC = 4 cm, MC = 6 cm and A F is a tangent to the circle through A BC D at A. The perpendicular drawn from the vertex of the right angle of a right DM = 2x. angled triangle to the hypotenuse divides the triangle into two similar right Questions: angled triangles, which are similar to the original triangle. (⊥ from rt ∠ vert. Questions: Prove: to hyp.) T 1 R 1. Prove △ C R T | | | △ D C M 1. △ BC D | | | △ BEC 3. △ BC D | | | △ B A F 2 2. Calculate x 2. B Ĉ D = B FA ̂ 4. BD ⋅ A F = BC ⋅ C D D 1 R T 1 D U 2 1 2 2 1 2 P A GIVEN: △ PR T with R̂ = 90∘ and RU ⊥ T P. 1 2 3 RTP: △ PR T | | | △ PU R | | | △ RU T A 2 12 E 14 3 PROOF: T ̂ + R̂1 = 90∘ (sum ∠ ’s of △) 3 R̂1 + R̂2 = 90∘ (given) C 2 F ∴ T ̂ = R̂2 NOTE: M 1 1 3 ∴ P̂ = R̂1 (sum ∠ ’s of △) G 2 In △ PU R and △ RU T Solutions: C B Û 2 = Û 1 = 90∘ J H J 2 = JG ⋅ J I 1. Ĉ3 = 90∘ (∠ in semi-circle) R̂2 = T ̂ (proven) In △ C R T and △ C DM P̂ = R̂1 (proven) R̂2 = Ĉ3 = 90∘ (given and proven) Solutions: ∴ △ PU R | | | △ RU T (AAA) 3. In △ BC D and △ BFA I T1̂ = M̂ (ext. ∠ of cyc. quad) 1. In △ BC D and △ BEC In △ PR T and △ PU R H Ĉ1 = D̂ 2 (sum ∠ ’s of △) i. B̂1 = B̂1 (common) i. B Ĉ D = B FA ̂ (proven) P R̂ T = Û 2 = 90∘ G ∴ △ R T C | | | △ C M D (AAA) ii. D̂ = A2̂ (∠ ’s same seg.) ii. A3̂ = Ĉ2 (tan-chord th.) P̂ = P̂ (common) H I 2 = I J ⋅ IG J RT TC RC but A2̂ = Ĉ2 (∠ ’s opp. = sides, A B = BC) but D̂ = Ĉ2 (proven) T ̂ = R̂2 (proven) ∴ = = ( | | | △ ’s) CM MD CD ∴ D̂ = Ĉ2 ∴ D̂ = A3̂ ∴ △ PR T | | | △ PU R (AAA) iii. B Ĉ D = E4̂ (sum ∠ ’s of △) iii. B̂1 = B̂3 (sum ∠ ’s of △) ∴ △ PR T | | | △ PU R | | | △ RU T ∴ △ BC D | | | △ BEC (AAA) ∴ △ C DB | | | △ FA B (AAA) I 2. T C 2 = 32 + 42 (Pythag.) PR PT H ∴ = ∴ PR 2 = P T ⋅ PU ∴ T C = 5 cm 2. F̂ = E4̂ (ext. ∠ ’s of cyc. quad) CD DB CB PU PR G RT TC 4. = = ( | | | △ ’s) UR PU FA AB FB = B Ĉ D = E4̂ (proven) ∴ = ∴ U R 2 = PU ⋅ U T HG 2 = GJ ⋅ GI CM MD CD DB UT UR J ∴ B Ĉ D = B FA ̂ ∴ = RT PT 3 5 FA AB = ∴ = ∴ RT2 = PT ⋅ U T 6 2x but A B = BC (given) UT RT 6x = 30 ∴ C D ⋅ BC = DB ⋅ FA ∴ x = 5 cm H I Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 47

Grade 12 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © Theorem 4: Theorem of Pythagoras EXAMPLE 1 EXAMPLE 2 (Pythag.) O is the centre of circle with tangents K I and I T. OE I is a straight line. O is the centre of the circle with tangent B A and secant BCE. In a right angled triangle, the area of the square on the OD = DE = D C and AOE is a straight line. hypotenuse is equal to the sum of the areas on the other Questions: two sides. (Pythag.) If E I is 15 cm and I T is 17 cm, calculate: Questions: E 1. E T Prove: T 1 R 2. OE 1. OD ∥ AC 2 3. T O 2. A1̂ = A2̂ K 3. C B = 2E D O 1 1 D 4. A E = 2 2OD 2 2 1 2 U 1 1 P E A 2 2 C O I GIVEN: △ PR T with R̂ = 90∘. RTP: T P 2 = PR 2 + T R 2 Solutions: 1. D1̂ = 90∘ (line from centre mid-pt. chord) PROOF: B T C1̂ = 90∘ (∠ in semi-circle) Construct RU ⊥ T P ∴ D1̂ = C1̂ △ PR T | | | △ PU R (⊥ from rt ∠ vert. to hyp.) Solutions: ∴ OD ∥ AC (corresp. ∠ ’s =) PR PT 1. K I = I T (tan from same pt.) ∴ = ∴ PR 2 = P T ⋅ PU PU PR OK = O T (radii) 2. A1̂ = Ô 1 (corresp. ∠ ’s, OD ∥ AC) △ PR T | | | △ T U R (⊥ from rt ∠ vert. to hyp.) ∴ K I T O is a kite (both pairs adj. sides =) O1̂ = Ê (∠ ’s opp = sides, OD = DE) RT PT ∴ O Ê T = 90∘ (diag. kite ⊥ ) ∴ = ∴ RT2 = PT ⋅ U T A2̂ = Ê (tan-chord) UT RT ∴ E T 2 = 172 − 152 (Pythag) ∴ A1̂ = A2̂ ∴ PR 2 + R T 2 = P T ⋅ PU + P T ⋅ U T ∴ E T = 8 cm = P T (PU + U T ) 3. In △ ACE : DE = D C (given) = P T (P T ) 2. O T ̂ I = 90∘ (tan⊥ rad) OA = OE (radii) ∴ T P 2 = PR 2 + T R 2 ∴ △ T OE | | | △ I T E (⊥ from rt ∠ vert. to hyp.) ∴ AC = 2OD (mid-pt. Th.) ∴ E T 2 = EO ⋅ E I △ A BE | | | △ C A E | | | △ C B A (⊥ from rt ∠ vert. to hyp.) 82 = OE ⋅ 15 ∴ AC 2 = CE ⋅ C B ∴ OE = 4,27 cm ∴ (2OD 2 ) = CE ⋅ C B 4OD 2 = (2OD) ⋅ C B (OD = DE = D C ) 3. △ T OE | | | △ IO T (⊥ from rt ∠ vert. to hyp.) 2OD = C B T O 2 = OE ⋅ OI ∴ 2DE = C B T O 2 = 4,27 ⋅ (4,27 + 15) ∴ T O = 9,07 cm 4. A E 2 = AC 2 + CE 2 (Pythag.) ∴ A E 2 = (2OD)2 + (2OD)2 ∴ A E 2 = 8OD 2 ∴ A E 2 = 2 2OD Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 48

