Mathematics Survival Kit 2

Directorate: Curriculum FET CUBIC GRAPHS The standard form of the cubic function is 𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 1. Shape: The value of 𝑎 influences the shape of the graph. If 𝑎 > 0 [positive value] If 𝑎 < 0[negative value] Shape 2. Intercepts: Points where the graph intersects the axes 𝑥 −intercept 𝑦 −intercept Let 𝑦 = 0 and solve for 𝑥 by using Let 𝑥 = 0 and solve for 𝑦. the factor theorem. 3. Turning Points, now referred to as STATIONARY POINTS 𝑥 −coordinate of stationary point 𝑦 −coordinate of stationary point ′ (𝑥) 𝑓 = 0 and solve for 𝑥 Substitute the 𝑥-coordinate of stationary point into the original function to determine 𝑦.

Directorate: Curriculum FET Cubic functions have a special point which is known as the Point of Inflection This is the point where the graph changes from concave up to concave down and vice versa. 𝑥 −coordinate of Point of Inflection 𝑦 −coordinate of Point of Inflection 𝑓 ′′ (𝑥) < 0 𝑓 ′′ (𝑥) = 0 and solve for 𝑥 Substitute the 𝑥-coordinate of the point of negative inflection into the original function to 𝑓 ′′ (𝑥) > 0 determine 𝑦. positive These concepts will be unpacked as we work through examples. CONCEPTS AND SKILLS Example 1: Given: Sketch 𝒇(𝒙) = 𝟐𝒙𝟑 − 𝟓𝒙𝟐 − 𝟒𝒙 + 𝟑 Factor theorem: (a) Sketch the graph of 𝒇(𝒙) Let 𝑥 = −1: (b) Determine the coordinates of the point of 𝑓(−1) = 2(−1)3 − 5(−1)2 − 4(−1) + 3 = −2 − 5 + 4 + 3 = 0 inflection Therefore 𝑥 + 1 is a factor Solution: (a) Shape: 𝑎 = 2 thus 1 (− ; 3,7) 3 Intercepts: Roots 𝑥 − intercepts: 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3 = 0 (𝑥 + 1)(2𝑥 2 − 7𝑥 + 3) = 0 (𝑥 + 1)(2𝑥 − 1)(𝑥 − 3) = 0 Let 𝑥 = 0: 1 𝑥 = −1 𝑜𝑟 𝑥 = 𝑜𝑟 𝑥 = 3 2 𝑦 − intercepts:𝑦 = 3 (2; −9)

Directorate: Curriculum FET Stationary Points 𝑦 −coordinates of stationary points 𝑓 ′ (𝑥) = 0 1 1 3 1 2 1 𝑓 (− ) = 2 (− ) − 5 (− ) − 4 (− ) + 3 6𝑥 2 − 10𝑥 − 4 = 0 3 3 3 3 2(3𝑥 2 − 5𝑥 − 2) = 0 100 = = 3,7 2(3𝑥 + 1)(𝑥 − 2) = 0 27 1 𝑓(2) = 2(2)3 − 5(2)2 − 4(2) + 3 = −9 𝑥=− 𝑜𝑟 𝑥 = 2 1 3 Stationary points:(− ; 3,7) 𝑎𝑛𝑑 (2; −9) 3 (b) Point of inflection: 𝑓 ′′ (𝑥) = 0 Concavity changes 𝑓 ′ (𝑥) = 6𝑥 2 − 10𝑥 − 4 Concave down 𝑓 ′′ (𝑥) = 12𝑥 − 10 𝑓 ′′ (𝑥) = 0 12𝑥 − 10 = 0 12𝑥 = 10 5 10 5 ( ; −2,6) Pnt of Inflection 6 𝑥= = 12 6 5 5 3 5 2 5 𝑓 ( ) = 2 ( ) − 5 ( ) − 4 ( ) + 3 = −2,6 6 6 6 6 Concave up We do not need this information to draw the graph

Directorate: Curriculum FET Example 2: CAN YOU? Sketch the function 𝑔(𝑥) = −𝑥 3 + 6𝑥 2 − 9𝑥 Solution: Sketch the following cubic functions: Shape: 𝑎 = −1 thus [Show all intercepts and stationary point] Intercepts: 𝑥 − intercepts: (a) 𝑦 = = 𝑥 3 + 14𝑥 2 − 49𝑥 + 36 −𝑥 3 + 6𝑥 2 − 9𝑥 = 0 −𝑥(𝑥 2 − 6𝑥 + 9) = 0 (b) 𝑦 = 𝑥 3 − 5𝑥 − 8𝑥 + 12 −𝑥(𝑥 − 3)(𝑥 − 3) = 0 𝑥 = 0 𝑜𝑟 𝑥 = 3 (c) 𝑦 = 𝑥 3 − 4𝑥 2 + 4𝑥 𝑦 −intercept: 𝑦=0 (d) 𝑦 = 𝑥(𝑥 − 1)(𝑥 + 1) Stationary Points: 𝑔′ (𝑥) = −3𝑥 2 + 12𝑥 − 9 𝑔′ (𝑥) = 0 −3𝑥 2 + 12𝑥 − 9 = 0 −3(𝑥 2 − 4𝑥 + 3) = 0 −3(𝑥 − 3)(𝑥 − 1) = 0 𝑥=3 𝑜𝑟 𝑥 = 1 𝑦 −coordinate of stationary point: 𝑔(3) = −(3)3 + 6(3)2 − 9(3) = 0 𝑔(1) = −(1)3 + 6(1)2 − 9(1) = −4 Coordinates of stationary points: (3; 0) and (1; −4)

Directorate: Curriculum FET Summary: There are 4 types of sketch options Type 1 Type 2 Type 3 Type 4 𝒚 = 𝒙𝟐 − 𝟑𝒙 − 𝒙 + 𝟑 𝑦 = 𝑥 3 + 3𝑥 2 𝑦 = 2𝑥 3 + 3𝑥 2 − 2 𝒚 = −𝒙𝟑 − 𝟐 Three 𝑥 −intercepts/roots Two 𝑥 −intercepts/roots One 𝑥 −intercept/root One 𝑥 −intercept Two turning points Two turning points Two turning points One stationary point( not turning point but Note: Point of inflection) We have 2 equal roots at 𝑥 = 0 Finding the equation of a cubic function Given three roots and one other point Given the turning point and two other points Use the formula: Start with: 𝑦 = 𝑎(𝑥 − 𝑟1 )(𝑥 − 𝑟2 )(𝑥 − 𝑟3 ) where 𝑟1 ; 𝑟2 and 𝑟3 are the roots. • Derivative = 0 at the turning point • Substitute the roots and simplify as far as possible. • Substitute other points • Substitute the other point to determine the value of 𝑎.

