Functional Notation: Quadratic, Exponential, and Trigonometric Functions Handout, Lecture notes of Mathematics

Download Functional Notation: Quadratic, Exponential, and Trigonometric Functions Handout and more Lecture notes Mathematics in PDF only on Docsity! PARTICIPANT HANDOUT – FET PHASE 1 Module 1: Functions Sub-topic 1: Introduction to functions CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 The concept of a function Relationships between variables using words, tables, graphs, and formulae Domain and range Definition of a function Introduction When introducing functions to learners in grade 10, it is important to refer to the mathematics which they have come across in their earlier grades. Some examples of these are:  Working with the different operations  Checking for relationships (between sets of numbers, in formulae, etc.)  Substitution in equations/formulae  Sketching, finding the equation of and interpreting the graph of (a straight line) The approach used in grade 10 is developmental. In grade 12, learners are introduced to a more formal approach to a function. In this module, we used both approaches, starting with a developmental approach, which makes use of learners‟ prior knowledge. Activity 1: Introduction to functions Group organisation: Time: Resources: Appendix: Groups of 6 10 min  Flip chart  Permanent markers. None In your groups you will: 1. Select a scribe and a spokesperson for this activity only.(Rotate from activity to activity) 2. Use the flipchart and permanent markers and answer the questions as per the activity. 2 PARTICIPANT HAND-OUT – FET PHASE 1. Complete the table below: x 0 1 2 3 4 2x 1.1 Write down some ordered pairs for the above table. 1.2 Write down the equation represented by the information in the table. 1.3 Draw the graph. 1.4 Is there any restriction on the domain and range? Why? 1.5 Write down the domain and range. 2. The number of diagonals in a polygon is given by the following table: Number of sides (x) 3 4 5 6 7 Number of diagonals (y) 0 2 5 2.1 Complete the table. 2.2 Draw the graph. 2.3 Is there any restriction on the domain and range? 2.4 Write down the domain and range. 2.5 Determine the equation showing y in terms of x. Consolidation & Terminology The situations/relationships described in activity 1 represent functions. Use this information to explain to your learners in very simple terms what a function is. Even though this is handled in detail and formally assessed in grade 12, for grade 10 learners this is intuitive. NB: On page 131 of the Report on the 2013 National Senior Certificate (Diagnostic report), it was noted that learners struggled to identify or define a function. They were not clear about terminology such as “one-to-many” and “many-to-many”. Definition: A function is a rule by means of which each element of a first set, called the domain, is associated with only one element of a second set, called the range. Each element of the range is an image of corresponding elements of the range. Look at the following Venn diagrams. PARTICIPANT HAND-OUT – FET PHASE 5 Worked Examples 1 & 2: Functional Notation (10 mins) The facilitator will now explain these suitable examples of functional notation. 1. Remember to make notes as the facilitator is talking 2. Ask as many questions as possible so as to clarify any misconceptions that may occur. 1. Given that  2( ) 4f x x Determine the following: 1.1 f(1) 1.2 f(-3) 1.3 The value(s) of x if ( ) 0f x Solution Once again, for 1.1 and 1.2 it is simple substitution: 1.1       2(1) (1) 4 1 4 3 f 1.2        2( 3) ( 3) 4 9 4 5 f 1.3 For this question please note that ( ) 0f x is not the same as (0)f . You may show learners this difference on a graph.             2( ) 0 4 0 ( 2)( 2) 0 2 or 2 f x x x x x x We note that in this question we used factorisation to obtain the values of x. 1. Given the following function:  2( ) 9f x x Determine: 2.1 (5)f 2.2 The value(s) of x if ( ) 2f x . 2.3 The domain of f. 6 PARTICIPANT HAND-OUT – FET PHASE 2.4 The range of f. Solution 2.1      2(5) (5) 9 25 9 16 4f 2.2             2 2 2 ( ) 2 9 2 9 4 (square both sides) 13 13 f x x x x x 2.3 The domain of f exists when  2 9 0x (why? – this is a restriction)           2 9 0 (this is a simple quadratic inequality) ( 3)( 3) 0 3 or 3 x x x x x NB: You may have to briefly revise how to solve quadratic inequalities. Thus, the domain is: {x: x  -3, x  R}  {x: x  3, x R} 2.4 Note that indicates positive square root. Thus, our range can never be negative. Range = {y: y  0, y R} Activity 2, 3 & 4: Functional Notation Group organisation: Time: Resources: Appendix: Groups of 6 10 min  Flip chart  Permanent markers. None In your groups you will: 1. Select a scribe and a spokesperson for this activity only. (Rotate from activity to activity) 2. Use the flipchart and permanent markers and answer the questions as per the activity. 2. Use  ( ) 2 1f x x } and g(x) =  2 4x to determine: 2.1 f(-3) 2.2 1 2 g(2) + 3 PARTICIPANT HAND-OUT – FET PHASE 7 3. Use   ( ) 2x 1f x ;   2 ( ) 3 x g x x and  ( ) 5h x x to determine: 3.1 h(-4) – f(-1) 3.2 x if g(x) = 0 3.3 the domain of g 4. Use    2 ( ) 3 x g x x to determine the domain of g. Activity 5: Types of Correspondence Group organisation: Time: Resources: Appendix: Individual 5 min  Participants Handout None 1. Study the graphs that follow and state whether they represent functions or not. Give your reason/s in terms of correspondence. 