Grade 12 Maths Essentials Trigonometry Grade 11 Recap SCIENCE CLINIC 2022 © BASIC DEFINITIONS BASIC CAST DIAGRAM FUNDAMENTAL TRIG REDUCTION FORMULAE Shows the quadrants where each trig ratio is + IDENTITIES Reducing all angles to acute angles. Memorise: 180° – θ 360° + θ 90⁰ S A θ sin θ = o soh sin A Opposite t h = tan A T C po S A 180° + θ 360° – θ Hy cos θ = a cah cos A e h sin + all + us EXAMPLES en tan θ = o toa 180⁰ II I 0⁰ a Reduce to an acute angle and simplify if possible (without III IV 360⁰ sin2 B + cos2 B = 1 a calculator): T C can be written as 1. sin 125o 2. cos 260 o θ cos + = sin(180 o − 55o) = cos(180 o + 80 o) Adjacent tan + sin2 B = 1 − cos2 B = sin 55o = − cos 80 o cos2 B = 1 − sin2 B (QII so sin is +) (QIII so cos is -) These are our basic trig ratios. 270⁰ EXAMPLE tan 660 o 3. = tan(360 o + 300 o) On the Cartesian Plane Special Angles 1. In which quadrant does θ lie if tanθ < 0 and cosθ > 0? = tan 300 o (QI so tan is +) Remember: y = tan(360 o − 60 o) r = 2 (x; y) 60 o is a = − tan 60 o (QIV so tan is -) special angle tanθ - cosθ + y 3 sin θ = cosθ + = − 1 r r (0; 2) y x tanθ - cos θ = r (1; 3) = − 3 y Quadrant IV tan θ = ( 2; 2) tan(180 o − β)cos(180 o + β)cos2 (360 o − β) θ x x 90o 4. + sin2(180 o + β ) x 60o sin(360 + β) 2. In which quadrant does θ lie if sinθ < 0 and ( 3; 1) (−tan β)(−cos β)(cos β) 2 cosθ < 0? 45o = sin β + (−sin β )2 30o cos3 β Remember: cosθ - = tan β . sin β + sin2 β sinθ - sinθ - cos3 β • x2 + y2 = r2 (Pythagoras) cosθ - = sin β . sin β + sin2 β (2; 0) cos β • Angles are measured upwards from the posi- 0o Remember: tive (+) x-axis (anti-clockwise) up to the hy- = cos2 β + sin2 β Identities potenuse (r). Quadrant III = 1 EXAMPLE 1 Pythagoras Problems 2 cos θ + tan θ 1 Steps: If 3sinθ – 2 = 0 and tanθ < 0, determine 2 cos θ + without using a calculator and using a diagram. − 5 1 tan θ = 2( )+ Remember: 1. Isolate the trig ratio 2 2 2 3 2 x x + y = r ( ) cos θ = r 2. Determine the quadrant 1. 3 sin θ − 2 = 0 2. 3. 4. − 5 3. Draw a sketch and use Pythagoras 2 y tanθ - r=3 x 2 + (2)2 = (3)2 and 4. Answer the question sin θ = 3 r sinθ + sinθ + = −2 5 − 2 5 y y=2 x2 = 5 3 tan θ = tanθ - x θ x = ± 5 −4 5 − 3 5 = 6 ∴ Q u a d r a nt I I ∴x = − 5 −7 5 = 6 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 49

Grade 12 Maths Essentials Trigonometry Grade 11 Recap SCIENCE CLINIC 2022 © NEGATIVE ANGLES CO-FUNCTIONS FULL CAST DIAGRAM Angles measured downwards (clockwise) from the B If A + B = 90∘ then sinA and cosB are Memorise the following diagram: positive x-axis, which can be seen as Quadrant IV. known as co-functions. 90° 90° + θ 90° 90° – θ sinA = sin(90⁰ – B) 90° + θ 90°θ– θ y A = cosB 180° – θ 360°θ + θ EXAMPLES 180° – θ II I S A + θ 0° 360° 180° 1. sin 30 o 2. cos 25o T S III IV C A 360° 0° 180° x = sin(90 o − 60 o) = cos(90 o − 65o) 180°T + θ III IV 360° C – θ 360° –A = cos 60 o = sin 65o θ – 180° 180° +θ – θ– θ 360° NOTE: θ – 180° 270º –θ θ –– 90° θ Look at the quadrant first, THEN use 270° θ 270º – + 90°θ the reduction/co-function formulae * Reductions Co-functions 270° Method 1: Q IV Method 2: Get rid of negative 3. sin(90 o − α) Q I, so sin + PROVING IDENTITIES sin(–A) = –sinA Add 360° to the angle to make = cos α 90 o ∴ sin ↔ cos Steps: cos(–A) = cosA it positive. 1. Separate LHS and RHS tan(–A) = –tanA 2. Start on the more complex side 4. cos(90 o + β ) Q II, so cos − 3. Prove that the sides are equal. EXAMPLES = − sin β 90 o ∴ sin ↔ cos EXAMPLES Simplify without the use of a calculator: sin(–330°) cos x 1 1. cos2 x . tan2 x = sin2 x 3. tan x + = o 1 + sin x cos x NB: Negative Angle 5. sin(θ − 90 ) Q I V, so sin − = − cos θ 90 o ∴ sin ↔ cos LHS = cos2 x . tan2 x cos x LHS = tan x + 2 1 + sin x 1) sin(−330∘ ) 2) + 360∘ 6. Simplify to a ratio of 10⁰: sin x sin x cos x = cos2 x . = + = − sin(330∘ ) = sin(−330∘ ) a) cos 100 o Q II, so cos − cos2 x cos x 1 + sin x = − sin(360∘ − 30∘ ) = sin(360∘ − 330∘ ) = cos(90 o + 10 o) 90 o ∴ sin ↔ cos = − sin(−30∘ ) = sin 30∘ = sin2 x = RHS sin x (1 + sin x) + cos x (cos x) = − sin 10 o = 1 cos x (1 + sin x) = sin 30∘ = 2 b) tan 170 o Q II, so tan − 1 sin x + sin2 x + cos2 x = = tan(180 o − 10 o) 180 o ∴ r ed u c t i on = 2 2. 1 − 2 sin x . cos x = (sin x − cos x)2 = − tan 10 o cos x (1 + sin x) PROBLEM SOLVING: RHS = (sin x − cos x)2 sin x + 1 = If cos25° = p, express the following in terms of p (i.e. get all angles to 25°): cos x (1 + sin x) = sin2 x + cos2 x − 2 sin x . cos x 1. cos(−385 ) o negative angle, so a) + 360° 3. sin(335 ) o 1 o = 1 − 2 sin x . cos x = LHS = = RHS = cos(−25 ) or b) Q IV o o = sin(360 − 25 ) Q IV, sin - cos x = cos 25o - 385 Q I = − sin 25o = p ∴ + cos REMEMBER: Correct angle is 25° BUT sin is wrong ratio. Thus draw sketch. 4. tan(155o) Method 1: Ratio Method 2: Sketch = tan(180 o − 25o) Q II, tan - −y −sin 25o −y 2. sin(65o) Given cos25⁰ = p x So, − sin 25o = = − tan 25o = = = sin(90 o − 25o) Q I, sin + r cos 25o x 1 r = cos(25o) 1 This can be solved in two ways: r = y = √1 –p2 − 1 − p2 − 1 − p2 − 1 − p2 = p 2 = = x=p (Pythag) = = − 1−p p p 1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 50

Grade 12 Maths Essentials Trig Equations Grade 11 Recap SCIENCE CLINIC 2022 © BASICS SQUARES CO-FUNCTIONS FACTORISING Steps: Hints: Hints: Steps: • Isolate trig ratios • Do all four quadrants (± means the ratio • sin and cos with different angles • Solve as you would a quadratic equation must be both + and -) • Reference angle (don’t put negative • Introduce the co-function with 90° - z EXAMPLES into calculator) EXAMPLE • The angle you change is the reference angle Solve for x: • Choose quadrants Solve for β: EXAMPLES ➡ sin or cos: 2 Quadrants 4 sin2 β − 3 = 0 Solve for x: 1. tan2 x − 2 tan x + 1 = 0 ➡ tan: 1 Quadrant 3 2 sin β = 1. cos x = sin(x − 10 o) (tan x − 1)(tan x − 1) = 0 • General solutions 4 cos x = cos(90 o − (x − 10 o)) tan x = 1 ➡ sin θ or cos θ + k 360∘; k ∈ℤ cos x = cos(100 o − x) 3 Reference∠ : 45∘ ➡ tan θ + k180∘; k ∈ℤ sin β = ± Reference∠ : 100∘ − x 4 REMEMBER: Only round off at the end Reference∠ : 60∘ QI: x = 45∘ + k180∘; k ∈ℤ QI: x = 100∘ − x + k 360∘; k ∈ℤ Common formulae: 2x = 100∘ + k 360∘ QI: β = 60∘ + k 360∘; k ∈ℤ x = 50∘ + k180∘ θ = sin−1 a + k 360∘ or QII: β = 180∘ − 60∘ + k 360∘ QIV: x = 360∘ − (100∘ − x) + k 360∘ 2. cos2 x + sin x . cos x = 0 θ = (180∘ − sin−1 a) + k 360∘ (k ∈ℤ) = 120∘ + k 360∘ x − x = 260∘ + k 360∘ QIII: β = 180∘ + 60∘ + k 360∘ 0 = 260∘ + k 360∘ cos x (cos x + sin x) = 0 θ = ± cos−1 a + k 360∘ (k ∈ℤ) = 240∘ + k 360∘ No real solution cos x = 0 OR cos x = − sin x θ = tan−1 a + k180∘ (k ∈ℤ) QIV: β = 360∘ − 60∘ + k 360∘ = 300∘ + k 360∘ cos x −sin x EXAMPLES 2. sin(x + 30 o) = cos 2x Use trig graph: = o o sin(x + 30 ) = sin(90 − 2x) cos x cos x Solve for θ: SINθ AND COSθ Reference∠ : 90∘ − 2x tan x = − 1 1. 3 sin θ − 1 = 0 90⁰ 270⁰ 1 Steps: Reference∠ : 45∘ sin θ = QI: x + 30∘ = 90∘ − 2x + k 360∘; k ∈ℤ 3 • sin and cos with the same angle 3x = 60∘ + k 360∘ x = 90∘ + k180∘; k ∈ℤ QII: x = 135∘ + k180∘ • Divide by cos to get tan x = 20∘ + k120∘ sin + in QI and QII QII: x + 30∘ = 180∘ − (90∘ − 2x) + k 360∘ Reference∠ : 19,47∘ EXAMPLE x + 30∘ = 90∘ + 2x + k 360∘ −x = 60∘ + k 360∘ 3. 2 cos2 x + 3 sin x = 0 Solve for α: QI: θ = 19,47∘ + k 360∘; k ∈ℤ x = − 60∘ − k 360∘ 2(1 − sin2 x) + 3 sin x = 0 QII: θ = 180∘ − 19,47∘ + k 360∘; k ∈ℤ 2 sin 2α − cos 2α = 0 2 sin2 x − 3 sin x − 2 = 0 = 160,53∘ + k 360∘ 2 sin 2α = cos 2α NOTE: Specific Solutions 2 sin 2α cos 2α If they ask for x ∈[−360∘; 360∘ ], choose integer (2 sin x + 1)(sin x − 2) = 0 = values for k cos 2α cos 2α −1 2. tan(3θ + 30 o) + 1 = 0 (...-3; -2; -1; 0; 1; 2; 3...) sin x = OR sin x = 2 o tan(3θ + 30 ) = − 1 2 tan 2α = 1 so that x falls in the given intervals. 2 1 tan 2α = Reference∠ : 30∘ No real solution 2 x = 30∘ + k120∘ tan − in QII x = − 330∘; − 210∘; − 90∘; 30∘; 150∘; 270∘ Reference∠ : 45∘ tan + in QI k = − 3; k = − 2; k = − 1; k = 0; k = 1; k = 2 QIII: x = 180∘ + 30∘ + k 360∘; k ∈ℤ Reference∠ : 26,57∘ OR x = 210∘ + k 360∘ QII: 3θ + 30∘ = 180∘ − 45∘ + k180∘; k ∈ℤ x = − 60∘ + k 360∘ QIV: x = 360∘ − 30∘ + k 360∘; k ∈ℤ 3θ = 105∘ + k180∘ QI: 2α = 26,57∘ + k180∘; k ∈ℤ x = − 60∘; 300∘ x = 330∘ + k 360∘ θ = 35∘ + k 60∘ α = 13,28∘ + k 90∘ k = 0; k = 1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 51