Directorate: Curriculum FET Example 3 Example 4 Given: The sketch below is defined by 𝒑(𝒙) = 𝒂𝒙𝟑 + 𝟓𝒙𝟐 + 𝟒𝒙 + 𝒅. Determine the 𝒇(𝒙) = −𝒙𝟑 + 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 values of 𝒂 and 𝒅. and the sketch, (a) find the values of 𝒂, 𝒃 (6; 0) B(2; 9) and 𝒄. (b) Determine the coordinates of A, the local minimum of the curve. -3 Solution: 𝑎 = −1 (a) 𝑦 = 𝑎(𝑥 − 𝑟1 )(𝑥 − 𝑟2 )(𝑥 − 𝑟3 ) Roots: 0; 6; 6 Solution: 𝑓(𝑥) = −(𝑥 − 0)(𝑥 − 6)(𝑥 − 6) Equal roots at 6 𝑓(𝑥) = −𝑥(𝑥 2 − 12𝑥 + 36) 𝑦 = 𝑎𝑥 3 + 5𝑥 2 + 4𝑥 + 𝑑 𝑓(𝑥) = −𝑥 3 + 12𝑥 2 − 36𝑥 𝑦=0 𝑑𝑦 ∴ 𝑎 = 12; 𝑏 = −36 and 𝑐 = 0 = 3𝑎𝑥 2 + 10𝑥 + 4 𝑑𝑥 (b) 𝑓(𝑥) = −𝑥 3 + 12𝑥 2 − 36𝑥 𝑑𝑦 𝐴𝑡 𝑥 = 2, =0 𝑓 ′ (𝑥) = −3𝑥 2 + 24 − 36 𝑑𝑥 𝑓 ′ (𝑥) = 0 3𝑎𝑥 2 + 10𝑥 + 4 = 0 𝑥 value at turning point is 2. The turning point at 𝑥 = −3𝑥 2 + 24𝑥 − 36 = 0 3𝑎(2)2 + 10(2) + 4 = 0 Substitute into derivative 6 was given. −3(𝑥 2 − 8𝑥 + 12) = 0 𝑥 −coordinate of A is 2. 12𝑎 + 20 + 4 = 0 −3(𝑥 − 6)(𝑥 − 2) = 0 12𝑎 = −24 𝑥=6 or 𝑥=2 𝑎 = −2 Graph cuts 𝑦 −axis at -3 𝑦 - value at 𝑥 = 2. ∴ 𝑦 = −2𝑥 + 5𝑥 2 + 4𝑥 + 𝑑 3 𝑓(𝑥) = −𝑥 3 + 12𝑥 2 − 36𝑥 𝑓(𝑥) = −(2)3 + 12(2)2 − 36(2) = −32 𝑑 = −3 A(𝟐; −𝟑𝟐) 𝑦 = −2𝑥 3 + 5𝑥 2 + 4𝑥 − 3

Directorate: Curriculum FET STATISTICS: CONCEPTS AND SKILLS: Maatstawwe van verspreiding Mean (𝑥) Kwartiele Mode • Divides an arranged data set into quarters. The value of the data set with the highest • The sum of all the values(𝑥) of • 1 Lower quartile (𝑄1 ) : The position of 𝑄1 = 4 (𝑛 + 1). frequency/ most common value. the data set dived by the number 1 Range • Median (𝑄2 ): The position of 𝑄2 = (𝑛 + 1) Maximum value – Minimum value. of values (𝑛). ∑ = 𝑠𝑢𝑚 𝑜𝑓 2 3 𝐼𝑄𝑅 = 𝑄𝟑 − 𝑄𝟏 ∑𝑥 • Upper quartile (𝑄3 ) : The position of 𝑄3 = (𝑛 + 1). • 𝑥= 4 Semi – Interquartile range 𝑛 𝑄𝟑 − 𝑄𝟏 2 Measures of central tendency (grouped data) Example 2: CAN YOU? Find the estimated mean, median and mode of the data Calculating the estimated mean. Find the estimated mean, median given below. 1. Create additional columns Midpoint of class and modal interval of the data interval and Frequency × Midpoint. given below. Class Frequency Midpoint Frequency interval of × class Midpoint 2. Calculate the Midpoint of class interval and Class interval Frequency Frequency × Midpoint. 0 < 𝑥 ≤ 10 1 interval 2.1 Midpoint of class interval = 10 < 𝑥 ≤ 20 2 0 ≤ 𝑥 ≤ 20 6 10 60 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑐𝑙𝑎𝑠𝑠+𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑐𝑙𝑎𝑠𝑠 20 < 𝑥 ≤ 40 8 30 240 2 20 < 𝑥 ≤ 30 11 2.2 Multiply the frequency column with the 30 < 𝑥 ≤ 40 9 40 < 𝑥 ≤ 60 10 50 500 midpoint column for each class interval. 60 < 𝑥 ≤ 80 3 70 210 40 < 𝑥 ≤ 50 14 80 < 𝑥 2 90 180 3. Calculate the sum of the Frequency column to find 50 < 𝑥 ≤ 60 3 ≤ 100 the “𝑛” value. 29 1190 4. Calculate the sum of the Frequency × Midpoint

Directorate: Curriculum FET Solution: column to find the “ ∑ 𝑥 ” value. Mean ∑𝑥 5. Substitute step 3 and 4 into 𝑥 = 𝑛 to find the ∑ 𝑥 1190 𝑥= = = 41,03 estimated mean. Answers: 𝑛 29 Median. Mean 35.5 Median 1 1 • Find the position of 𝑄2 . Median 35 Position of 𝑄2 = (𝑛 + 1) = (29 + 1) = 15𝑡ℎ value. • Determine in which class interval the position of Modal interval: 40 < 𝑥 ≤ 50 2 2 The 15𝑡ℎ value falls in the 40 < 𝑥 ≤ 60 class interval. 𝑄2 falls. ∴ 𝑄2 = 50 (midpoint of class interval.) • Use the class midpoint as the value of 𝑄2 . Mode The class interval with the highest frequency. Mode: 40 < 𝑥 ≤ 60 Example 2: Determine the lower quartile, upper quartile, range, interquartile range, semi – interquartile range, and 80𝑡ℎ percentile of the data below. 20, 21, 25, 25, 26, 26, 26, 30, 32, 32, 33, 33, 35, 36, 37, 38, 41, 42, 43, 45 Solution Lower quartile (𝑸𝟏 ) Upper quartile(𝑸𝟑 ) 𝟖𝟎𝒕𝒉 percentile 1 3 80 The position of 𝑄1 = (𝑛 + 1) 4 The position of 𝑄3 = (𝑛 + 1) 4 The position of the 80𝑡ℎ = 100 (20 + 1) = 16.8𝑡ℎ position. 1 3 38+41 = (20 + 1) = 5,25𝑡ℎ position. = (20 + 1) = 15,75𝑡ℎ position. ∴ 80 𝑡ℎ percentile = 2 = 39,5 4 4 26 + 26 37 + 38 ∴ 𝑄1 = = 26 ∴ 𝑄3 = = 37,5 2 2 Range = 45 − 20 = 25 𝑰𝑸𝑹 = 𝑄𝟑 − 𝑄𝟏 = 37,5 − 26 = 11,5 𝑄𝟑 −𝑄𝟏 37,5−26 Semi - 𝑰𝑸𝑹 = 2 = 2 = 5,75 The five number summary Outliers The five number summary consists of the following measures of A value that "lies outside" (is much smaller or larger than) most of the dispersion: other values in a set of data. ❖ The minimum value of the data set ❖ 𝑄1 ….. 25% ❖ 𝑄2 50% median ❖ 𝑄3 ……75% ❖ The maximum value of the data set