2. Write your answers in the space provided. (The circle and semi-circle is not prescribed in CAPS for this section. It is merely used to show the vertical line test.) 10 PARTICIPANT HAND-OUT – FET PHASE Misconceptions Teachers should pay careful attention when working with functions which have restrictions as these affect the domain and range of the said function. Learners tend to ignore restrictions on certain functions. Examples 1.  ( ) 3g x x is a restricted function as x – 3 lies under the square root so x – 3  0, that is x  3 2.   1 h( ) 2 x x is also a restricted function as x + 2 cannot be 0 so  2.x 3. The domain of   2f( ) 2 3 1x x x is not restricted but its range is restricted. Please make learners aware of the following:  The difference between g(0) and g(x)=0  The difference between g(2) and g(x)=2 Conclusion When working with functions it is important to explain to learners how to get the correspondence. They should be able to look at any table, set of ordered pairs, equation or graph and state what the correspondence is. Functions are a major portion of the Mathematics paper 1 curriculum. NB: In each lesson please ensure that the lesson incorporates the following:  Kinaesthetic learning – learners are given hands-on activities/exercises to work through.  Auditory learning – using one‟s voice effectively without confusing learners.  Visual learning – learners must be able to see. In the cases of graphs, there should be some colour and the various features shown. PARTICIPANT HAND-OUT – FET PHASE 11 Sub-topic 2: Introduction to quadratic functions   2{( ; ) : ( ) }x y y a x p q CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 Point-by point plotting of  2y x Shape of functions, domain & range; axis of symmetry; turning points, intercepts on the axes The effect of a and q on functions defined by  . ( )y a f x q where  2( )f x x Sketch graphs and find the equation of graphs and interpret graphs Revise the effect of the parameters a and q and investigate the effect of p and the function defined by   2( )y a x p q Sketch graphs and find the equation of graphs and interpret graphs Introduction It is important to note that the quadratic function is also known as the parabola. Study the picture below: This is an example of a parabola in real life. It is the St Louis Arch, which is located in the state of Missouri in the United States of America. Accessed from: http://www.tattoodonkey.com/parabolas-in-real-life-tattoo/2/ 12 PARTICIPANT HAND-OUT – FET PHASE Another example of a parabola in real-life is to be found in car headlights. Accessed from http://www3.ul.ie/~rynnet/swconics/UP.htm The general quadratic function The general formula for the quadratic function is:   2 ( )y a x p q , where a, p and q are all real numbers. However  0a . Why? We can simplify the quadratic function as follows: 2 2 2 2 2 2 2 ( ) ( 2 ) (2 ) +(ap )) ( 2 ) y a x p q a x px p q ax ap x q ax bx c where b ap and c ap q                   2 y ax bx c is another way of writing the general formula of a quadratic function. Quadratic functions in the form   2 ( )y a x p q or   2 y ax bx c represent the parabola. Let b = c = 0 and let a = 1 Then we have a very simple parabola  2y x PARTICIPANT HAND-OUT – FET PHASE 15 1.1 Graphs: 1.2 If a increases, the graph becomes narrower or stretches. As a decreases the graph becomes wider or flatter. Example 2 2.1 Sketch the graphs of: 2 2 2 2 2 2 3 1 2 1 4 y x y x y x y x y x           2.2 What do you observe? 16 PARTICIPANT HAND-OUT – FET PHASE Solution 2.1 We can bring down the table and extend the number of rows: x 0 1 -1 -2 2 -3 3   2y x 0 -1 -1 -4 -4 -9 -9   22y x 0 -2 -2 -8 -8 -18 -18   23y x 0 -3 -3 -12 -12 -27 -27   21 2 y x 0  1 2  1 2 -2 -2 -4,5 -4,5   21 4 y x 0  1 4  1 4 -1 -1 -2,25 -2,25 2.2 As a decreases the graph becomes narrower or stretches. As a increases the graph becomes wider or flatter. Some properties of the function   2{( ; ) : }f x y y ax  When  0a , it has a minimum value. When  0a , it has a maximum value  The y-axis (x = 0) is the axis of symmetry  The turning point is (0;0)  Domain = : ∈ 𝑅  When  0a , the range = * : ≥ 0; ∈ 𝑅+  When  0a , the range = : ≤ 0; ∈ 𝑅 PARTICIPANT HAND-OUT – FET PHASE 17 Example 3 3.1 Sketch the graphs of 2 2 2 4 4 y x y x y x      Activity 1-3: Basic exercise Group organisation: Time: Resources: Appendix: Pairs 30 min  Graph Paper None 1. Read each question carefully. 2. Discuss the question with your partner and proceed to answer the questions. You may draw the graphs on the graph paper supplied and answer the questions alongside the graphs. 1. Given the function: 2{( ; ) : 2 }f x y y x   1.1 Sketch the graph of f. 1.2 If f is moved 3 units downwards, to form the graph of g, determine the equation defining graph g and its maximum value. 1.3 Describe how the graph of y x2 may be transformed firstly to f and then to g. 20 PARTICIPANT HAND-OUT – FET PHASE Question: The graphs of   2 4 and y x y d are drawn. For what value(s) of d will there be: (a) Two points of intersection? (two roots) (b) One point of intersection? (one root) (c) No points of intersection? (no roots) Solution We get the value(s) of d from the graphs (a)  4d (b)  4d (c)  4d This question can be phrased in another way. See below: Use your graph to determine the value(s) of d for which the roots of  2 4 x d (a) Real and unequal (two points of intersection). (b) Real and equal (one point of intersection). (c) Non-real or imaginary (no points of intersection). Misconceptions Learners must know the following:  If the y-intercept of a graph is 6, then the coordinate of the point is (0;6).  