Grade 12 Maths Essentials Trig Graphs Grade 11 Recap SCIENCE CLINIC 2022 © VERTICAL SHIFT IMPORTANT! When sketching trig graphs, you need to label the Notes for sin x and cos x : • y = sin x + q or y = cos x + q or y = tan x + q following: ❖ Key points (intercepts/turning pts) every 90° If q > 0 : upwards (e.g: y = sin x + 1) ! both axes ! x- and y-intercepts If q < 0 : downwards (e.g: y = cos x − 2) ❖ Period (1 complete graph): 360° ! turning points ! endpoints (if not on the axes) EXAMPLE y ! asymptotes (tan graph only) ❖ Amplitude (halfway between min and max): 1 y = cos x − 1 x ∈[0∘; 360∘ ] (solid line) BASICS y = cos x (dotted line - for comparison) Notes for tan x : • y = sin x for x ∈[−360∘; 360∘ ] ❖ Key points every 45° 360° y 0° x ❖ Period (1 complete graph): 180° x (90°;-1) (270°;-1) ❖ No amplitude can be defined (-270°; 1) (90°; 1) ❖ Asymptotes at x = 90o + k180o, k ∈ℤ (180°; -2) -360° -180° 180° 360° 0° x AMPLITUDE CHANGE PERIOD CHANGE (-90°; -1) (270°; -1) • y = a . sin x or y = a . cos x or y = a . tan x • y = sin b x or y = cos b x or y = tan b x If a > 1 : stretch upwards The value of b indicates how many graphs are completed in the ‘regular’ period of 0 < a < 1 : compress downward that graph (i.e. sin x /cos x : 360∘ and tan x : 180∘) a < 0 : reflection in x-axis EXAMPLES • y = cos x for x ∈[−360∘; 360∘ ] EXAMPLES y 1. y = cos 3x x ∈[0∘; 360∘ ] y (90°; 2) 1. y = 2 sin x (solid line) * Normal period: 360° (-360°; 1) (360°; 1) 1 * New period: 120° (3 graphs in 360°) y = sin x * Critical points every 90/3 = 30° (dotted line - 180° 360° -270° -90° 90° 270° y for comparison) 0° x 0° x (120°; 1) (240°; 1) (360°; 1) (-180°; -1) (180°; -1) * Amplitude = 2 30° 150° 270° (270°; -2) 0° 90° 210° 330° x • y = tan x for x ∈[−360∘; 360∘ ] (180°;3) (60°;-1) (180°;-1) (300°;-1) 2. y = − 3 cos x y y (solid line) y y = cos x 1 (dotted line - 1 2. y = tan x x ∈[0∘; 360∘ ] (45°; 1) for comparison) 2 (90°; 1) 0° 90° 270° xx x * Normal period: 180° -360° -180° 0° 180° 360° * Range: y ∈[−3; 3] * New period: 360° -360° 360° x (180°; –1) (½ graph in 180°) * Critical points: every 45/0,5 = 90° -3 (360°; -3) x = -270° x = -90° x = 90° x = 270° x = -180° x = 180° Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 52

Grade 12 Maths Essentials Trig Graphs Grade 11 Recap SCIENCE CLINIC 2022 © HORIZONTAL SHIFT EXAMPLE Solutions: • y = sin(x − p) or y = cos(x − p) or y = tan(x − p) Given f (x) = cos(x + 60∘ ) and g (x) = sin 2x 1. cos(x + 60∘ ) = sin 2x If p > 0 : shift right (e.g: y = sin(x − 30∘ )) Questions: cos(x + 60∘ ) = cos(90∘ − 2x) p < 0 : shift left (e.g: y = cos(x + 45)) Reference ∠ : 90∘ − 2x 1. Determine algebraically the points of intersection of How to plot a horizontal shift: f (x) and g (x) for x ∈[-90°;180°] QI: x + 60∘ = 90∘ − 2x + k 360∘; k ∈ℤ • Plot the original curve 2. Sketch f (x) and g (x) for x ∈[-90°;180°] 3x = 30∘ + k 360∘ • Move the critical points left/right x = 10∘ + k120∘ • Label the x-cuts and turning points 3. State the amplitude of f (x) QIV: x + 60∘ = 360∘ − (90∘ − 2x) + k 360∘; k ∈ℤ • Calculate and label the endpoints and y-cut 4. Give the period of g (x) x + 60∘ = 270∘ + 2x + k 360∘ −x = 210∘ + k 360∘ EXAMPLES 5. Use the graphs to determine the values of x for which: x = − 210∘ + k 360∘ 1. y = cos(x + 45∘ ) for x ∈[−360∘; 360∘ ] (dotted line) a. g (x) is increasing and positive but x ∈[−90∘; 180∘ ] y = cos x (solid line - for comparison) b. f (x) is increasing and positive ∴ x = 10∘; 130∘; 150∘ y c. f (x) ≥ g (x) - i.e. f (x) is above g (x) 2. g (x) = sin 2x (315°; 1) d. f (x) ⋅ g (x) ≥ 0 - i.e. product is + or 0 (-45°; 1) y f (x) = cos(x + 60∘ ) (-360°; ½2 ) 2 6. Explain the transformation that takes y = s i nx to 2 2 (360°; 2 2) -315° -135° 225° y = sin(2x − 60∘ ) (-60°; 1) (45°; 1) 45° x 3 (-90°; ½)2 -90° ½ 30° 90° 180° (-225°; -1) (135°; -1) x 0° (180°; -½) Endpoints: (-45°; -1) (120°; -1) (135°; -1) 2 2 cos(−360∘ + 45∘ ) = and c os (−360∘ + 45∘ ) = 2 2 y-cut: For f(x): 2 3 1 cos(0∘ + 45∘ ) = Endpoints: cos(−90∘ + 60∘ ) = and cos(180∘ + 60∘ ) = − 2 2 2 1 y-cut: cos(0∘ + 60∘ ) = 2 2. y = sin(x − 30∘ ) for x ∈[0∘; 360∘ ] (dotted line) y = sin x (solid line - for comparison) 3. Amplitude = 1 y (120°; 1) 4. Period = 180∘ 30° 210° x 5. a. x ∈(0∘; 45∘ ) (360°; -½) -½ b. x ∈[−90∘; − 60∘ ) (300°; -1) c. x ∈[−90∘; 10∘ ] ∪ [130∘; 150∘ ] d. x ∈[0∘; 30∘ ] ∪ [90∘; 180∘ ] or x = − 90∘ Endpoints: 1 1 sin(0∘ + 45∘ ) = − and sin(360∘ − 30∘ ) = − 2 2 y-cut: 6. Rewrite y = sin(2x − 60∘ ) in the form y = sin b (x − p) = sin(2(x − 30∘ )) The y-cut is one of the endpoints Transformation: b = 2 ∴ period is halved p = 30 ∴ shifted 30 to the right° Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 53