Directorate: Curriculum FET Box and whisker diagram The Box and Whisker diagram is a graphical representation of the five number summary. Drawing tips Whisker Box Diagram must always have a box and whiskers. Clearly show five number summary on axis. Whisker Min Max 𝑄1 𝑄2 𝑄3 value value Important conclusions from box and whisker diagram. • 25% of data lies between the minimum value and 𝑄1 . • 25% of data lies between 𝑄1 and 𝑄2 . Answer • 25% of data lies between 𝑄2 and 𝑄3 . • 25% of data lies between 𝑄3 and the maximum value. CAN YOU? Draw a box and whisker diagram of the data below. 12, 12, 13, 14, 16, 22, 23, 24, 30, 31, 31, 31, 32, 34, 35, 35, 35, 36 Distribution of data Positive skew Symmetrical distribution Negative skew

Directorate: Curriculum FET Example 4: CAN YOU? The box and whisker diagram below show the marks (out of 90) obtained by a class of The data below shows the number of people visiting a local nine learners. clinic per day for 6 days. 12, 29, 13, 63, 12, 3 1. Draw a box and whisker diagram. 2. Use your calculator to determine the mean of the 20 30 40 70 90 data Answer the following questions. 3. Calculate the mean – median 1. Comment on the skewness of the data. 4. Comment on the skewness/ spread of the data 2. Why will the mean not be a good measure of central tendency? Answer 3. Do you think the test was too easy or too difficult? Give a reason for your 2. mean ≈ 14,1 answer. 3. mean - median = 14,1 − 17,5 = −3,4 Solution: 4. The data is clustered more closely on the right and is 1. Skew to the right or positively skew. slightly more spread out on the left. It is negatively skewed. 2. The mean will be too high. The difference between the mean and the median is 3. Too difficult 50% of learners had less than 40%. negative which also indicates that the data is negatively skewed Cumulative frequency graph (ogive) Definition: The total of a frequency and all frequencies so far in a frequency distribution. It is the “running total” of frequencies. What must I be able to do? https://youtu.be/ETeQxOHX3xc • Use a cumulative frequency table to draw an ogive. • Determine values by using the ogive. • Use a given ogive and complete a cumulative frequency table from it. • Use the ogive to determine percentiles and quantiles. When drawing the curve, you may NOT What is important when you draw an ogive? use a ruler. It is an S – shaped curve. 1. It must have a heading. 2. Mark and label axes. 3. Plot points. (Upper boundary of class ; cumulative frequency of class) GROUND ogive at (lower boundary of first interval ; 0)

Directorate: Curriculum FET Example 5: Guidelines on drawing an ogive. The following frequency table shows the amount of goals scored by the Protea 1. Create a cumulative frequency column. netball team in the 50 matches they played in 2019. 2. Create a coordinates column. Amount of Frequency Cumulative Coordinates 3. Ground ogive at (lower boundary ; 0). goals frequency 4. Plot points of coordinates column. 5 < 𝑥 ≤ 10 6 6 (10 ; 6) 5. Connect points with smooth curve. 10 < 𝑥 ≤ 15 8 14 (15 ; 14) 15 < 𝑥 ≤ 20 13 27 (20 ; 27) How to find quartiles and percentiles graphically. 20 < 𝑥 ≤ 25 9 36 (25 ; 36) 1. Find the position. (on y-axis) 25 < 𝑥 ≤ 30 7 43 (30 ; 43) 2. Find this position on cumulative frequency axis. 30 < 𝑥 ≤ 35 5 45 (35 ; 45) 3. Draw a dashed line towards the ogive. 4. Draw a dashed line from the ogive towards the 𝑥 − axis. 35 < 𝑥 ≤ 40 2 50 (40 ; 50) 5. Read the value off 𝑥 − axis. 1. Draw the cumulative frequency graph(ogive) of the given data. 2. Use the graph to determine the median. CAN YOU? Solution: A survey was done on 50 people to determine the distances they 1. Goals scored in 2019 travel to work daily. The following table shows the results of the 60 survey: Cumulative frequency 50 Distance in km Frequency 40 0<𝑑≤5 2 5 < 𝑑 ≤ 10 7 30 10 < 𝑑 ≤ 15 4 20 15 < 𝑑 ≤ 20 13 10 20 < 𝑑 ≤ 25 16 0 25 < 𝑑 ≤ 30 8 0 10 20 30 40 50 1. Draw the cumulative frequency graph(ogive) of the given data. Amount of goals 2. Use the graph to determine the 60𝑡ℎ percentile 1 1 2. Position of 𝑄2 = 2 (𝑛 + 1) = 2 (50 + 1) = 25,5𝑡ℎ value. ∴ 𝑄2 = 19 3. Use your graph to estimate the median number of kilometers travelled per week. Answers: (2) ≈ 54 (3) Median between 25th and 26th learner ≈ 50 𝑘𝑚 𝑝𝑒𝑟 𝑤𝑒𝑒𝑘

Directorate: Curriculum FET CONCEPTS AND SKILLS Scatterplots A scatterplot is a graph used to determine whether there is a relationship between paired data. A scatter plot diagram is a powerful tool for researchers to determine if there is any association between two variables. The data on the scatterplot could follow the following trends: linear, quadratic or exponential. Linear Quadratic Exponential Regression line How to determine the regression Line of best fit line with a SHARP (EL-W535HT) The regression line is basically an accurate line of best fit. It uses the least Refers to a line through a calculator squares method to calculate the gradient and y intercept of the regression scatter plot of data points that line. best represents the trend of the Standard form of regression line: 𝑦 = 𝐴 + 𝐵𝑥 1.< 𝑀𝑂𝐷𝐸 >[1:STAT][1:LINE] data. The line of best fit is not How to determine the equation of the regression line with a Casio(fx-82 2. Enter x-values and the y-values that accurate, but it helps us to ZA) calculator: together. see if there is a trend in the data 1. Press MODE and select STAT. 3. Press CHANGE after each (x; y) set. 2. Select 𝐴 + 𝐵𝑥. 4. Find the value for a , the y- How to draw a line of best fit: 3. Enter the data points. intercept [ALPHA][( ] [=] Try to have the same amount of Column(X) type = after each data point 5. Find the value for b , the gradient data points above and below the Column(Y) type = after each data point [ALPHA] [ ) ] [=] line. 4. After entering all the data points press AC. Exclude any outliers 5. Press SHIFT then press 1. 6. Press 5: Reg 7. Press 1 then =: to find 𝐴 8. Press SHIFT then press 1. 9. Press 5: Reg 10. Press 2 then =: to find 𝐵.