If the x-intercepts are -2 and 4, then the coordinates are (-2;0) and (4;0). Conclusion The work done in this module only included examples of parabolas which had the y-axis as the axis of symmetry. The graphs drawn were translated up and down the y-axis. These examples provide more than suitable preparation for the ones that follow in the next module. PARTICIPANT HAND-OUT – FET PHASE 21 Sub-topic 3: Sketching quadratic functions Introduction We have seen from the previous module that the graph of  2{( ; ) : }x y y ax q can be stretched or flattened depending on the value of a. The value of q will dictate whether the graph moves up or down the y-axis. For example the graph of   2g {( ; ) : 3 6}x y y x tells us that  2y x has been stretched by a factor of 3 and moved 6 units down on the y-axis. We can see this in the graph below: We could also move  2y ax along the x-axis (left or right) The graph of   2( )y a x p A stretch by a factor of 3 A move or translation of 6 units downward 22 PARTICIPANT HAND-OUT – FET PHASE Worked Examples 1-3: Simple sketches of quadratic Functions(30 Mins) Pay careful attention as the facilitator discusses the various sketches of quadratic functions. Don’t forget to ask questions and make notes. Example 1 1.1 Sketch the following graphs on the same system of axes:  2y x   2( 2)y x   2( 3)y x 1.2 What do you notice? Solution 1.1 Graphs PARTICIPANT HAND-OUT – FET PHASE 25 Example 3 3. Given the following:        2 2{( ; ) : 2 4 6} {( ; ) : 2( 1) 8}f x y y x x x y y x 3.1 Determine the x and y-intercepts. 3.2 Write down the coordinates of the turning point. 3.3 Write down the equation of the axis of symmetry. 3.4 Does the graph have a minimum or maximum value? Explain your answer. 3.5 Draw the graph of f. Solution 3.1 y intercept is f(0) = -6 x-intercepts:               2 2 2 4 6 0 2 3 0 (Divide by 2) ( 1)( 3) 0 1 or 3 x x x x x x x 3.2 Turning point (1;-8) 3.3 1 x 3.4 The graph has a minimum value since a = 2 (positive). 3.5 Graph In general, the following will apply: If f = 2{( ; ) : }x y y ax bx c   then: (1) Axis of symmetry: 2 b x a   (2) Minimum or maximum value: 24 4 ac b y a   26 PARTICIPANT HAND-OUT – FET PHASE (3) Turning point 24 ; 2 4 b ac b a a       (4) y- intercept occurs when x = 0. Thus; y = c (is the y-intercept). (5) x-intercepts occur when y = 0, so 2 0ax bx c   . This is solved by factorisation or using the formula. NB: Gradients and average gradients We know learners should be familiar with straight line graphs from grade 9. The general form of a straight line function is   ( ; ) :x y ax by c 0 ; If ax by c then by ax c a c y x b b a c y mx k m k b b                       where is the gradient and is the intercept.m k y  Thus, in an equation such as  2 3y x the gradient (m = 2) is the same throughout. However, in a parabola, the gradient is changing all time. We can, however, find the average gradient between two points on the parabola. Given:   2( ) 2 3 1f x x x The points A (-1;4) and B(0;-1) lie on the graph. The average gradient of the line joining A to B is calculated as follows:             2 1 2 1 ( ) ( ) (0) ( 1) 1 4 5 0 ( 1) 1 f x f x f f x x X PARTICIPANT HAND-OUT – FET PHASE 27 Activity 1: Graphs and Inequalities Group organisation: Time: Resources: Appendix: Individual 5 min  Participant Handout None 1. Read each question carefully. 2. Discuss the question with your partner and proceed to answer the questions in the space provided below. The graphs of      2( ) 5 6 and ( ) 6f x x x g x x are drawn below. The intercepts with the axes (-6;0); (1;0); (0;6) and the turning point ( 2,5;12,25) are shown. Determine the following: The value(s) of x for which: (a) ( ) ( )f x g x (b) ( ). ( ) 0f x g x ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 30 PARTICIPANT HAND-OUT – FET PHASE      2Given g( ) with 0; 0 and 0 and x-intercepts of different signs. Draw a sketch graph of g x ax bx c a b c Draw a rough sketch of f(x) Possible graph In module 2 we saw how quadratic functions and a straight line graph can be used to determine roots of quadratic equations. Let us do another example: PARTICIPANT HAND-OUT – FET PHASE 31 Suppose the following question was given: Use your graph(s) to find real value(s) of k for which    2 5 6x x k has two negative, unequal roots. We draw the graph of    2 5 6y x x The maximum value is found in two ways: We find the axis of symmetry and then substitute in the function to find the y value.               2 ( 5) 2,5 2 2( 1) ( 2,5) 5( 2,5) 6 12,25 b x a y X 32 PARTICIPANT HAND-OUT – FET PHASE The maximum value is y = 12,25. Another way of find the maximum value is:            2 24 4( 1)(6) ( 5) 49 12,25 4 4( 1) 4 ac b y a We bring the graph down: The value of k must lie between the y- intercept (y = 6) and the maximum value y = 12,25 for there to be two negative, unequal roots. Thus,  6 12,25y X PARTICIPANT HAND-OUT – FET PHASE 35 Now that we know what shape it is, let us find out more about the shape and properties of the hyperbola. We start with the general form of the hyperbola: The general form of the hyperbola is given by: ℎ {( ; ): } ℎ ∈ 𝑅 ∈ 𝑅 Let p = 0 and q = 0 So we have:  a y x If a = 6 so    6 6y xy x We can use a table to get both x and y values which satisfy this function: x 1 2 3 6 -1 -2 -3 -6 y 6 3 2 1 -6 -3 -2 -1 We now use point-by-point plotting to draw the graph of   6 6 or xy y x . Please note that both negative and positive values are used. We also join the points. 