B Chap02 023-041.qxd 14/2/05 1:04 pm Page 23 CHAPTER 2 C3:Transformations of graphs and the modulus function Learning objectives After studying this chapter, you should be able to: ■ transform simple graphs to produce other graphs ■ understand the effect of composite transformations on equations of curves and describe them geometrically ■ understand what is meant by a modulus function ■ sketch graphs of functions involving modulus functions ■ solve equations and inequalities involving modulus functions. 2.1 Review of simple transformations of graphs Some simple transformations of graphs were introduced in chapter 5 of C2. The basic results are reviewed below. For instance, the graph of y x2 can be transformed into the graph 3 . of y (x 3)2 4 by a translation of 4 In general, a translation of a transforms the graph of b Although you may use a graphics calculator to draw graphs, it is important to see how the graph of one curve can be obtained from the graph of a simpler curve using a sequence of transformations. y y 2 x3 y f(x) into the graph of y f(x a) b. You learnt how to find the equations of new curves after a reflection in one of the coordinate axes. For example, the graph of y 2 x3 is sketched opposite. After reflection in the x-axis, the new curve will have equation y 2 x3. The general result is given below. The graph of y f(x) is transformed into the graph of y f(x) by a reflection in the line y 0 (the x-axis). When the curve y 2 x3 is reflected in the y-axis, the new curve has equation y 2 x3. The graph of y f(x) is transformed into the graph of y f(x) by a reflection in the line x 0 (the y-axis). 2 O x 2 B Chap02 023-041.qxd 24 14/2/05 1:04 pm Page 24 C3: Transformations of graphs and the modulus function The graph of y sin x is transformed into the graph of y 3 sin x by a stretch of scale factor 3 in the y-direction. y y 3 1 x 1 x 3 The general result is: The graph of y f(x) is transformed into the graph of y d f(x) by a stretch of scale factor d in the y-direction. A stretch of scale factor 2 in the x-direction transforms the x graph of y sin x into the graph with equation y sin . 2 y y 1 1 x 1 1 x The graph of y f(x) is transformed into the graph of x y f by a stretch of scale factor c in the x-direction. c Worked example 2.1 Find the equation of the resulting curve when the curve y 2 tan x is transformed by: (a) a reflection in the y-axis, (b) a translation of 3 5 , (c) a stretch of scale factor 0.5 in the x-direction. You are transforming the original curve in each case. Successive transformations will be discussed in the next section. B Chap02 023-041.qxd 14/2/05 1:04 pm Page 25 C3: Transformations of graphs and the modulus function 25 Solution (a) The graph of y f(x) is transformed into the graph of y f(x) by a reflection in the y-axis. Hence the new curve has equation y 2 tan (x). However, since tan(x) tan x, the equation of the new curve can be written as y 2 tan x. a transforms the graph of (b) Recall that a translation of b y f(x) into the graph of y f(x a) b. 2 After translation through the vector 3 5 , the curve y 2 tan x has equation y 2 tan x 5 3 or y 7 tan x . 3 (c) The graph of y f(x) is transformed into the graph of x y f by a stretch of scale factor c in the x-direction. c x Hence y 2 tan x is transformed into y 2 tan 0.5 or y 2 tan 2x. Worked example 2.2 Describe geometrically how the first curve is transformed into the second curve in each of the following cases: (a) y x2, y (x 3)2 2 (b) y cos x, y cos 3x (c) y 2x, y 2x Solution (a) The first curve has been translated through the vector to give the second curve. 3 2 (b) A one-way stretch in the x-direction has taken place. The scale factor is 13. (c) The curve has been reflected in the y-axis (or x 0). EXERCISE 2A 1 Find the equation of the resulting curve after each of the 2 following has been translated through . 