Directorate: Curriculum FET How to draw the regression line: How to determine the mean point Outliers using a calculator: If your scale stars with the origin of (0;0) A value that "lies outside" (is ● Make sure your data is much smaller or larger than) ● Plot the 𝐴 value of the regression line i.e the y-intercept. entered in your calculator. most of the other values in a set ● Determine the mean point(𝑥 ; 𝑦 ) and plot the mean point. of data. Do not add an outlier to ● Draw a line from point 𝐴 through the mean point. ● Press SHIFT then Press 1. the dataset when drawing a line How to draw a regression line if the scale of the graph do not start with ● Press 4: Var of best fit or calculating the (0; 𝟎) regression line. The reason for ● Press 2 then =: 𝑥 this is that the outlier is not part In order to draw the regression line, substitute any two x-values that lie ● Press SHIFT then press 1. of the trend and will influence between the minimum and maximum x- values into the equation of the the trend lines and hence any regression line, plot the two points and then join them up Press 4: Var Press 5 then = : 𝑦 future predictions from these trend lines. Example 1: At a clinic where Covid19 tests were conducted the following data CAN YOU? was recorded for 7 days in June The table below shows the time taken by 12 athletes to run 100m sprint and their best distance for long jump Number of COVID Time for 49 65 82 60 65 94 88 100m 19 positive results sprint 10,1 10,2 10,5 10,9 11 11,1 11,2 11,5 12 12 12,2 12,5 (sec) Number of tests 156 176 183 153 163 192 180 Distance 1. Draw a scatterplot of the given data. of long 8 7,7 7,6 7,3 7,6 7,2 6,8 7 6,6 6,3 6,8 6,4 jump (m) 2. Determine the equation of the least squares regression line. 3. Draw the least squares regression line. 4. How many tests needs to be done to record approximately 100 COVID 19 positive results per day.

Directorate: Curriculum FET 1. Represent the data as a scatter plot. Scatter Plot 2. Use the data to calculate the equation of the least squares 200 regression line. 3. Draw the least squares regression line on your scatter plot. 190 4. An athlete runs the 100m in 11,7 seconds, use the formula to Number of tests 180 predict the distance of this athlete’s jump. 170 160 Scatter Plot 150 8.5 Time for 100m sprint (sec) 40 50 60 70 80 90 100 8 7.5 Number of COVID 19 positive results) 7 6.5 2. 𝑦 = 𝐴 + 𝐵𝑥 6 𝑦 = 113,47 + 0,81𝑥 10 10.5 11 11.5 12 12.5 Distance of long jump (m) 3. 𝑦 = 113,47 + 0,81(𝟒𝟗) = 𝟏𝟓𝟑. 𝟏𝟔 ∴ (49; 153.16) 𝑦 = 113,47 + 0,81(𝟗𝟒) = 𝟏𝟖𝟗, 𝟔𝟏. ∴ (94; 189,61) 2. 𝑦 = 14,34 − 0,64𝑥 3. (10,1; 7,8) 4. 𝑦 = 113,47 + 0,81(100) 𝑦 ≈ 195 tests (12,5; 6,3) 4. 𝑦=6,852m Correlation The correlation coefficient is a statistical measure of the strength of the relationship between two variables. Correlation coefficient is denoted by (𝑟) and is between -1 and 1. The closer the data points are to the regression line the stronger the relationship. That means the closer 𝑟 is to 1 or -1. The further the data points are away from the regression line, the weaker the relationship, and the closer 𝑟 is to 0. If the gradient of the regression line is positive, then the data set has a positive correlation and if the gradient of the regression line is negative, then the data has a negative correlation. If the correlation coefficient is greater than 0,9 we say there is a very strong positive correlation. If the correlation coefficient is smaller than -0,9 we say there is a very strong negative correlation. The strength of the association is determined by the correlation coefficient (𝑟).

Directorate: Curriculum FET Perfect positive linear association. Very Strong positive linear association. Very Strong negative association. No correlation. Perfect negative association. Example 2: How to calculate the correlation coefficient using your calculator: The data in the table represents the results of the trial examination and the ● Make sure your data is entered correctly in your calculator. See final examination mark of 11 Gr 12 learners page 2. ● Press shift then press 1. Trial exam 80 68 94 72 74 83 56 78 65 75 88 ● Press the number next to “Reg” Final exam 78 71 96 77 77 80 58 83 71 80 92 ● Press the number next to 𝑟 then press =. ● Use the given table to find the correlation coefficient 𝑟 and describe the relationship of the trial exam and the final examination. Comment on the strength of the correlation of the two variables. Solution: = 0,96 ∴ strong positive linear correlation between variables..

Directorate: Curriculum FET QUADRATIC INEQUALITIES AND SIMULTANEOUS EQUATIONS Worked Example 1 Solve the following Linear inequalities 𝑥+1<3 3 −4(𝑥 − 3) ≥ 2(𝑥 − 12) − 𝑥+3≥1 Remember to 2 𝑥 <3−1 3 reverse the −4 + 12 ≥ 2𝑥 − 24 𝑥<2 − 𝑥 ≥ −2 −4𝑥 − 2𝑥 ≥ 24 − 12 2 inequality sign if you −3𝑥 ≥ −4 −6𝑥 ≥ −36 divide by a negative ∴𝑥≤6 4 𝑥 ≤ 3 CAN YOU? Solve the following linear inequalities 2𝑥+4 3(𝑥−3) 2𝑥 + 3 > 1 2. ≥ 3. 4(𝑥 − 3) ≥ 2(𝑥 − 10) 7 3 Answers 1. 𝑥 > −1 2. 𝑥 ≤ 5 3. 𝒙 ≥ −𝟒 Quadratic Inequalities A quadratic inequality involves determining the values of x for which the graph of a parabola lies either above or below the x- axis. In the second term (on functions) you will learn how to sketch the graphs of parabola in much more detail. All we need now is a basic understanding of how to determine the x- intercepts of a parabola as well as its shape. Worked examples 2 Solve for x in the following inequalities Remember 𝑥2 ≤ 4 𝑥2 − 4 ≤ 0 As in solving quadratic equations (𝑥 + 2)(𝑥 − 2) ≤ 0 factorise to find the critical values ∴ −2 ≤ 𝑥 ≤ 2 The graph of this parabola is positive because the coefficient of 𝑥 2 is positive The x-intercepts can be determined as follows: (𝑥 + 2)(𝑥 − 2) = 0 The critical values are included in the final solution. Therefor we will use ∴x= −2 𝑜𝑟 𝑥 = 2 Draw a graph of this parabola closed dots to draw the number line of the inequality (focus on x-intercepts) The critical values are −2 𝑎𝑛𝑑 2 Now determine the values of x for which 𝑥 2 − 4 ≤ 0 This is where the graph cuts the x- axis and where it lies below the x-axis. The solution is clearly between the 2 intercepts: −2 ≤ 𝑥 ≤ 2