Grade 12 Maths Essentials Trig Graphs Grade 11 Recap SCIENCE CLINIC 2022 © USING TRIG GRAPHS TO FIND RESTRICTIONS ON IDENTITIES i.e. answering the question “for which values of x will this identity be undefined?” Identities are undefined if: • the function is undefined t a nx has asymptotes at x = 90∘ + k180∘; k ∈ℤ • any denominator is zero Reminder: A is undefined 0 EXAMPLES 1. For which values of x will cos2 x ⋅ tan2 x = sin2 x be defined? • t a nx is undefined at x = 90∘ + k180∘; k ∈ℤ ∴ will be defined at x ∈ℝ and x ≠ 90∘ + k180∘; k ∈ℤ • no denominators that could be zero cos x 1 2. For which values of x will tan x + = be undefined? 1 + sin x cos x • t a nx is undefined at x = 90∘ + k180∘; k ∈ℤ • fractions are undefined if the denominator = 0 ∴ if 1 + sin x = 0 or if cos x = 0 * 1 = sin x = 0 ∴ sin x = − 1 y y = sinx Use trig graphs for 0; ± 1 x -1 (270°; -1) ∴ x = 270∘ + k 360∘; k ∈ℤ * cos x = 0 y y = cosx Use trig graphs for 0; ± 1 90° 270° xx ∴ x = 90∘ + k180∘; k ∈ℤ x = 90∘ + k180∘; k ∈ℤ x = 270∘ + k 360∘; k ∈ℤ can be summarised as: x = 90∘ + k180∘; k ∈ℤ x = 90∘ + k180∘; k ∈ℤ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 54

Grade 12 Maths Essentials Trigonometry SCIENCE CLINIC 2022 © CAST DIAGRAM GENERAL SOLUTIONS 90 ∘ • sin A = sin B → ref angle ∘ 1. sin(180 − θ ) = sin θ ∘ 1. sin(θ + 360 ) = sin θ ‣ A = B + k.360∘ or A 180∘ − θ θ + 360∘ 2. cos(180∘ − θ ) = − cos θ 2. cos(θ + 360∘ ) = cos θ ‣ A = 180∘ − B + k.360∘; k ∈ℤ 3. tan(180∘ − θ ) = − tan θ 3. tan(θ + 360∘ ) = tan θ • cos A = cos B ‣ A = ± B + k.360∘; k ∈ℤ ∘ ∘ 90∘ + θ 1. sin(90 + θ ) = cos θ 90∘ − θ 1. sin(90 − θ ) = cos θ • tan A = tan B ∘ 2. cos(90 + θ ) = − sin θ 2. cos(90∘ − θ ) = sin θ ‣ A = B + k.180∘; k ∈ℤ • Other/ Co-function Double 1. sin 2A = 2 sin A ⋅ cos A ‣ sin A = cos B ∴ sin A = sin (90∘ − B ) → ref angle Angles ‣ cos A = sin B ∴ cos A = cos (90∘ − B ) → ref angle 1. sin(A + B ) = sin A ⋅ cos B + cos A ⋅ sin B EXAMPLE 1 Compound Angles Solve for x if cos 2x = − sin x 2. sin(A − B ) = sin A ⋅ cos B − cos A ⋅ sin B 1 − 2 sin2 x = − sin x 0 = 2 sin2 x − sin x − 1 Sin + y + y All 0 = (sin x − 1)(2 sin x + 1) −x θ θ+ x 180∘ θ θ 0∘ 1 sin x = − Tan −y −y Cos sin x = 1 or 2 x = 90∘ + k .360∘ x = 330∘ + k .360∘ or 1. sin(180∘ + θ ) = − sin θ 1. cos(−θ ) = cos θ x = 210∘ + k .360∘; k ∈ℤ 180∘ + θ −θ 2. cos(180∘ + θ ) = − cos θ 2. sin(−θ ) = − sin θ EXAMPLE 2 ∘ 3. tan(−θ ) = − tan θ 3. tan(180 + θ ) = tan θ Solve for x if cos(90∘ − x) ⋅ sin x − cos 2x = 0 1. sin(θ − 90∘ ) = − cos θ sin x ⋅ sin x − (1 − 2 sin2 x) = 0 θ − 90∘ sin2 x − 1 + 2 sin2 x = 0 2. cos(θ − 90∘ ) = sin θ 3 sin2 x = 1 1 Double 1. cos 2A = cos2 A − sin2 A sin2 x = 3 Angles 2. cos 2A = 2 cos2 A − 1 ± 3 sin x = 3. cos 2A = 1 − 2 sin2 A 3 x = 35,26∘ + k .360∘ or x = − 35,26∘ + k .360∘ Compound 1. cos(A + B ) = cos A ⋅ cos B − sin A ⋅ sin B or or Angles 2. cos(A − B ) = cos A ⋅ cos B + sin A ⋅ sin B x = 180∘ − 35,26∘ + k .360∘; k ∈ℤ x = 180∘ − (−35,26∘ ) + k .360∘ 270∘ x = 144,74∘ + k .360∘ x = 215,26∘ + k .360∘ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 55

Grade 12 Maths Essentials Trigonometry SCIENCE CLINIC 2022 © TRIG IDENTITIES EXAMPLE 2 MIXED EXAMPLE 1 sin x Show\Prove that: sin 3x = 3 sin x − 4 sin x 3 If sin 54∘ = p, express the following in terms of p: • tan x = cos x 1 cos x 1. cos 36∘ 2. sin 108∘ 3. sin 84∘ • tan x = sin x LHS = sin(2x + x) Solutions: 1 (h ) • sin 2x = 2 sin x ⋅ cos x p o = sin 2x ⋅ cos x + cos 2x ⋅ sin x sin 54∘ = 36° 1 • sin 3x = sin(2x + x) p = 2 sin x ⋅ cos x ⋅ cos x + (1 − 2 sin2 x) ⋅ sin x 3. sin 84∘ = sin(54∘ + 30∘ ) = sin 2x ⋅ cos x + cos 2x ⋅ sin x (to be expanded further) 54° = sin 54∘ ⋅ cos 30∘ + cos 54∘ ⋅ sin 30∘ = 2 sin x ⋅ cos2 x + sin x − 2 sin3 x 1. cos 36∘ = p 1 – p2 • sin 4x = sin 2(2x) ( )( 2 ) 3 1 = 2 sin x (1 − sin2 x) + sin x − 2 sin3 x 2. sin 108∘ = sin 2(54∘ ) = p⋅ + 1 − p2 = 2 sin 2x ⋅ cos 2x 2 = 2 sin 54∘ ⋅ cos 54∘ = 4(sin x ⋅ cos x)(cos2 x − sin2 x) = 2 sin x − 2 sin3 x + sin x − 2 sin3 x 3p+ 1 − p2 = 4 sin x ⋅ cos3 x − 4 sin2 x ⋅ cos x (can be expanded further) = 2( p) ⋅ ( 1 − p 2 ) = = 3 sin x − 4 sin3 x 2 2 2 • cos 2x = cos x − sin x MIXED EXAMPLE 2 = 1 − 2 sin2 x ∴ LHS = RHS Find the value of k if: cos 75∘ ⋅ sin 25∘ − sin 75∘ ⋅ sin k = sin(−50∘ ) = 2 cos2 x − 1 cos 75∘ ⋅ sin 25∘ − sin 75∘ ⋅ sin k = sin(−50∘ ) HINTS FOR PROVING IDENTITIES EXAMPLE 3 1. Start on the side with the least number of “terms” and simplify sin 25∘ ⋅ cos 75∘ − cos(90∘ − k) ⋅ sin 75∘ = sin(25∘ − 75∘ ) 1 + sin x ( cos x ) 2 if possible. 1 − sin x 1 Show\Prove that: = − tan x sin 25∘ ⋅ cos 75∘ − cos(90∘ − k) ⋅ sin 75∘ = sin 25∘ ⋅ cos 75∘ − cos 25∘ ⋅ sin 75∘ 2. Go to the other side and simplify until you get the same answer. 90 − k = 25∘ 3. Look for a conjugate and multiply with the “opposite” sign ∴ k = 65∘ (to make a difference of squares in the denominator of your fraction) 1 − sin x 1 − sin x LHS = × 4. Always try to factorise where possible 1 + sin x 1 − sin x MIXED EXAMPLE 3 2 1 − 2 sin x + sin x Express the following in terms of p if cos 73∘ ⋅ cos 31∘ + sin 73∘ ⋅ sin 31∘ = p EXAMPLE 1 = 1 − sin2 x 1. cos2 21∘ − sin2 21∘ + 7 2. sin 42∘ sin 2x Show\Prove that: = 2 tan x cos 2x + sin2 x Solutions: ( cos x cos x ) 2 1 sin x RHS = − cos 73∘ ⋅ cos 31∘ + sin 73∘ ⋅ sin 31∘ sin x = cos(73∘ − 31∘ ) RHS = 2 ⋅ 48° 1 cos x 1 − 2 sin x + sin x 2 = cos 42∘ = 1 – p2 2 sin x ⋅ cos x 1 (h ) cos2 x p a 42° LHS = ∴ cos 42∘ = (2 cos2 x − 1) + (1 − cos2 x) 1 − 2 sin x + sin2 x p = 2 sin x ⋅ cos x 1 − sin2 x = cos2 x 1. cos 2(21∘ ) + 7 = cos 42∘ + 7 ∴ LHS = RHS 2 sin x = p+ 7 = cos x 2. sin 42∘ = 1 − p2 ∴ LHS = RHS Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 56