1 (b) y x2 4x 5 (a) y x3 (c) y tan x (d) y 3x Notice that x has been divided by 1 to give 3x. 3 B Chap02 023-041.qxd 26 14/2/05 1:04 pm Page 26 C3: Transformations of graphs and the modulus function 2 Find the equation of the resulting curve after each of the following has been stretched in the x-direction by scale factor 12. (a) y x3 (b) y x2 4x 5 (c) y tan x (d) y 3x 3 Describe geometrically how the curve y 1 sin x is transformed into the following curves: (a) y sin x (b) y 1 sin x (c) y 5 sin (x 2) (d) y 4 4 sin x (e) y 1 sin x (f) y 1 sin 5x 4 Describe geometrically how the curve y 2x is transformed into the following curves: x (a) y 25x (b) y 2x 3 (c) y 23 (d) y 2x 7 (e) y 2x 5 Describe a geometrical transformation which maps the graph of y 3x onto: 5x (a) y 3x (b) y 3x 4 (c) y 3 2 (d) y 2 3x (e) y 4 3x (f) y 9x 2.2 Composite transformations You can perform each of the transformations described above in sequence so as to produce a composite transformation. Worked example 2.3 Find the equation of the resulting curve when the following transformations are performed in sequence on the curve with equation y x3: (a) a stretch by a factor 2 in the y-direction, 1 , (b) a translation through 3 (c) a reflection in the x-axis. Solution (a) After a stretch by a factor 2 in the y-direction, the curve y x3 becomes the curve y 2x3. (b) Applying a translation of 1 3 to the curve y 2x the new curve y 2(x 1)3 3. 3 gives B Chap02 023-041.qxd 14/2/05 1:04 pm Page 27 C3: Transformations of graphs and the modulus function (c) Finally the effect of a reflection in the x-axis is to transform y f(x) into the curve y f(x). The curve y 2(x 1)3 3 therefore becomes the curve with equation y 3 2(x 1)3. It is important to perform the transformations in the given order or you will not obtain the correct final equation. 2 Worked example 2.4 Describe geometrically a sequence of transformations that transforms y sin x into y 2 5 sin 3x. Solution The curve y sin x is stretched by scale factor 13 in the x-direction to give the curve y sin 3x. Next the curve y sin 3x is stretched by scale factor 5 in the y-direction to give y 5 sin 3x. 0 to give the final Finally y 5 sin 3x is translated through 2 curve with equation y 2 5 sin 3x. EXERCISE 2B 1 Describe a sequence of transformations that would map the graph of y x3 onto the graph of y 3(x 5)3. 2 Find the resulting curve when the following transformations are applied in sequence to the curve y cos x. (a) a reflection in the x-axis, 0 (b) a translation through , 2 (c) a stretch by a scale factor 3 in the x-direction. 3 Find the equation of the resulting graph when the graph of 1 then reflected in the y-axis. y 3x is translated through 5 4 Find the equation of the new curve when the y sin x is transformed by the following sequence of transformations: , (a) a translation through 0 (b) a reflection in the x-axis, (c) a stretch by a scale factor 4 in the y-direction. 5 Express 2x2 12x 19 in the form 2(x a)2 b. Hence decribe geometrically how the graph of y x2 can be transformed into the graph of y 2x2 12x 19. 27 The two stretches could have been done in any order to map y sin x onto y 5 sin 3x. B Chap02 023-041.qxd 28 14/2/05 1:04 pm Page 28 C3: Transformations of graphs and the modulus function 6 Describe geometrically how the first curve can be transformed into the second curve by a sequence of transformations: (a) y x2, y 4(x 2)2 (b) y x2, y 4 3(x 1)2 (c) y x3, y (2x 1)3 (d) y x3, y (x 3)3 5 x (e) y x4, y (3x 5)4 (f) y x5, y 4 2 3 7 Describe geometrically how the curve y 3x2 5 can be transformed into the curve y x2 by a sequence of transformations. 8 Describe geometrically how the curve y 5 sin(x 3) can be transformed into the curve y sin(x 1) by a sequence of transformations. 9 Describe geometrically how the curve y 3 cos 2x can be transformed into the curve y 5 cos x by a sequence of transformations. 10 Describe geometrically how the curve y 4x 3 can be 2x transformed into the curve y by a sequence of 5 stretches. 