Directorate: Curriculum FET 2. Solve for x in the inequality 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟖 ≤ 𝟎 (𝒙 + 𝟑)(𝒙 − 𝟔) ≤ 𝟎 𝒙 + 𝟑 ≤ 𝟎 𝒐𝒓 𝒙 − 𝟔 ≤ 𝟎 The inequality sign tells us whether the answer must be read above the x-axis 𝒙 ≤ −𝟑 𝒐𝒓 𝒙 ≤ 𝟔 (where y is positive) or below the x-axis(where y-is negative) ∴ −𝟑 ≤ 𝒙 ≤ 𝟔 We can represent the result on a number line The following tables will refresh your memory on knowledge that is important as we proceed with gr 11 content 3.. Solve for x in the inequality Inequality words Included/ Number line 𝑥 2 − 𝑥 − 12 > 0 sign excluded (𝑥 − 4)(𝑥 + 3) > 0 > Greater than excluded (open) Critical values is 4 and −3 ≥ Greater than or equal to included (closed) 𝑥 < −3 𝑜𝑟 𝑥 > 4 < Less than excluded (open) Remember that (𝑥 − 4)(𝑥 + 3) > 0 ≤ Less than or equal to included (close) 4..Solve for x in the inequality 𝒙 − 𝟐𝒙𝟐 ≥ 𝟎 Number line representation −2𝑥 𝑥 + 𝑥 ≥ 0 write in standard form 2𝑥 2 − 𝑥 ≤ 0 make the coefficient of 𝑥 2 positive, multiply by−1 𝑥(2𝑥 − 1) ≤ 0 reverse sign if multiply or divide by a negative The graph of the parabola 𝑦 = 2𝑥 2 − 𝑥 is concave The x-intercepts can be determined as follows 𝑥(2𝑥 − 1) = 0 1 ∴ 𝑥 = 0 𝑜𝑟 𝑥 = 2 critical values Now determine the values of x for which 𝑥(2𝑥 − 1) ≤ 0 This is where the graph intersect the x-axis and where it lies below the x-axis. The solution is clearly between 1 The two intercepts:0 ≤ 𝑥 ≤ 2 CAN YOU? Solve the following inequalities 𝑥 2 < 16 2. 𝑥 2 + 5𝑥 − 6 ≥ 0 3. 𝑥 2 ≤ 4𝑥 4 . – 𝑥 2 + 3𝑥 + 4 ≤ 0

Directorate: Curriculum FET Simultaneous Equations Answers: In gr 10 you solved simultaneous linear equations to find the point of 𝟏. −4 < 𝑥 < 4 2. 𝑥 ≤ −6 𝑜𝑟 𝑥 ≥ 1 3. 0 ≤ 𝑥 ≤ 4 4. −1 ≤ 𝑥 ≤ 4 intersection of 2 straight lines. (one x- value and one y-value) Let us see if you remember how to solve linear equations simultaneously. Worked example 3 We can solve the equations of other pairs of graphs to find the points of Solve the following equations simultaneously 2𝑦 = 2𝑥 − 1 𝑎𝑛𝑑 2𝑦 + 𝑥 − 2 = 0 intersection e.g. a straight line and a parabola, two parabolas or a straight line and a hyperbola etc. 2𝑦 + 𝑥 − 2 = 0 get one of the variables by itself in Remember we can represent each equation as a graph. If we draw graphs of 𝑥 = 2 − 2𝑦 one of the equations both equations on the same set of axes, the solution of the simultaneous equation represents the points where the two graphs intersect. 2𝑦 = 2𝑥 − 1 Use this information to substitute into 2𝑦 = 2(2 − 2𝑦) − 1 the second equation. You should now have We will now learn how to solve equations simultaneously when one of the 2𝑦 = 4 − 4𝑦 − 1 an equation with only one variable equations is a quadratic equation. 6𝑦 = 3 When solving simultaneous equations we can: 1 𝑦= Solve it graphically, by drawing both graphs to scale and find the points of 2 Now substitute the variable into the first intersection or 𝑥 = 2 − 2𝑦 equation and solve for the second variable Solve it algebraically. 1 In this lesson we will explain how to solve the equations algebraically. 𝑥 = 2 − 2( ) Worked example 4 2 𝑥=1 Solve for x and y simultaneously 1 ∴ 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 2 1 (1 ; 2) (quadratic equation) 𝑦 = 𝑥 2 − 1 𝑎𝑛𝑑 𝑦 − 𝑥 = 5 (linear equation) Solution 𝑦 − 𝑥 = 5 make y the subject of the formula in the linear equation ∴𝑦 =𝑥+5 Simultaneous Now substitute 𝑥 + 5 in the quadratic equation and then solve for x solution 𝑦 = 𝑥2 − 1 ∴ 𝑥 + 5 = 𝑥2 − 1 ∴ 0 = 𝑥2 − 𝑥 − 6 ∴ 0 = (𝑥 − 3)(𝑥 + 2) ∴ 𝑥 = 3 𝑜𝑟 𝑥 = −2 The corresponding values of y can be determined by substituting the values for x in the linear equation 𝑦 = 𝑥 + 5 For 𝑥 = 3 for 𝑥 = −2 𝒚=𝟑+𝟓 𝒚 = −𝟐 + 𝟓 = 𝟑

Directorate: Curriculum FET 1. Solve for x and y simultaneously: 𝒚=𝟖 𝒚=𝟑 3𝑥 − 𝑦 + 2 = 0 𝑎𝑛𝑑 𝑦 = −𝑥 2 + 2𝑥 + 8 3𝑥 + 2 = 𝑦 The solution can also be written in the form (3; 8) and (-2; 3) 3𝑥 + 2 = −𝑥 2 + 2𝑥 + 8 0 = −𝑥 2 − 𝑥 + 6 0 = (𝑥 + 3)(𝑥 − 2) Worked example 5 𝑥 = −3 𝑜𝑟 𝑥 = 2 Determine algebraically the points of intersection of the graphs indicated 𝑦 = 3(−3) + 2 𝑦 = 3(2) + 2 below 𝑦 = −7 𝑦=8 𝒙𝟐 − 𝟐𝒙 − 𝟑 𝒂𝒏𝒅 𝒚 − 𝒙 + 𝟑 = 𝟎 CAN YOU? Solution Solve simultaneously for x and y in the following equations 𝑦 = 𝑥 2 − 2𝑥 − 3 (equation 1) 𝑦−𝑥+3 = 0 (equation 2) 1. 𝑦 = 𝑥 2 − 2 𝑎𝑛𝑑 𝑦 − 2𝑥 = 1 𝑦 =𝑥−3 2. 𝑦 = 3𝑥 + 7 𝑎𝑛𝑑 𝑦 = 2𝑥 2 + 8 𝑥 2 − 2𝑥 − 3 = 𝑥 − 3 3. 𝑥 − 3𝑦 = 1 𝑎𝑛𝑑 𝑥 2 + 𝑥𝑦 + 9𝑦 2 = 17 𝑥 2 − 3𝑥 = 0 Calculate the points of intersection of the graphs of : 𝑥(𝑥 − 3) = 0 ∴ 𝑥 = 0 𝑜𝑟 𝑥 = 3 4. 𝑦 = 2𝑥 𝑎𝑛𝑑 𝑦 = 3𝑥 2 − 1 (Substitute for x in the equation 2) Answers 𝑦 =𝑥−3 𝑦 = 0 − 3 𝑜𝑟 𝑦 = 3 − 3 1. 𝑥 = 3 𝑜𝑟 − 1 𝑎𝑛𝑑 𝑦 = 7 𝑎𝑛𝑑 − 1 𝑎𝑙𝑠𝑜 (3; 7)(−1; −1) 1 17 1 17 ∴ 𝑦 = −3 𝑜𝑟 𝑦 = 0 2. 𝑥 = 2 𝑜𝑟 1 𝑎𝑛𝑑 𝑦 = 2 𝑎𝑛𝑑 10 𝑎𝑙𝑠𝑜 (2 ; 2 )(1; 10) 4 1 2 Points of intersection 3. (−3; − ) or (4; 1) 4. (− ; − ) 𝑜𝑟 (1; 2) 3 3 3 (0, -3) and (3; 0) Note: When we solve two equations simultaneously, we are looking for the ordered pairs that will satisfy both equations. Consolidation • You should now be able to solve quadratic inequalities in one variable and interpret the solution graphically. • You should also know how to solve equations in two unknowns, one of which is a linear and the other quadratic, algebraically.