36 PARTICIPANT HAND-OUT – FET PHASE Worked Examples 1- 4: Drawing a hyperbola(30 Mins) Pay careful attention as the facilitator discusses the sketches of the various hyperbolas. Don’t forget to ask questions and make notes. Example 1 When p = 0 in the function h we have: a y q x   1.1 Sketch on the same system of axes, the graphs of: 6 6 2 6 2 y x y x y x      1.2 What do you observe? Solution 1.1 We use Geogebra to sketch the graphs: 1.2 We note that   6 2y x is a translation of  6 y x two units upwards (along the y-axis). We also note that   6 2y x is a translation of  6 y x two units downwards (along the y- axis). PARTICIPANT HAND-OUT – FET PHASE 37 Example 2 Let 0q  in the function h. We have   a y x p 2.1 Sketch (on the same system of axes): 6 6 1 6 2 6 1 6 2 y x y x y x y x y x          2.2 What do you observe? Solution 2.1 Once again we use Geogebra to draw these graphs 2.2 We note the following:  The graph of 6 y x  does not cut the x-or y-axis.  The others cut the y-axis but not the x-axis.  The x-axis is a horizontal asymptote for all the graphs; the others have vertical asymptotes at different places (x = -2; x = -1; x = 0; x = 1; x = 2).  The graph of 6 𝑥:1 is the graph of 6 𝑥 shifted 1 unit to the left.  The graph of 6 𝑥;2 is the graph of 6 𝑥 shifted 2 units to the right.  In general if a y x p   , then the horizontal asymptote is the x-axis (y=0); while the vertical asymptote is when x p  40 PARTICIPANT HAND-OUT – FET PHASE Activities 1-4: The Hyperbola Group organisation: Time: Resources: Appendix: Pairs(Not the same as before) 60 min  Graph paper and A4 Sheets None Together you will: 1. Study the questions and answer them on the graph and A4 paper that is provided. Activity 1 1. Determine the asymptotes for each of the following functions:         4 1.1 4 1.2 2 4 1.3 2 4 1.4 2 2 y x y x y x y x Activity 2 2. Given 8 ( ; ) : 2 2 f x y y x         2.1 Determine the domain of f. 2.2 Determine the coordinates of the x-and y-intercepts of f. 2.3 Write down the equations of the asymptotes. 2.4 Draw the graph of f. 2.5 Now use your graph to determine the range of f. Activity 3 3. The graph of 10 ( ; ) :g x y y x        is translated 3 units to the right and 4 units upwards to form the graph of h. Determine the equation of h Activity 4 4. Draw the graph of    2 1 x y x PARTICIPANT HAND-OUT – FET PHASE 41 Misconceptions Learners should be able to calculate the value of p and q from a drawn graph of    a y q x p Learners should know that another form of the hyperbola is              2 5 3 5 eg) can be written as 1 3 3 3 x b y x c x x y y x x x Conclusion Summarise key points of this module:  The value of a in    a y q x p determines the location of the hyperbola. If a > 0, it will be in the “first” and “third” quadrants; if a < 0 it will be in the “second” and “fourth” quadrants.  The vertical asymptote is calculated as     0x p x p  The horizontal asymptote is y q 42 PARTICIPANT HAND-OUT – FET PHASE Sub-topic 5: The exponential function CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 Point-by point plotting of y xb Shape of functions, domain & range; axis of symmetry; turning points, intercepts on the axes The effect of a and q on functions defined by  . ( )y a f x q where ( ) xf x b Sketch graphs and find the equation of graphs and interpret graphs Revise the effect of the parameters a and q and investigate the effect of p and the function defined by    . 0; 1x py a b q b b Sketch graphs and find the equation of graphs and interpret graphs Introduction Exponential growth and decay is the rate of change in a graph, whether it is increasing or decreasing. Real life examples of exponential growth are population, plants, animals and, bacteria. http://www.ecofuture.org/pop/facts/exponential70.html The graph below shows the compound growth on an investment of R5000,00 for n years at 6% per annum, compounded annually. PARTICIPANT HAND-OUT – FET PHASE 45 Some observations:   Domain { : }x x R    Range { : 0, }y y y R  We note that the y-intercept is 1 and occurs at (0;1). There is no x-intercept  The x-axis is an asymptote (horizontal)  The graph represents a one-to-function 46 PARTICIPANT HAND-OUT – FET PHASE Worked Examples 1-2: Simple sketches of exponential functions(30 Mins) The facilitator will now take you through a number of sketches of exponential functions. Example 1 1.1 Sketch the graphs of: 2 3.2 3.2 2 3.2 2 x x x x y y y y       1.2 What do you observe? Solution 1.1 Graphs PARTICIPANT HAND-OUT – FET PHASE 47 1.2 What do you observe?            2 asymptote is negative -axis; intercept is 1; no x-intercept 3.2 asymptote is negative -axis; intercept is 3; no x-intercept 3.2 2 asymptote is 2; intercept is 3; no x-interce x x x y x y y y x y y y y y y       pt 3.2 2 asymptote is 2; intercept is 1; x-intercept is -0,6 (approximately)xy y y y The asymptotes above are all horizontal asymptotes. Please note: 2𝑥 ℎ 𝑓 2𝑥 𝑓 𝑓 Example 2 2.1 Sketch the graphs of: 1 1 1 1 3 3 2.3 2.3 1 2.3 2 x x x x x y y y y y            2.2 What do you observe? Solution 2.1 Graphs 50 PARTICIPANT HAND-OUT – FET PHASE Activity 4 4. Answer the questions that follow on the graphs below: 4.1 Which function f(x) or g(x) has the form y x. 4.2 What are the roots of f(x) =0? 4.3 What is the y-intercept of g(x)? 4.4 What is the range of g(x)? 4.5 Give the equation of the asymptote of g(x). 4.6 Give the equation of the axis of symmetry of f(x)? Misconceptions Emphasise to learners the following:  Using point-by-point plotting of exponential functions.  