Grade 12 Maths Essentials Analytical Geometry Grade 11 Recap SCIENCE CLINIC 2022 © What is Analytical Geometry? EXAMPLE 3. Midpoint formula x + x1 y2 + y1 M(x ; y) = ( 2 2 ) Analytical Geometry (Co-ordinate Geometry): ; Given: A(−2; 3) and C( p; − 5) are points on a Cartesian Plane. Application of straight line functions in conjunction 2 with Euclidean Geometry by using points on a 1. If AC = 10 units determine the value(s) of p. −2 + 4 3 + (−5) Cartesian Plane. = ( ; ) 2 2 2. If C(4; − 5), determine the equation of the line AC. FLASHBACK M(1; − 1) 3. Determine the co-ordinates of M, the midpoint of AC. Straight line parallel to the x-axis: m = 0 5 Straight line parallel to the y-axis: m = undefined 4. If B( − 1; ) determine if A, B and C are collinear. 4. Prove collinearity by proving that the points share a common gradient. 3 5. Determine the equation of the line perpendicular to AC passing Straight line equation: Δy Δy through B. m = m = y = mx + c Δx Δx 5 5 3− − (−5) 3 3 mAB = mBC = Gradient formula: −2 − (−1) −1 − 4 Δy y − y1 SOLUTION 4 m= = 2 mAB = − 4 mBC = − Δx x 2 − x1 1. Draw a sketch diagram. C has two potential x-coordinates for p. 3 3 Parallel gradients: ∴ A, B and C are collinear d = (x 2 − x1)2 + (y2 − y1)2 m1 = m 2 10 = ( p − (−2))2 + (−5 − 3)2 5. Line equation requires solving m 2 and c w.r.t. B. Perpendicular gradients: 100 = ( p + 2)2 + 64 mAC × m 2 = − 1 m1 × m 2 = − 1 100 = p 2 + 4p + 4 + 64 4 − × m2 = − 1 0 = p 2 + 4p − 32 3 Distance: 0 = ( p − 4)( p + 8) 3 m2 = 2 2 p = 4 or p = − 8 4 d = (x 2 − x1) + (y2 − y1) 2. Line equation requires solving m and c. y = mx + c Co-linear: 5 3 mAB = mBC OR dAB + dBC = dAC m = Δy y = mx + c ( 3 ) = 4 (−1) + c Δx 4 Collinear points A, B and C lie on the same line (3) = − (−2) + c 29 y − y1 c = mAC = 2 3 12 x 2 − x1 1 Midpoint formula: c = 3 − (−5) 3 x + x1 y2 + y1 = M(x ; y) = ( 2 2 ) −2 − 4 ; 4 29 2 4 ∴y = x+ = − 3 12 Midpoint Theorem: If two midpoints on adjacent 3 sides of a triangle are joined by a straight line, the 4 1 line will be parallel to and half the distance of the ∴y = − x+ 3 3 third side of the triangle. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 57

Analytical Geometry Grade 11 Examples: Grade 12 Maths Essentials Grade 11 Recap SCIENCE CLINIC 2022 © Grade 11 Examples: Converting gradient (m) into angle of inclination (θ) Converting a positive gradient into an angle Example 1: EXAMPLE m> 0 Example 1: Δy Given: straight line with the equation 3y −Given: 4x = −straight 5. Determine the the equation 3 mAB = y tan−1(m) = θ line with Δx The reference angle is equal to the angle of inclination. angle of inclination correct to two decimal places. Determine the Given: angle straight lineof inclination with the cor equation B and Determine𝑦the angle of inclination co Given: A(−1; − 6) and B(3; 5) are two points on a straight line. 3𝑦 − 4 o Δy 3y − 4x = − 5 𝑦 tan θ = = Determine the angle of inclination. 3𝑦3𝑦=−4 a Δx - make y the subject 3y = 4x − 5 3𝑦 4= m = tan θ 4 5 ! 𝑦= therefore; y = x− 𝑥 3 θ x y2 − y1 - note that m>0 𝑦= = tan θ 3 3 x 2 − x1 𝑥 ∴ mAB = tan θ θ 𝑚=𝑡 A 5 − (−6) = tan θ m = tan θ 4𝑚 = 3 − (−1) ! - sub. m and solve θ = 𝑡𝑎 4 34 11 = tan θ = The angle of inclination (θ) is always in relation to a horizontal plane in tan−1( ) = θ 3 3 −1 tan an anti-clockwise direction. 4 4 ∴ θ = 70∘ tan−1( ) = θ tan− 53,13° y 3 B 53,1 ∴ θ = 53,13∘ - m>0; ref. ∠ = angle of inclination Positive gradient: Angle of Example 2: inclination Example 2: m> 0 Converting a negative gradient into an angle EXAMPLE Given: straight line with the equation 3 tan−1(m) = θ m< 0 Given: Determine Given: straight line with the equation 3x + 5y straight the line angle = 7. Determine with the theof equation inclination cor θ tan−1(m) = ref ∠ x Determine the angle of inclination co The reference angle is equal angle of inclination correct to two decimal places. Angle of inclination: 𝑦 to the angle of inclination. θ 𝑦 3𝑥 + 5 A θ + ref ∠ = 180∘ (∠‘s on str. line) 𝐵(3; 5) 𝑦 5𝑦3𝑥=+− 3x + 5y = 7 5𝑦 = - make y the subject Given: C (−5; 3) and D (7; − 2) are two points on a straight line. 𝑦=− 5y = − 3x + 7 Determine the angle of inclination. 𝑦= 3 7 𝑦 y = − x− - note that m<0 A y 5 5 Negative gradient: 𝑥 𝑚=𝑡 Angle of 11 ! 𝑥 3𝑚 = 𝐶(−5; 3) 𝑥 = 𝑡𝑎 m< 0 inclination tan−1(m) = ref ∠ m = tan θ - sub. m as a positive value 53 = 3 5 5 ! − = tan θ to determine the ref. ∠ tan−1 Angle of inclination: 𝑥 5 4 tan− θ + ref ∠ = 180∘ (∠‘s on str. line) Ref ∠ θ 3 𝐴(−1; −6) 30,96° tan−1( ) = θ x 12 𝐷(7; −2) 30,9 5 𝜃= 18 Ref ∠ 𝜃 𝜃==14 The angle of inclination must be B m = tan θ θ + ref . ∠ = 180∘ ∴ ref . ∠ = 30,96∘ calculated from the reference angle. 𝜃= 5 θ = 180∘ − 22,6∘ − = tan θ 12 = 157,4∘ θ + ref ∠ = 180∘ - m<0; ref. ∠ + θ = 180° 5 tan−1( ) = θ θ = 180∘ − 30,96∘ 12 ∴ ref . ∠ = 22,6∘ θ = 149,04∘ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of 𝑦 these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 58 𝐹(4; 2) 𝑦