SEQUENCES AND SERIES An Arithmetic sequence is a sequence where the difference between consecutive numbers or terms remain constant. In the sequence: 5; 9; 13; 17; 21; … We see that 𝒂 = 𝟓 and 𝒅 = 𝟒 𝑻𝟏 = 5 = 𝒂 𝒂 = 𝐭𝐡𝐞 𝐟𝐢𝐫𝐬𝐭 𝐭𝐞𝐫𝐦 = 𝑻𝟏 𝑻𝟐 = 9 = 5 + 4 = 𝒂+𝒅 𝒅 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐜𝐞 𝑻𝟑 = 13 = 5 + 2(4) = 𝒂 + 𝟐𝒅 𝒅 = 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 𝑻𝟒 = 17 = 5 + 3(4) = 𝒂 + 𝟑𝒅 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 𝑻𝒏 = 5 + (𝑛 − 1)(4) = 𝒂 + (𝒏 − 𝟏)𝒅 Example 1: Solution: Constant difference: 𝒅 = 𝑻𝟐 − 𝑻𝟏 = 𝟓 − 𝟐 = 𝟑 Given the sequence: 2; 5; 8; … 𝒂 = 𝑻𝟏 = 𝟐 a) Determine the general term of 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 the sequence. = 2 + (𝑛 − 1)3 = 2 + 3𝑛 − 3 = 3𝑛 − 1 ∴ 𝑇𝑛 = 3𝑛 − 1 b) Use the general rule to 𝑇𝑛 = 3𝑛 − 1 determine the 40th term. ∴ 𝑇40 = 3(40) − 1 𝑇40 = 119 The position of the term is 40. Therefore, 𝑛 = 40 c) Which term in the pattern will be equal to 2012. 𝑇𝑛 = 2012 ∴ 𝑇𝑛 = 3𝑛 − 1 = 2012 The value of the term is 2012. Therefore, 𝑇𝑛 = 2012 3𝑛 = 2012 + 1 3𝑛 = 2013 𝑛 = 671 ∴ term number 671 of the sequence is equal to 2012

Example 2: Solution: Example 3: Solution: Find the number of terms in 𝒅 = −6 − (−2) = −𝟒 Determine the first three Constant difference: 𝒅 = 𝟏𝟎 the arithmetic sequence 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 terms of an arithmetic 𝑻𝟒 = 𝟑𝟗 -2; -6; -10; … ; -150 𝑇𝑛 = −2 + (𝑛 − 1)(−4) sequence if the constant 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑇𝑛 = −4𝑛 + 2 difference is 10 and the 𝑇4 = 𝑎 + (4 − 1)𝑑 𝑇𝑛 = −4𝑛 + 2 = −𝟏𝟓𝟎 fourth term is 39. 39 = 𝑎 + 3𝑑 ∴ −4𝑛 = −152 39 = 𝑎 + 3(10) ∴ 𝑛 = 38 9=𝑎 There are 38 terms Hence the sequence is 9; 19; 29 Example 4: 𝑇2 = 𝑎 + 𝑑 = 9 (1) 2𝑛𝑑 term Example 5: Solution: In an arithmetic sequence the 𝑇5 = 𝑎 + 4𝑑 = 21 (2) 5𝑡ℎ term 2𝑝 − 3; 3𝑝 − 1; 5𝑝 − 2 a) 𝒅 = 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 𝑛𝑑 𝑡ℎ (3𝑝 − 1) − (2𝑝 − 3) = (5𝑝 − 2) − (3𝑝 − 1) 2 term is 9 and the 5 term 3𝑑 = 12 (2) − (1) are the first three terms of is 21. Determine 𝒅=𝟒 an arithmetic sequence. 3𝑝 − 1 − 2𝑝 + 3 = 5𝑝 − 2 − 3𝑝 + 1 ∴ 𝑇2 = 𝑎 + 𝑑 = 9 𝑝 + 2 = 2𝑝 − 1 a) The first three terms of 𝑎+4 =9 a) Determine the value of 𝒑=𝟑 the sequence. 𝒂=𝟓 𝑝. First three terms are 5; 9; 13; b) Replacing 𝑝 = 3 in the sequence we b) The first three terms of have the first three terms as 3; 8; 13 𝑇60 = 𝑎 + 59𝑑 = 5 + 59(4) = 241 the sequence. b) The 60𝑡ℎ term c) 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 2013 Hence the 60𝑡ℎ term is 241 c) Determine the term 3 + (𝑛 − 1)(5) = 2013 equal to 2013 3 + 5𝑛 − 5 = 2013 5𝑛 = 2015 𝑛 = 403 CAN YOU? 1) Given the following sequence: 3; 8; 13; 18; … Solutions: Determine: 1) a) 𝑇𝑛 = 5𝑛 − 2 a) The general term. b) 98 b) The 20th term. c) 𝑛 = 45 c) Which term of the sequence is equal to 223? 2) In an arithmetic sequence, 𝑇3 = −2 and 𝑇8 = 23. Determine the first 2) 𝑑 = 5 term and the constant difference. 𝑎 = −12 3) Find the number of terms in the arithmetic sequence 3) 82 −5; −11; −17; … ; −491 4) The first three terms of an arithmetic sequence are, 4) a) 13 𝑥 − 8; 𝑥; 2𝑥 − 5 . Determine b) 𝑇𝑛 = 8𝑛 − 3 a) The value of 𝑥. c) 917 b) The general term. c) The value of the 115𝑡ℎ term.