Knowing that the exponential function does not cut the x-axis unless the graph is translated down the –y-axis.  How to interpret the features of drawn graphs. Conclusion Summarise key points of the lesson  The features of the exponential function.  The shape of the exponential function. PARTICIPANT HAND-OUT – FET PHASE 51 Sub-topic 6: Inverse of selected functions CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 General concept of the inverse of a function and how the domain of the function may be restricted (in order to obtain a one-to-one function) to ensure that the inverse is a function Determine and sketch graphs of the inverses of functions defined by:  y ax q ;  2y ax ;   ;( 0; 1)xy b b b Focus on the following: Domain & range; intercepts on the axes; turning points; minima; maxima; shape and symmetry; average gradient; intervals on which the function increases/decreases. Revision of exponential laws and the exponential graph The definition of a logarithm The graph of the function defined by    log for cases 0 1 1by x b and b Introduction To find the inverse of functions, we interchange the x-and-y values in the given function: This can be seen in the table below: Function Inverse  3y x  3x y  22y x  22x y  6 y x  6 x y  2xy  2yx 52 PARTICIPANT HAND-OUT – FET PHASE The inverse of linear functions Consider: Let f(x) = 2x – 4, we may write this function as y = 2x – 4 Thus, the inverse of f written as 1f  will be: x = 2y – 4. Writing this inverse in the y-form we get: y = ( 4) 2 x  So  1 ( 4) 2 x f x   Remark We have to be able to denote the difference in notation between the function and its inverse. The standard notation is f(x) for the function and f -1 (x) for the inverse. Do not confuse f -1 (x) with 1 f(x) as they do not represent the same thing. Please note:  Only one-to-one functions have a unique inverse.  If the function is not one-to-one, the domain of the function must be restricted so that a portion of the graph is one-to one. You can find a unique inverse over that portion of the restricted domain.  The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse. Worked Example 1,2,3&4: Inverse functions-all types (60 Mins) The facilitator will now discuss the various inverse functions. Don‟t forget that any question is an important question! Example 1 Determine the inverses of the following functions and sketch both the function and its inverse on the same system of axes. Indicate whether the given function and its inverse is a function or not, and state the domain and range of the inverse. 1.1 y = x +3 1.2 3x + 2y = 6 PARTICIPANT HAND-OUT – FET PHASE 55 Example 2 Let f = {(x; y): 29 y x } 29 y x Just like the linear function, interchange x and y: Thus, the inverse is: 29 x y Making y the subject of the formula, we have : 2 1 9 3 : ( ; ) : 3 x y x y x So f x y y              We may now draw both 1 and f f  . We see that this inverse is not a function (Why?). In order for the inverse to be a function we have to restrict the range, either taking positive values or negative values. 56 PARTICIPANT HAND-OUT – FET PHASE If y  0 we have the following sketch graph, which will be a function If y  0 we have the following sketch graph, which will also be a function Example 3   Let ( ; ) : 2xf x y y We draw this graph as follows: x PARTICIPANT HAND-OUT – FET PHASE 57 We see that the graph approaches the x-axis but does not cut the x-axis. We say the graph is asymptotic at y = 0 (the x-axis). Although it is not possible to state what the exact minimum value of the graph is, we know that the graph will not go below the x-axis. We also see that the graph is increasing. Now if we interchange x and y in 2xy  then the inverse is: 2yx  If we make y the subject of the formula: 2logy x  1 2Hence, ( ; ) : logf x y y x   We sketch this graph as follows: We see that the graph cuts the x-axis at 1. The above graph gets close to the y-axis but does not cut the y-axis. We say the graph is asymptotic at x = 0 (the y-axis). For all x >0 we see that the graph is increasing Domain = {x: x > 0; x R} Range = {y: y  R} The inverse is a one-to-one function 60 PARTICIPANT HAND-OUT – FET PHASE (2)   1 2 {( ; ) : log x}g x y y            1 2 2 log log Now log x log 1 log2 log 2 x x y x We see that:  1 2 2 log x log x PARTICIPANT HAND-OUT – FET PHASE 61 Features of the logarithmic graphs Graph   2{( ; ) : log x}f x y y   1 2 {( ; ) : log x}g x y y Domain  { : R; 0}x x x  { : R; 0}x x x Range { : R}y y { : R}y y Asymptotes The x-axis The x-axis Coordinates of x-intercepts (1;0) (1;0) Coordinates of y-intercepts Nil Nil Increasing or decreasing function Increasing Decreasing If we draw the graphs together We note that  2 1 2 y log and log x are reflections of each other with respect to the positive x-axis. x y 62 PARTICIPANT HAND-OUT – FET PHASE Misconceptions Learners should be able to determine the following features from inverse functions:  The domain and range.  The asymptotic behaviour of the exponential function and its inverse, the logarithmic function.  Work with functions and inverse functions which are transformed.  The correspondence of both the function and inverse function. Conclusion Summarise key points of the inverse functions:  How to determine the inverse of a given function.  How to modify the domain of the inverse so the inverse is a function.  