Grade 12 Maths Essentials Analytical Geometry Grade 11 Recap 𝑦 SCIENCE CLINIC 2022 © Finding an angle that is not in Converting an angle EXAMPLE 𝐶(−5; 3) relation to a horizontal plane into a gradient Given: In the diagram: Straight line with the equation 2y − x = 5, which passes through A and B. Straight line Construct a horizontal plane, parallel to the x-axis. 5 ref. ∠ into m = tan θ. Sub. the with the equation y + 2x = 10, which passes through B and C. M is the midpoint of BC. A, B and C are vertices 𝑥 This will allow you to use the ‘sum of adjacent ̂ = θ. A and M lie on the x-axis. Remember to12add the – sign to answers for of Δ A BC. M AC y 𝐷(7; −2) angles on a straight line’ in order to calculate the negative gradients. value of the angle. Questions: y B Given: E and F (4; 2) are points on a straight 1. Determine the following: J(–4; 3) line with an angle of inclination of 36,9°. a. The co-ordinates of A A M θ x Determine the value of m correct to two K(6;2) b. The co-ordinates of M. 2y – x = 5 decimal places. C x c. The co-ordinates of B. 𝑦 𝐹(4; 2) 2. What type of triangle is A BC? Give a reason for your answer. y + 2x = 10 θ L(–2; –3) 3. If A(−5; 0) and B(3; 4), show that A B = BC (leave your answer in simplest surd 36,9° 𝑥 form). y 4. If C(7; − 4), determine the co-ordinate of N, the midpoint of AC. J(–4; 3) 𝐸 5. Hence, or otherwise, determine the length of MN. K(6;2) 6. If A BCD is a square, determine the co-ordinates of D. m = tan θ x 7. Solve for θ correct to one decimal places. m = tan(36,9∘ ) m = 0,75 Solutions: α θ β 1 5 a. 2y − x = 5 x − cut : 0 = x+ 3. dAB = (−5 − 3)2 + (0 − 4)2 dBC = (3 − 7)2 + (4 − (−4))2 L(–2; –3) 2y = x + 5 2 2 = 4 5 = 4 5 1 5 0= x+ 5 HELPFUL HINTS: y = x+ ∴ A B = BC 6 5 2 2 −5 = x mJL = − = −3 mKL = ( ) 2 8 ∴ A(−5; 0) −5 + 7 0 + (−4) 1. Make a quick rough sketch if you are 4. N (x ; y) = ; m = tan α m = tan β b. y + 2x + 10 x − cut : 0 = − 2x + 10 2 2 given co-ordinates without a drawing. 5 N (1; − 2) −3 = tan α = tan β y = − 2x + 10 2x = 10 8 2. Always make y the subject if you are tan−1(3) = α x = 5 5. M N = 2 5 (Midpt theorem) given straight line equations. (8) 5 71,6∘ = α tan−1 = β ∴ M(5; 0) 1 5 6. If A BC D is a square, then AC is the diagonal, which y = − 2(3) + 10 3. Know your types of triangles and quad- c. x + = − 2x + 10 makes N the midpoint for both diagonals ∴ D (−3; − 8) 32∘ = β 2 2 rilaterals. Proving them or using their y = 4 x + 5 = − 4x + 20 properties is a common occurrence. ∴ B(3; 4) Δy 7. mAC = m = tan θ 5x = 15 Δx 4. The angle of inclination is ALWAYS in θ = 180∘ − (α + β ) x = 3 0 − (−4) 1 = − = tan θ relation to the horizontal plane. 2. A BC is a right-angled triangle: −5 − 7 3 = 180∘ − (71,6∘ + 32∘ ) ( 3) 1 1 = 76,4∘ (sum ∠ ‘s of Δ) mAD × mBC = − 1 = − tan−1 − = θ 3 ∴ b = 90∘ θ = 18,4∘ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 59

Grade 12 Maths Essentials Analytical Geometry SCIENCE CLINIC 2022 © What is Analytical EQUATION OF A CIRCLE EXAMPLE 1 EXAMPLE 2 Geometry? The equation of a circle is given by the equation: Find the equation of the circle with its centre through Determine the equation of the circle in the form Analytical Geometry (also called the origin and passing through the point (−5; 12). A x 2 + B x + C y 2 + D y + E = 0 that has its centre at (x − a)2 + (y − b)2 = r 2, Co-ordinate Geometry) is the study 𝑃(−3; 1) and passes through 𝑅(5; 7). where (a; b) is the centre of the circle, and x and y and application of straight line r 2 = (x − a)2 + (y − b)2 functions, trigonometry and Euclidean are coordinates of a point on the circle. y r = (0 − (−5))2 + (0 − 12)2 PR = (x − a)2 + (y − b)2 Geometry by using points on a Cartesian Plane. In Grade 12 we = 13 PR = (5 − (−3))2 + (7 − 1)2 x combine this knowledge and find the (a;b) equation of circles. = 10 r (x − a)2 + (y − b)2 = r 2 (x − a)2 + (y − b)2 = r 2 (x − 0)2 + (y − 0)2 = 132 Prior Knowledge: (x + 3)2 + (y − 1)2 = 10 2 ∴ x 2 + y 2 = 169 1. All analytical formulae (distance, x 2 + 6x + 9 + y 2 − 2y + 1 = 100 midpoint, gradients, and straight With radius as the subject: x 2 + 6x + y 2 − 2y − 90 = 0 line functions) 2. Euclidean Geometry (types of r = (x − a)2 + (y − b)2 triangles and quadrilaterals, circle Given two circles with centres A and B respectively, EXAMPLE 3 geometry theorems) it can be determined if the circles are: Two circles are given: a) Touching externally (one point of intersection) 3. Trigonometry (general applications, If the centre of the circle is at the origin (0;0): x 2 − 2x + y 2 + 6y − 6 = 0 with centre A, and sine and cosine rules, sine area y (x + 4)2 + (y − 2)2 = 9 with centre B. rule, double and compound angle R r AB = R + r identities) A B a) Express circle with Centre A in standard centre-radius form. (0;0) b) Determine whether the circles are touching externally, x r internally, neither, or cutting. Glossary of Terms: b) Cutting (two points of intersection) Concentric a) x 2 − 2x + y 2 + 6y = 6 Two or more circles that share the R r AB < R + r same centre. A B x 2 − 2x + 1 + y 2 + 6y + 9 = 6 + 1 + 9 (x − a)2 + (y − b)2 = r 2 (x − 1)2 + (y + 3)2 = 16 Median Line from the vertex of a triangle to (x − 0)2 + (y − 0)2 = r 2 c) Not touching at all the midpoint of the opposite side. b) A(1;–3) and R=4 ; B(–4;2) and r=3 ∴ x2 + y2 = r 2 Centroid OR R r AB > R + r Point of intersection for all the A B AB = (1 − (−4))2 + (−3 − 2)2 medians. 2 2 r = x + y = 5 2 Altitude d) Touching internally (one point of intersection) Perpendicular line drawn from a side = 7,07 of a triangle to the opposite vertex. The standard centre-radius equation can be expanded to give; AB = R − r Orthocentre A AB > R + r ∴ circles are NOT touching Point of intersection for all the A x 2 + Bx + C y 2 + D y + E = 0 B r R altitudes. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 60