SERIES: A series is created by adding the terms of a sequence. Arithmetic Sequence 2; 5; 8; 11; 14; Arithmetic Series 2 + 5 + 8 + 11 + 14 Hence the sum of the first 𝒏 terms of an arithmetic series is given by the formula: 𝒏 𝒏 𝑺𝒏 = 𝑺𝒏−𝟏 + 𝑻𝒏 𝑺𝒏 = [𝟐𝒂 + (𝒏 − 𝟏)𝒅] or [𝒂 + 𝒍] ; where 𝒍 , is the last term 𝟐 𝟐 −3 Example 6: Consider the arithmetic series (−1) + ( ) + (−2) + ⋯ + (−16). 2 a) Determine the number of terms in this series. b) Calculate the sum of the series. Solution: Solution: 𝑛 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑆𝑛 = 2 [𝑎 + 𝑙] 1 𝑇𝑛 = −1 + (𝑛 − 1) (− 2) = −16 1 1 31 ∴ −1 − 𝑛 + = −16 ∴ 𝑆31 = [−1 + (−16)] 2 2 2 1 1 ∴ − 2 𝑛 = −16 + 2 527 ∴ 𝑛 = 31 ∴ 𝑆31 = − 2 Example 7: Example 8: How many terms of the arithmetic series 1 + 4 + 7 + ⋯ will add up to 145? Consider the arithmetic series −4 − 1 + 2 + ⋯ Solution: Calculate the smallest value of 𝑛 for which 𝑆𝑛 > 300 𝑎 = 1; 𝑑 = 3; 𝑛 =? ; 𝑆𝑛 = 145 Solution: 𝑎 = −4; 𝑑 = 3; 𝑛 =? ; let 𝑆𝑛 = 300 𝑛 𝑛 𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑] 𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑] = 300 𝑛 [2(−4) + (𝑛 − 1)3] = 300 𝑛 2 145 = 2 [2(1) + (𝑛 − 1)3] ∴ 𝑛[3𝑛 − 11] = 600 290 = 𝑛(2 + 3𝑛 − 3) ∴ 3𝑛2 − 11𝑛 − 600 = 0 290 = 𝑛(3𝑛 − 1) 0 = 3𝑛2 − 𝑛 − 290 −(11)±√(−11)2 −4(3)(−600) ∴ 𝑛= 0 = (3𝑛 + 29)(𝑛 − 10) 2(3) 29 ∴ 𝑛 = 16,09 or 𝑛 = −12,43 𝑛=− or 𝑛 = 10 3 ∴ 𝑛 = 10 ; 𝑛 𝜖 𝑁 ∴ The smallest possible value of 𝑛 is 17.

SIGMA NOTATION: The Greek letter ∑ Sigma means the sum of. It is used to denote the sum of a set of consecutive terms of a sequence or series. In this notation we have to indicate the position of the first and last term of the sequence/series which are added. This is read as, “the sigma or sum of 𝑇𝑘 , from 𝒏 𝑘 = 1 to 𝑘 = 𝑛. 𝑛 is the number or position of ∑ 𝑻𝒌 = 𝑻𝟏 + 𝑻𝟐 + 𝑻𝟑 + ⋯ + 𝑻𝒏 = 𝑺𝒏 the last term of the set of 𝒌=𝟏 This means, take the sum of the terms of a numbers which is being sequences/series from the first term to the n’th added. Lets refer to it as term of the sequence/series. last/top value Or it is read as the sum of the first 𝑛 terms of the sequence/series. 𝒏 General term expressed in 𝑛, the number of terms which are added: ∑ 𝑻𝒌 terms of 𝑘 𝒏 = 𝒕𝒐𝒑 𝒗𝒂𝒍𝒖𝒆 − 𝒃𝒐𝒕𝒕𝒐𝒎 𝒗𝒂𝒍𝒖𝒆 + 𝟏 𝒌=𝒎 𝒏 𝑚 is the number or position of the first 𝒏 𝑺𝒏 = [𝟐𝒂 + (𝒏 − 𝟏)𝒅] = ∑[𝒂 + (𝒌 − 𝟏)𝒅] term of the set of numbers which is being 𝟐 added. Lets refer to it as bottom value 𝒌=𝟏 Example 9: Determine the value of: Example 10: Determine the value of: 𝟓 𝟕 ∑(𝟑𝒏 + 𝟐) Substitute, 𝑛 = 1, in ∑ 𝟐𝒌 𝒏=𝟏 general term up to, 𝒌=𝟒 𝑛 = 5. Solution: Number of terms: Solution: Method 1 Top – bottom+1 Method 1 = (7 − 4 + 1) = 4 𝟓 𝟕 𝑆𝑛 = ∑ (𝟑𝒏 + 𝟐) ∑ 𝟐𝒌 𝒏=𝟏 𝒌=𝟒 𝑺𝟓 = (3.1 + 2) + (3.2 + 2) + (3.3 + 2) + (3.4 + 2) + (3.5 + 2) 𝑺𝟒 = 2(4) + 2(5) + 2(6) + 2(7) = 5 + 8 + 11 + 14 + 17 = 8 + 10 + 12 + 14 𝑆5 = 55 𝑆4 = 44

Method 2: Method 2: 𝟓 Substitute, 𝑛 = 1, 2, 3 𝟕 ∑(𝟑𝒏 + 𝟐) To establish the pattern, is ∑ 𝟐𝒌 𝒏=𝟏 it an Arithmetic or not 𝒌=𝟒 Substitute, 𝑛 = 1, 2, 3 (3.1 + 2) + (3.2 + 2) + (3.3 + 2) + ⋯. 2(4) + 2(5) + 2(6) + ⋯ To establish the pattern, is = 5 + 8 + 11 + ⋯ … .. = 8 + 10 + 12 + ⋯ … .. it an Arithmetic or not 8 − 5 = 3 and 11 − 8 = 3 ∴ it is Arithmetic. 𝑛 10 − 8 = 2 and 12 − 10 = 2 ∴ 𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑] ∴ it is Arithmetic. Number of terms is 5 5 ∴ 𝑆5 = 2 [2(5) + 4(3)] or 4 𝑇𝑜𝑝 – 𝑏𝑜𝑡𝑡𝑜𝑚 + 1 = (5 − 1 + 1) ∴ 𝑆4 = 2 [2𝑎 + 3𝑑] Number of terms: ∴ 𝑆5 = 55 Top – bottom+1 ∴ 𝑆4 = 2[2(8) + 3(2)] = (7 − 4 + 1) = 4 ∴ 𝑆4 = 44 Note the two methods done above. If the sum of a set of terms of a sequence/series is for many terms say a hundred terms, it will make more sense to use the second method as you will be able to determine the sum without needing to calculate the value of all the terms and then to add them. Example 11: Write the following series in sigma notation: 5 + 8 + 11 + 14 + 17 Solution: 1) First calculate the general term for the series where 𝒂 = 𝟓 𝐚𝐧𝐝 𝒅 = 𝟑 . Hence 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑇𝑛 = 5 + (𝑛 − 1)3 𝑇𝑛 = 5 + 3𝑛 − 3 … 𝑇𝑛 = 3𝑛 + 2 ∑ (𝟑𝒏 + 𝟐) 2) Write the formula now in sigma notation 𝒏=⋯ Bottom value is the first term which is equal to 5: Top value is the last term which is equal to 17: 3𝑛 + 2 = 𝟓 3𝑛 + 2 = 𝟏𝟕 3𝑛 = 3 3𝑛 = 15 𝑛=1 𝑛=5 𝟓 3) ∑(𝟑𝒏 + 𝟐) 𝒏=𝟏