Key steps in drawing the inverse functions. PARTICIPANT HAND-OUT – FET PHASE 65 We can select a few values using the calculator to sketch this graph: x 0 0 0 0 20 0 0 2 0 240 2 0 00 0 0 sin x 0 0,5 0,866 1 0,866 0,5 0 -0,5 -,0,866 -1 -0,866 -0,5 0 We can select a few points to draw this graph for [0 ;360 ] We can now join these points: We can do the same for  cosy x 66 PARTICIPANT HAND-OUT – FET PHASE Draw your own table of values: We will select a few: x 0 0 0 2 0 0 cos x 0 0,5 0,866 1 0,866 We can join the points: PARTICIPANT HAND-OUT – FET PHASE 67 We can do the same for y=tan x We join the points. 70 PARTICIPANT HAND-OUT – FET PHASE Note: the range [-1;1] can be written as -1≤ y ≤ 1         sin 1 graph of sin "moved" or translated 1 unit upward along the y-axis 2sin 1 graph of sin stretched by a factor of 2 then translated down 1 unit down the y-axis y x y x y x y x Graphs of the type  ( ; ) : cosx y y a x q  Example 2 Sketch the following graphs on the same system of axes for [0 ;360 ] . cos cos + 1 2cos 2cos + 1 2cos 1 y x y x y x y x y x       PARTICIPANT HAND-OUT – FET PHASE 71 Activity 1: Trigonometric functions Group organisation: Time: Resources: Appendix: Individual 15 min  Participants Handout None You will complete the table as explained by the facilitator. Activity 1 Use the graphs drawn to complete the table below: Function cos cos + 1 2cos 2cos + 1 2cos 1 Amplitude Period Zeros Range y x y x y x y x y x       Comment on the transformation that resulted in: cos 1 ; 2cos 1 ; 2cos 1y x y x y x      . ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ ______________________________________________________________________ 72 PARTICIPANT HAND-OUT – FET PHASE Please note the following: The period above 360 is very useful when finding the general solutions of trigonometric functions of sine and cosine. It means that the angle (solution) is repeated after 360 . If we were to find the general solution of the equation  1 sin 2 x Then we have:          1 sin 2 30 or 150 (specific solutions) and the general solutions are: 30 .360 or 150 .360 0;1;2;3;.... x x x k k k Worked Example 3: Graphs of the type:  ( ; ) : tanx y y a x q  Functions (5 Mins) The facilitator will now discuss these functions with you. Don‟t forget that any question is an important question! Example 3 Sketch the following graphs on the same system of axis. tan tan + 1 tan 1 y x y x y x     PARTICIPANT HAND-OUT – FET PHASE 75 Complete the following table: Function 2tan 2tan + 1 2tan 1 Amplitude Period Zeros Range Asymptotes y x y x y x     76 PARTICIPANT HAND-OUT – FET PHASE Worked Examples: Various types Functions (40 Mins) The facilitator will now discuss these functions with you. Don‟t forget that any question is an important question! Graphs of the type  ( ; ) : sin( )x y y kx Example 4 Draw the following graphs on the same system of axes for [0 ;360 ] . sin sin2 sin3 y x y x y x    PARTICIPANT HAND-OUT – FET PHASE 77 Completed table:         Function sin 1 360 0 ;180 ;360 [ 1;1] 0 ;90 ;180 ;360 sin2 1 180 [ 1;1] 2 270 ;360 0 ;60 ;120 ; 360 sin3 1 120 180 ;240 ; [ 1;1] 3 300 ;360 Amplitude Period Zeros Range y x y x y x Graphs of the type  ( ; ) : cos( )x y y kx Example 5 Draw the following graphs on the same system of axes for [0 ;360 ] . cos cos2 cos3 y x y x y x    80 PARTICIPANT HAND-OUT – FET PHASE         : sin( 30 ) graph of sin translated or "moved" 30 to the right (on the x-axis) sin( 30 ) graph of sin translated or "moved" 30 to the left (on the x-axis) Note y x y x y x y x Graphs of the type  ( ; ) : cos( )x y y x p  Draw the following graphs on the same system of axes for [0 ;360 ] . cos cos( 30 ) cos( 30 ) y x y x y x      Completed table:            Function cos 1 360 90 ;90 ;270 [ 1;1] cos( 30 ) 1 360 60 ;120 ;300 [ 1;1] cos( 30 ) 1 360 120 ;60 ;240 [ 1;1] Amplitude Period Zeros Range y x y x y x         : cos( 30 ) graph of cos translated 30 to the right on the x-axis cos( 30 ) graph of cos translated 30 to the left on the x-axis Note y x y x y x y x PARTICIPANT HAND-OUT – FET PHASE 81 Graphs of the type  ( ; ) : tan( )x y y x p  Sketch the following graphs on the same system of axes for [0 ;360 ] . tan tan( 30 ) tan( 30 ) y x y x y x      Completed table: Function Amplitude Period Zeros Range  tan y x - 180 0 ;180 ;360  [ ; ]  tan( 30 ) y x - 180 30 ;150 ;330  [ ; ]  tan( 30 ) y x - 180 30 ;210  [ ; ]         : tan( 30 ) graph of tan translated 30 to the right (on the x-axis) tan( 30 ) graph of tan translated 30 to the left (on the x-axis) Note y x y x y x y x 82 PARTICIPANT HAND-OUT – FET PHASE Activity 3-5: Interpretation of graphs Group organisation: Time: Resources: Appendix: Groups of 6 60 min  Whiteboard markers  Flipchart paper None In your groups you will: 1. Select a new scribe and a spokesperson for these activities. 