Grade 12 Maths Essentials Analytical Geometry SCIENCE CLINIC 2022 © EXAMPLE 4 EXAMPLE 5 In the diagram below: Given: Circle with centre M passes through A, B and C(4; 4), PQ is a tangent to the Circle with Centre P is drawn. The line given by the equation 3x − 2y − 44 = 0 circle at A. Q lies on the x-axis. A lies on the y-axis. AC is parallel to the x-axis. passes through its centre. Points Q(4; −2) and R(12; −18) are points on the A Ĉ M = x; B Ĉ M = 2x; B Â M = 30∘; C AQ ̂ = θ circumference. Determine the co-ordinates of P. a) Determine the co-ordinates of A. b) Prove that 𝑥 = 20°. c) Prove that 𝜃 = 70°. Round off your answers to TWO decimal places: d) Determine the equation of the tangent PQ. e) Determine the equation for radius AM. f) Determine the co-ordinates of M. g) Express the equation of the circle in centre-radius form. 3x − 2y − 44 = 0 a) (0;4) e) mAM × mPQ = − 1 −2 = − 3x + 44 mAM = 0,36 y = 3 x − 22 b) M ̂ AC = x (∠’s opp equal sides; AM=CM) 2 A ̂ MC = 180∘ − 2x (int ∠’s of Δ) y = 0,36x + 4 A ̂ BC = 90∘ − x (∠ at centre = 2 x ∠ at circumference) 3 ∴ P (x ; y) is P (x ; x − 22) f) M(x ; y) is M(x ; 0,36x + 4) 2 PQ 2 = PR 2 (radii) Â + B̂ + Ĉ = 180∘ (int ∠ of Δ) A M2 = C M2 3 2 3 2 30∘ + x + 3x + 90 − x = 180∘ (x − 4)2 + ( x − 22 − (−2)) = (x − 12)2 + ( x − 22 − (−18)) (x − 0)2 + (0,36x + 4 − 4)2 = (x − 4)2 + (0,36x + 4 − 4)2 2 2 3x + 120∘ = 180∘ x 2 + 0,1296x 2 = x 2 − 8x + 16 + 0,1296x 2 3x − 40 2 3x − 8 2 ( 2 ) ( 2 ) x 2 − 8x + 16 + = x 2 − 24x + 144 + 3x = 60∘ 8x = 16 x = 20∘ x = 2 9x 2 − 240x + 1600 9x 2 − 48x + 64 x 2 − 8x + 16 + = x 2 − 24x + 144 + 4 4 c) M ̂ AQ = 90∘ (tan ⟂ radius) ∴M’s x-value = 2 (⟂ line from centre to chord) 4x 2 − 32x + 64 + 9x 2 − 240x + 1600 = 4x 2 − 96x + 576 + 9x 2 − 48x + 64 M ̂ AC = 20∘ (proven above) −272x + 1664 = − 144x + 640 y = 0,36(2) + 4 ∴ θ = 70∘ = 4,72 1024 = 128x ∴ M(2; 4,72) 8= x ∘ d) tan 110 = mPQ 3 = − 2,75 g) MC = (x − 2)2 + (y − 4,72)2 y = x − 22 2 MC 2 = (4 − 2)2 + (4 − 4,72)2 3 = (8) − 22 y = − 2,75x + 4 ∴ MC 2 = 4,52 2 = − 10 (x − 2)2 + (y − 4,72)2 = 4,52 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 61

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © REMINDER REPRESENTING DATA Ungrouped data = discrete Discrete data: Data that can be counted, e.g. the Grouped data = continuous number of people. NB: Always arrange data in ascending order. Continuous data: quantitative data that can be measured, e.g. temperature range. FREQUENCY TABLE STEM AND LEAF PLOTS BAR GRAPH HISTOGRAM Measures of central tendency: a descriptive summary of a dataset through a single value that Mark Tally Frequency Stem Leaf 7 reflects the data distribution. 4 || 2 0 1, 1, 2, 2, 3, 4 6 Measures of dispersion: The dispersion of a data set is the amount of variability seen in that data 5 || 2 1 0, 0, 0, 1, 1, 1 5 set. 4 6 |||| 4 2 5, 5, 7, 7, 8, 8 Cumulative frequency: The total of a frequency 3 and all frequencies so far in a frequency distribution 7 |||| 5 3 0, 1, 1, 1, 2, 2 2 Variance: measures the variability from an 8 |||| 4 4 0, 4, 8, 9 average or mean. a Small change in the numbers of 1 9 || 2 5 2, 6, 7, 7, 8 a data set equals a very small variance 0 10 | 1 0~10 10~20 20~30 30~40 Standard Deviation: the amount the data value 6 3, 6 or class interval differs from the mean of the data set. FREQUENCY POLYGON OGIVES INDICATORS OF POSITION Outliers: Any data value that is more than 1,5 IQR Quartiles to the left of Q1 or the right of Q3, i.e. 8 100 The three quartiles divide the data Outlier < Q1 – (1,5×IQR) or into four quarters. 80 Cumulative frequency 6 Q1 = Lower quartile or first quartile Outlier > Q3 + (1,5×IQR) Q2 = Second quartile or median Frequency Regression: a measure of the relation between 60 Q3 = Upper quartile or third quartile the mean value of one variable (e.g. output) and 4 corresponding values of other variables (e.g. time Percentiles 40 and cost). Indicates which percentage of data is 2 below the specific percentile. Correlation: interdependence of variable 20 Q1= 25th percentile quantities 0 Q2= 50th percentile Causation: the action of causing something [0–3] [3–6] [6–9] [9–12] [12–15] 0 Q3= 75th percentile Mark 3 6 9 12 15 Univariate: Data concerning a single variable Mark All other percentiles can be Bivariate: Data concerning two variables MEASURES OF DISPERSEMENT calculated using the formula: Interpolation: an estimation of a value within two Range Interquartile range Semi-Interquartile range p i= (n) known values in a sequence of values. 1 100 range = max value − min value IQR = Q 3 − Q1 semi − IQR = (Q − Q1) Extrapolation: an estimation of a value based on 2 3 where; Note: range is greatly influenced Note: spans 50% of the data set Note: good measure of dispersion extending a known sequence of values or facts i = the position of the pth percentile by outliers for skewed distribution beyond the area that is certainly known p = the value of the ith position Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 62

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © MEASURES OF CENTRAL TENDENCY FOR MEASURES OF DISPERSION AROUND THE MEAN UNGROUPED DATA Variance Standard deviation Mean Variance measures the variability from an average or mean. Standard deviation is the amount the data value or class interval differs sum of all values x̄ = from the mean of the data set. total number of values The variance for a population is calculated by: Σx 1. Calculate the mean(the average). x̄ = 2. Subtracting the mean from each number in the data set and then Σ f (x − x̄ )2 n σ = where; squaring the result. The results are squared to make the negatives n positive. Otherwise negative numbers would cancel out the positives x̄ = mean where; in the next step. It’s the distance from the mean that’s important, Σx = sum of all values n = number of data values not positive or negative numbers. n = number of values x = midpoint of interval 3. Averaging the squared differences. Mode x̄ = estimated mean The mode is the value that appears most frequently in a EXAMPLE: set of data points. Continuous data is grouped into class intervals which consist of an upper class boundary (maximum value) and lower class values (minimum value). Bimodal: a data set with 2 modes Trimodal: a data set with 3 modes Midpoint Class interval frequency ( f × x) (x − x̄ )2 f (x − x̄ )2 Median uper class barrier + lower class barrier (f ) x = The median is the middle number in a set of data points. 2 1 0 ≤ x ≤ 10 3 10 + 0 3 × 5 = 15 (5 − 15,71)2 = 114,7 3(114,7) = 344,11 position of median = (n+ 1) = 5 2 2 Where; 10 ≤ x ≤ 20 7 20 + 10 7 × 15 = 105 (15 − 15,71)2 = 0,5 7(0,5) = 3,53 = 15 n = number of values 15 If n = odd number, the median is part of the data set. 20 ≤ x ≤ 30 4 30 + 20 4 × 25 = 100 (25 − 15,71)2 = 88,3 4(88,3) = 354,22 = 25 If n = even number, the median will be the average 2 between the two middle numbers. total : 14 14 220 Σf(x − x)2 = 692, 86 FIVE NUMBER SUMMARY Mean: 1. Minimum value sum of all frequency × mean value 2. Lower quartile Q1 mean(x ) = total frequency 3. Median CALCULATOR STEPS: 4. Upper quartile Q3 220 = Mode 2: STAT 1: VAR 5. Maximum value 14 BOX AND WHISKER PLOT = 15,71 Shift: Setup screendown. 3: STAT A box and whisker plot is a visual representation of the Frequency? 1: ON 2: OFF Standard deviation: five number summary. Grouped data Σ f (x − x )2 rem rt i rt i re σ = Enter data (x = mid pt) e le le me n qu qu ex t a a ex t AC ed Lowe Lowe ia n Up Up 692,86 r r pe r pe r = Shift 4: VAR M 14 A small Standard deviation tells us the numbers 2: x̄ (mean) = 7,03 are clustered closely around the mean, a larger standard deviation indicates more scattered data. 3: σ x (std deviation) 10 20 30 40 50 60 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 63