Western Cape Government FOR YOU Education Curriculum: FET ANALYTICAL GEOMETRY TOOLKIT FORMULAE Consider two points A(x;; y:) and B(x2; y2): DISTANCE . Thm. of Pythagoras = (2 — x4)? + (y2 — yr)? MIDPOINT ¥ Ye *Bix2) ig mist, by) Ye aes: 91) . The co-ordinates of the midpoint, M. are the averages of the co-ordinates of the endpoints, A and B. GRADIENT 8 ‘ m = Shangeiny . yoy y2-¥1 change inx X2 = x; A, X2~Xq . the gradient of the line ° 8, tan @ = opposite _ ¥2—¥4 yo-¥1 adjacent Xp -— X; y - m where @ is the angle of inclination of the line The Gradient of a line Values 4 | POSITIVE NEGATIVE ZERO UNDEFINED ye .-1 1 mee rs . =~ - Parallel lines B Parallel lines have os equal gradients. c AB||CD & mis=mco Perpendicular lines f2) If the gradient of line C4) is z. then the gradient of line C2] will be -2. Note: mg x mg = | 3) 3) 2) Le. The product of the gradients of | lines is -1. Collinear points A B Three points A, B & C are collinear if the gradients of AB & AC are equal. (Note: Point A is common.) ("as = Mae # AB&C are collinear The Inclination of a line Angles co and [i below are angles of inclination. The inclination of a line is the angle which the line makes with the positive direction of the x-axis. gradient of line gradient of line is positive is negative Gradient, m= tana or tan fi where a and # are the /* of inclination. Graphs in general 3 Basic facts Axis intercepts Every point on the y-axis has x = 0. Every point on the x-axis has y = 0. The equation The equation of a graph is true for all points on the graph e The equation of the y-axis is x = 0; & the equation of the x-axis is y = 0. Types of graph Different types/patterns are indicated by various equations. eg. y = mx+c indicates a straight line x? + y? = ¢? indicates a circle

Western Cape —_ FOR YOU Important Facts FACT (1) : Points on Graphs If a point lies on a graph, the equation is true for its coordinates, i.e. the coordinates of the point satisfy the equation 80, substitute! and, conversely, Ifa point (i.e. its coordinates) satisfies the equation of a graph (i.e. 'makes it true’), then it lies on the graph FACT ©: Point(s) of Intersection The coordinates of the point(s) of intersection of two graphs ‘obey the conditions’ of both graphs, ie. they satisfy both equations simultaneously. They are found * ‘algebraically’ by solving the 2 equations, or ® ‘graphically* by reading from the graph. THESE 2 FACTS ARE CRUCIAL! STRAIGHT LINE GRAPHS & their equations Standard forms Standard forms of the equation of a straight line: "y=mx+c: where m = the gradient & ¢ = the y-intercept Whenm=0: y=o Whenc=0: y=mx Also: x =k ... aline || x-axis ... a line through the origin .-a line || yeaxis = y—ys = m(x ~ x4)e we where m = the gradient & (x13 y1) is a fixed point. General form The general form of the equation of a straight line is ax+by+c=0. eg 2r+3y+6=0 CIRCLES & their equations Circles with the origin as centre True of any point (x; y) ty) y on a circle with centre (0; 0) and radius r is that: (xi y) x+y? = Thm. of Pythag.! Circles with any given centre True of any point (3 y) ‘y on a circle with centre (a; b) and radius r is that: o (x - a)? + (yb)? = r? Distance formula! (Thm. of Pythag.) Converting the equation of a circle General form: Ax? + Bx + Cy?+Dy+E=0 to Standard form: (x-a)’ + (yb =r {using completion of squares) x?-6x + y?+By- 25 = 0 x? -6x + y?+ By = 25 x= 6x +3? + y+ By +4? = 2549416 ~@-3P + (y+4)? = 50 This is the equation of a circle with: centre (3;-4) & radius, r=¥150 (= 52) units) Curriculum: FET A Tangent to acircle... is perpendicular to the radius of the circle at the To find the equation of a tangent, use ‘m and 1 point’ in the straight line equation: Y~¥s = m(x- x) eg.myp=29 meo = -3 (radius MP 1 tangent PQ) Point(s) of intersection of a Line and a Circle or C2) ‘touch’ (once!) [tangent] (1 point in common) or C3) don't cut or touch WA (no points in common) ©) if we substitute y = mx + ¢ into the equation of the ©, there will either be: €) 2 solutions or © 1 solution or C3] no solutions for .x, resulting in one of the above scenarios. FINAL ADVICE : Use common sense & ALWAYS DRAW A PICTURE !!! Aline and a circle either © ‘cut (twices) [secant] (2 points in common)

Western Cape Curriculum: FET Government Education FOR YOU ANSWER CALCULATOR INSTRUCTIONS 2); ANSWER MEAN & STANDARD DEVIATION REGRESSION & CORRELATION There are 3 phases for each calculator procedure: ‘4 \ STEP 4: how to get there ny) The Equation of the Regression Line STEP 2: how to enter the data ay ‘ Casio Saenie paved ao ENA Dae = PressiSelect: MODE ; 2 (STAT); 2(A+ Bx) STEP 2: Ungrouped Data Grouped Data/Frequency Tables Enter the x and y values, each followed by =. Press AC Casio You'll see; | Casio You'll see: STEP 3: © [wove] [2 : STAT] [1 : 1 - VAR} Le eee | | © fmooe) [2 : Star] [1 : 1 - VAR] | %: |e Press sift] STAT (on'1'); 5 (Reg); then: entered 2 [Sif T] [SETUP] Scroll down (use arrow) 2 * Enter each value, followed by [=] 3 papgtllar at a For A,press 1] =| or ForB, press 2/ =| after the last value: [=] [AC] «= * ©] Enter the int of interval To fir after A, press AC * To find the mean: [sHiFT] [STAT] [4 : VAR] [2 :¥] [=] = (or each score if provided) followed by [=], then. . . L and then do the whole of STEP 3. After the last value: [=] - and not [AC] - then... af © Tofind the $.D.: — {sHiFt] [STAT] [4 : VAR] [3 : cx] [=] a “a Use {_® to move to the top of the right-hand column, The Correlation Coefficient ) “= Note for Casio Type in the correct frequencies followed by [=] each Casio Itis essential to use (AC) at the end of entering all the data. time; after the last frequency: [=] [AC] = * Just like for A and B in the regression function: 5 e ; “x STEP 1: If this is not done, the mean and standard deviation * To find the mean: [sHiFT] [Stat] [4 : VAR] [2 :X] [=] ; will be added as entries in the data table. © Tofind the $.D.: [Siar] [STAT] [4 : van] [3 :xon} [=] o=— | | Press/Select: MODE; 2 (STAT); 2 (A+ Br) (These would then need to be deleted) STEP 2: an a at Enter the x and y values, each followed by =. wenn nnn e +--+ --------------------- Sharp Press AC Sharp You'll see: | © [Move] (1: STAT] - fo get into stats mode ... STEP 3: © [wove] [1 : STAT] - 10 get into stats mode ... then [0 : so] Press SHIFT) STAT (on't'); 5 (Reg): then [0 : so} © | Enter midpoint of each interval (or each score if . provided), followed by [To]. but, now: For r press 3] =| * Enter each value, followed by [M+] nome Then enter the frequency followed by [M+] f If you have found A and B first, press AC | © Tend te rc 7) a) (Enter alll the data like this.) before going back to the beginning of STEP 3. i fb @ To find the mean: [rci][T] «= * To find the $.0.; [RC] [ox] B35 To find the $.D.: [Rct] [ox] = To clear completely: Press SHIFT] CLR (on's); 1] =| ac