2. Use the flipchart and permanent markers and answer the questions as per the activity. Activity 3 Make a study of all the trigonometric graphs in this module. Now complete the following table for [ 360 ;360 ] int / ( int ) sin 1 3 cos 2 tan sin( 20 ) cos( 45 ) zeros Function Amplitude y ercept period asymptote s range x ercept y x y x y x y x y x           Activity 4 The diagram below shows the graphs of f(x) = sin ax and g(x)=cos(x+bo ) PARTICIPANT HAND-OUT – FET PHASE 85 Module 2: Probability Sub-topic 1: What is probability? Grade 10 Grade 11 Grade 12 Compare the relative frequency of an experimental outcome with the theoretical probability of the outcome. Venn diagrams as an aid to solving probability problems. Mutually exclusive events and complementary events. The identity for any two events A and B: 𝑃 𝐴 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 and 𝐵) Dependent and independent events. Venn diagrams or contingency tables and tree diagrams as aids to solving probability problems (where events are not necessarily independent). Generalisation of the fundamental counting principle. Probability problems using the fundamental counting principle Introduction Probability is defined as a likelihood that a particular event would occur. Before we go any further, why do we learn probability? Probability is useful to humankind in that it helps make decisions by providing information on the likelihood of events. What every day events use Probability? Name a few. The weather report; insurance; lotto; medical research; law enforcement; and political science Terminology Trial: a systematic opportunity for an event to occur. Experiment: one or more trials. Sample space: the set of all possible outcomes of an event. Event: an individual outcome or any specified combination of outcomes. Complement: all outcomes that are not in the event but are in the universal set. Complementary events: when two mutually exclusive events together contain all the outcomes in the sample space. For an event called A, we write the complement as “not A ”. Another way of writing the complement is as A′. Dependent events: if one event affects the occurrence of the other, then the events are dependent. Independent events: two events are independent if the occurrence of one event has no effect on the likelihood of the occurrence of the other. 86 PARTICIPANT HAND-OUT – FET PHASE Frequency: The number of times a particular outcome occurs in a particular period or given time or is measured against a number of trials is the frequency of that outcome. Incompatible events or mutually exclusive events: Two or more events are said to be mutually exclusive or incompatible when only one of those events can occur at a time. Basic probability calculations Probability is the study of how likely it is that a particular event will happen. For example, these are questions about the probability of something happening.  What is the chance that it will rain tomorrow?  If you buy a Lotto ticket, what is the chance that you will win the Lotto? Probability scale The scale of chance, or likelihood, starts with „no chance at all‟ and passes through various phases before getting to „certain‟. Some of those in-between phases could be „almost impossible‟, „unlikely‟, „even chance‟ or „highly likely‟. Scale Never Unlikely Equal Chance Likely Always No Chance Probably not Even chance Probably Certain Impossible Small chances of happening Fifty-fifty More chances of an event happening Definite Events Getting a 12 from a pack of playing cards Winning lotto Heads when tossing a coin Rain will fall in December in Free state The sun will rise tomorrow Relative frequency is approximated by performing and recording the ratio of the number of occurrences of the event to the number of trials. As the number of trials increases, the approximation of the experimental probability improves. PARTICIPANT HAND-OUT – FET PHASE 87 Consider this example: What is the probability that when you look out of a window at a street, the next car that passes will be red? You can no longer use equally likely outcomes for this. You need to use relative frequency. Theoretic probability is based on the assumption that all outcomes in the sample space occur randomly. Consider this example: When you toss a coin, there is an equal chance of obtaining heads or a tails. When you throw a die, the probability of getting a 6 is 1/6. In your own words, what is the difference between relative frequency and theoretic probability? For relative frequency in an event A: 𝑃 ℎ 𝐴 𝑓 𝑓 𝐴 ℎ 𝑓 In theoretic probability the probability of an event A, denoted by P(A) is defined by: 𝑃(𝐴) 𝑓 𝐴 𝑓 ℎ Worked Examples 1-3: Basic probability questions (25 Mins) The facilitator will now guide and support you through these worked examples. Don‟t forget to raise your concerns and ask questions! Example 1: Grade 10(R) A bag contains 3 white cards, 2 black cards and 5 red cards. Find the probability of each event for one draw. 1.1 a white card 1.2 a black card 1.3 a red or black card Solution: 𝑃 𝑓 𝑓 𝑓 1.1. P(a white card) 3 10 1.2. P(a black card) 2 10 1 5 1.3. P(a red or a black) 5:2 10 7 10 90 PARTICIPANT HAND-OUT – FET PHASE Solution: 3.1. The events “oppose” and “no opinion” are mutually exclusive events P (oppose or no opinion)= P(oppose) +P(no opinion) 00 00 00 0 3.2. The events “man and “oppose” are inclusive events P (man or oppose) = P(man)+P(oppose) –P(man and oppose) 0 00 00 − 2 00 00 4 0 Activity 1&2: Mutually exclusive and complementary events Group organisation: Time: Resources: Appendix: Individual 15 min  Participants Handout None Write the answers to the questions in the space provided. Activity 1: Grade 10 1.1. Explain the meaning of mutually exclusive events and of inclusive events. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ PARTICIPANT HAND-OUT – FET PHASE 91 1.2. Give 3 examples of each __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ Activity 2: Grade 10 2.1 Describe how to find the complement of the event „rolling 1” or “rolling 2” on a die. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ Conclusion Probability is a section with few marks in our curriculum but these marks can make a difference between distinctions and passing. The focus should be on learners understanding terms and noticing the difference. Use real examples to teach this topic as it is applicable to real life. 92 PARTICIPANT HAND-OUT – FET PHASE Sub-topic 2: Venn Diagrams CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 Compare the relative frequency of an experimental outcome with the theoretical probability of the outcome. Venn diagrams as an aid to solving probability problems. Mutually exclusive events and complementary events. The identity for any two events A and B: 𝑃 𝐴 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 and 𝐵) Dependent and independent events. Venn diagrams or contingency tables and tree diagrams as aids to solving prob bility problems (where events are not necessarily independent). Generalisation of the fundamental counting principle. Probability problems using the fundamental counting principle Introduction A Venn diagram is a diagram that shows all possible logical relations between finite collections of sets. Venn diagrams were conceived around 1880 by John Venn. Venn diagrams were a way of picturing relationships between different groups of things. They are used to teach elementary set theory, as well as illustrate simple set relationships in probability. A universal set in a Venn diagram represents all the elements in consideration e.g. learners in a class; cars in the school car park; flowers in the flowerbed, etc. Elements in the universal set are then further classified into smaller sets or groups. Worked Examples 1-3: Venn (30 Mins) With your assistance, the facilitator will now demonstrate the use of some of these Venn diagrams. PARTICIPANT HAND-OUT – FET PHASE 95 Activities 1-3: Venn Diagrams Group organisation: Time: Resources: Appendix: Groups of 6 30 min  Participants Handout None In your groups you will: 1. Select a new scribe and a spokesperson for these activities. 2.2.1.1 Use the flipchart and permanent markers and answer the questions as per the activity. Activity 1: Grade 10 The band at Morningside High School has 50 members, and the student council has 20 members. If x members are common to both groups and those in student council only are 15. 1.1 Find the value of x 1.2 Illustrate this information in a Venn diagram 1.3 Find the probability that a learner chosen at random is not in the student council Activity 2: Grade 11 & 12 2.1 How many integers from 1 to 600 are divisible by 2 or by 3? Now find the probability that a random integer from 1 to 600 is divisible by neither 2 nor 3. 2.2 How many integers from 1 to 3500 are divisible by 5 or 7? Now find the probability that a random integer from 1 to 3500 is divisible by neither 5 nor 7 Conclusion Venn diagrams are easy to use and they make information easy to work with. First establish if the events are mutually exclusive or not and then determine the number of elements in the intersection of the two sets. Make sure you don‟t count elements twice, subtracting the number of elements that are in the intersection from the number of elements in that set to get the number of elements that are exclusive to that set. 96 PARTICIPANT HAND-OUT – FET PHASE Sub-topic 3: Dependent and independent events CAPS extraction indicating progression from grades 10-12 Grade 10 Grade 11 Grade 12 Compare the relative frequency of an experimental outcome with the theoretical probability of the outcome. Venn diagrams as an aid to solving probability problems. Mutually exclusive events and complementary events. The identity for any two events A and B: 𝑃 𝐴 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 and 𝐵) Dependent and independent events. Venn diagrams or contingency tables and tree diagrams as aids to solving probability problems (where events are not necessarily independent). Generalisation of the fundamental counting principle. Probability problems using the fundamental counting principle. Introduction If one event does affect the occurrence of the other event, the events are dependent. Two events are independent if the occurrence or non-occurrence of one event has no effect on the likelihood of the occurrence of the other event. For example, tossing two coins is an example of a pair of independent events. Worked Examples 1-2: Dependent and independent events (30 Mins) Allow the facilitator some time to guide and support you through the following worked examples. Don‟t forget to correct him/her if they err. Probability of Independent Events Events A and B are independent events if and only if P(A and B)= P(A)× P(B). Otherwise, A and B are dependent events. PARTICIPANT HAND-OUT – FET PHASE 97 Example 1: Grade 11(C) Suppose that the probability of Kevin coming to a party is 80% and the probability of Judy coming to a party is 95%. Assuming that these events are independent, what is the probability that: 1.1 They both come to the party. 1.2 Kevin comes but Judy does not come to the party. 1.3 They both don‟t come to the party. Solution: 1.1 P(they both come) 0 × 0 1.2 P(Kevin comes judy does not) 0 × 4 0 04 1.3 P(They both don′t come) 20 × 0 0 Example 2: Grade 12(C) The integers 1 through 15 are written on slips of paper and placed into a box. One slip is selected at random and put back into the box, and then another slip is chosen at random. 2.1 What is the probability that the number 8 is selected both times? 2.2 What is the probability that the number 8 is selected exactly once? Solution:  P(8 and 8)= 1 15 × 1 15 1 225  P(8 and not 8 or not 8 and 8)= 1 15 × 14 15 14 15 × 1 15 28 225