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © MEASURES OF CENTRAL TENDENCY EXAMPLE: FOR GROUPED DATA Step 1: Determine cumulative frequencies form a frequency table. We conduct a survey on the ages of people who visit the corner shop, 80 people partake in the survey. Estimated mean sum of all frequency × mean value Class interval Frequency Cumulative frequency Interpretation Graph points mean(x ) = total frequency 0 ≤ x < 15 0 0 0 participants are younger than 15. (15;0) 15 ≤ x < 30 14 0 + 14 = 14 14 people were younger than 30. (30;14) Σf ⋅ x x̄ = 30 ≤ x < 45 22 14 + 22 = 36 36 people were younger than 45. (45;36) n 45 ≤ x < 60 30 36 + 30 = 66 66 people were younger than 60. (60;66) where; x̄ = estimated mean 60 ≤ x < 75 14 66+ 14 = 80 All participants were younger than 75. (75;80) n = number of values Step 2: Represent information on a cummulative frequency/ogive curve Coordinates (x;y) Modal class interval The graph takes on the The x-coordinate represents the upper boundary of the class interval. 100 f shape of an S-curve. The modal class interval is the class interval that y-coordinate represents the cumulative frequency. 90 contains the greatest number of data points. 80 Median class interval Interpretations from the graph: 70 Median Cum. Freq The median class interval is the interval that contains 60 There is an even nr of data items in our set (80) so the median liesmidway between the two !# the middle number in a set of data points. middle values. The median is halfway between the 40th and 41st term. Find the value on the 50 y-axis and draw a line from that point to determine the value on the x-axis. 40 M 30 Quartiles Similar to the method used to find the median you can determine the upper or lower quartiles 20 !" from the graph. 10 100 f 100 f Percentiles x 90 100 f 90 The median and quartiles divide the data into 50% and 25% respectively, should you need to 15 30 45 60 75 80 calculate a different percentile this can be done by calculation 80 90 or read from the graph. Age 80 70 Calculation of the 90th percentile: 0,9 × 80 = 72 70 nd 70 60 So 90% of the data is below the 72 value which will be int the 60 last class interval. 60 50 50 SYMMETRIC AND SKEWED DATA 50 40 40 Symmetric Skewed 40 30 30 Symmetric data has a balanced shape, with mean, median and mode close Skewed data is data that is spread more towards one side or the other. 30 20 20 together. 20 10 10 x x 10 mode = mean = median x 15 30 45 60 75 15 30 45 60 75 15 30 45 60 75 mean median mode mode median mean Skewed left: Negatively skewed if the tail extends to the left Skewed right: Positively skewed if the tail extends to the right Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 64

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © BIVARIATE DATA CORRELATION COEFFICIENT Bivariate data can be represented EXAMPLE: Draw a scatter plot for the following data and calculate the correlation coefficient, then write down a by scatter plots: Positive linear correlation nΣx y − (Σx)(Σy) conclusion about the type of correlation. Number of hours a sales person spends with his client vs the r = value of the sales for that client. nΣx 2 − (Σx)2 × nΣy 2 − (Σy)2 NUMBER OF HOURS 30 50 80 100 120 150 190 220 260 Where r indicates the strength of the relationship between the two VALUE OF SALES 270 275 376 100 420 602 684 800 820 variables (x and y). (IN THOUSANDS OF RANDS) Properties: Step 1: Scatter plot • The correlation coefficient is a number between –1 and 1 SCATTER PLOT (–1 ≤ r ≤ 1) 900 800 Negative correlation • Strong positive linear correlation, r is close to 1. 700 • Strong negative linear correlation, r is close to –1. 600 • No linear correlation or a weak linear correlation, –0,3 < r < 0,3 500 400 300 Value of r Meaning 200 r=1 Perfectly positive correlation 100 0,9 ≤ r < 1 Very strong positive linear correlation 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 No correlation 0,7 ≤ r < 0,9 Significant positive linear correlation Number of hours 0,3 ≤ r < 0,7 Weak positive linear correlation Step 2: Using your calculator to find r –0,3 ≤ r < 0,3 No significant linear correlation Once you understand the reasons for the process you can use your calculator to streamline the process. –0,7 ≤ r < –0,3 Weak negative linear correlation r can be found by completing the following steps on your calculator: –0,9 ≤ r < –0,7 Significant negative linear correlation (steps may vary slightly for different calculators) • Press [MODE] and [2:STAT] to enter stats mode –1 ≤ r < –0,9 Very strong negative linear correlation • [2:a+bx] r = –1 Perfect negative correlation • Enter the number of hours in the x-column and the value of sales in the y-column Non linear correlation • [AC], this will clear the screen, but the data remains stored • [SHIFT][1] to get the stats computation screen • [5:REG] and then [3:r] r = 0,893 ∴ a significant positive linear correlation linear correlation exists between the number of hours a sales person as can be seen on the scatterplot- with the exception of the outlier. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 65

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © THE LEAST SQUARES REGRESSION LINE EXAMPLE: A line can be drawn through the points to determine if there is a significant negative or Determine the equation of the least square regression line in the form ŷ = a + bx positive correlation on a scatter plot. A line of best fit will not represent the data per- fectly, but it will give you an idea of the trend. The line of best fit is also called a regres- Step 1: Complete the table as indicated by the column headings: sion line. This line will always pass through (x ; y ) where x is the mean of the x-values and y is the mean of the y-values. x y x −x y −y (x − x )2 (x − x )(y − y ) 1 2 –2 –2 4 4 2 4 –1 0 1 0 3 5 0 1 0 0 4 4 1 0 1 0 5 5 2 1 4 2 2 x=3 y=4 Σ(x − x ) =10 Σ(x − x )(y − y )=6 Step 2: Calculate Slope/gradient b by substituting the totals of your table as needed: Once you understand the process, these values can be found Σ(x − x )(y − y ) faster by completing the following steps on your calculator: (steps b= Σ(x − x )2 may vary slightly for different calculators) error • Press [MODE] and [2:STAT] to enter stats mode 6 b= • [2:a+bx] 10 • Enter the values in the x-and y-column respectively error • [AC], this will clear the screen, but the data remains stored b = 0,6 • [SHIFT][1] to get the stats computation screen error • [5:REG] error Step 3: Calculate the y-intercept a by substituting • To find a choose the option [1:a] OR [2:B] to find the value of b the values of the point (x ; y ) : a = y − bx error a = 4 − (0,6)(3) a = 2,2 The aim of the least squares regression line is to make the total of the square of the errors as small as possible. The straight line minimizes the sum of squared errors. Step 4: Assemble the equation of a line when we square each of those errors and add them all up, the total is as small as possible. ŷ = a + bx ŷ = 2,2 + 0,6x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 66

Grade 12 Maths Essentials Statistics SCIENCE CLINIC 2022 © EXAMPLE: 2. An athlete runs the 100m in 11,7 seconds, use the formula to predict the distance The table below shows the time taken by 12 athletes to run 100m sprint and their best distance for log jump. of this athlete’s jump. TIME FOR 100 M SPRINT 10,1 10,2 10,5 10,9 11 11,1 11,2 11,5 12 12 12,2 12,5 (IN SECONDS) ŷ = 14,34 − 0,64x ŷ = 14,34 − 0,64(11,7) DISTANCE OF BEST LONG JUMP 8 7,7 7,6 7,3 7,6 7,2 6,8 7 6,6 6,3 6,8 6,4 ŷ = 6,852 m (IN METRES) SCATTER PLOT 3. Another athlete completes the 100m sprint in 12,3 seconds and his best jump is 8,5 7,6m. If this is included in the data will the gradient of the least squares regres- sion line increase or decrease? Motivate your answer without using calculations. Distance of best long jump (im metres) The gradient will increase, the distance point will be much higher than the ones 8 around that time. 4. Calculate the mean time and standard deviation for the data set. 7,5 This can be found by completing the following steps on your calculator: (steps may vary slightly for different calculators) 7 • Press [MODE] and [2:STAT] to enter stats mode • [1:Var] • Enter the time values in the x-column • [AC], this will clear the screen, but the data remains stored 6.5 • [SHIFT][1] to get the stats computation screen • [4:Var] • [2:x] and = to find the mean time, which is 11,27 seconds OR 6 • [3:σ x] and = to find the standard deviation for the sample. The standard 10 10,5 11 11,5 12 12,5 deviation from the mean time is 0,755 seconds. Time for 100 m sprint (in seconds) 1. Determine the a and b values for the least squares regression line. This can be found by completing the following steps on your calculator: 5. Find the correlation coefficient (steps may vary slightly for different calculators) • Press [MODE] and [2:STAT] to enter stats mode • Press [MODE] and [2:STAT] to enter stats mode • [2:a+bx] • [2:a+bx] • Enter the time values in the x-column and the distances in the y-column • Enter the time values in the x-column and the distances in the y-column • [AC], this will clear the screen, but the data remains stored • [AC], this will clear the screen, but the data remains stored • [SHIFT][1] to get the stats computation screen • [SHIFT][1] to get the stats computation screen • [5:REG] • [5:REG] and then [3:r] • To find a choose the option [1:a] OR [2:B] to find the value of b r = –0.926 a = 14,34 ; b = − 0,64 ∴This is a significant negative linear correlation. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 67