11 (a) Describe the geometrical transformation that transforms the graph of y 3x into the graph of 3y x. (b) Find the equations of the new graphs after each of the folowing has been reflected in the line y x: (i) y 3x 2, (ii) y x2, (iii) (x 1)2 y2 4. 12 The graph of y f(x) is reflected in the x-axis and then the y-axis to produce the graph with equation y g(x). (a) Find g(x) in terms of f and x. (b) Describe geometrically the single transformation that maps the graph of y f(x) onto the graph of y g(x). 13 (a) Find the new equation resulting from reflecting the curve y x2 in the line y 1. (b) Describe a sequence of simple geometrical transformations that will map y x2 onto your answer to (a). 14 (a) Find the new equation resulting from reflecting the curve y 2x in the line x 5. (b) Describe a sequence of simple geometrical transformations that will map y 2x onto your answer to (a). B Chap02 023-041.qxd 14/2/05 1:04 pm Page 29 C3: Transformations of graphs and the modulus function 29 15 (a) Describe a sequence of geometrical transformations that will transform the graph of y f(x) into the graph of y 6 f(x). (b) Describe a single geometrical transformation that will transform the graph of y f(x) into the graph of y 6 f(x). 2 (c) Describe a single geometrical transformation that will transform the graph of y f(x) into the graph of y 2p f(x). 16 (a) Describe a sequence of geometrical transformations that will transform the graph of y f(x) into the graph of y f(4 x). (b) Describe a single geometrical transformation that will transform the graph of y f(x) into the graph of y f(4 x). (c) Describe a single geometrical transformation that will transform the graph of y f(x) into the graph of y f(2q x). 2.3 Modulus function The modulus function finds the absolute value of a number. Any negative sign in front of a number is disregarded and a positive answer is returned. Consider the function box below. input Take the modulus Often the key on a calculator which finds the modulus is denoted by ABS since it finds the absolute value of a number. output When the value 3 is input then the output is 3, whereas when 7 is input the output is 7. An input of zero gives an output of zero. y The modulus of x is written as |x| and is usually read as ‘mod x’. By taking a set of values of x it is easy to see that the graph of y |x| would have the appearance of the V shape below. y 3 2 1 3 2 1 O 1 2 3 x O x B Chap02 023-041.qxd 30 14/2/05 1:04 pm Page 30 C3: Transformations of graphs and the modulus function The modulus function is actually defined as follows: x when x 0 |x| x when x 0 Worked example 2.5 Sketch the graph of y |x 3|. Solution Method 1 When x 3, |x 3| x 3 since x 3 0. y 3 O x graph of yx3 3 So the graph for x 3 is the same as the graph of y x 3. However when x 3, x 3 is negative and so |x 3| (x 3). y 3 graph of y3x O 3 x This means that when x 3, the graph of y |x 3| is the same as the graph of y (x 3) 3 x. The graph of y |x 3| is therefore as drawn below. y 4 3 2 1 3 2 1 O 1 2 3 4 5 x B Chap02 023-041.qxd 14/2/05 1:04 pm Page 31 C3: Transformations of graphs and the modulus function 31 Method 2 An alternative approach is to draw the graph of y x 3 and then to reflect the section of the graph that lies below the x-axis in the x-axis. y 2 4 3 2 1 3 2 1 O 1 1 2 3 4 5 x 2 3 4 Worked example 2.6 Sketch the graphs of y (a) y |x 4| for 3 x 3, 8 2 (b) y |sin x| for 0 x 2. 4 Solution 2 (a) Draw the graph of y x2 4. O Now reflect in the x-axis all the parts of the graph which lie below the x-axis. 2 x 4 The resulting graph is that of y |x2 4|. y There are now some ‘sharp’ corners on the graph called cusps. Do not be tempted to smooth these out. 4 2 O 2 x (b) Draw the graph of y sin x for 0 x 2. Reflect in the x-axis the portion of the graph where y is negative so as to produce the graph of y |sin x|. y 1 O π 2π x B Chap02 023-041.qxd 32 14/2/05 1:04 pm Page 32 C3: Transformations of graphs and the modulus function Worked example 2.7 Sketch the graphs of: (a) y 3 |x|, (b) y |x 1| |x 2|. Solution (a) For x 0 , the graph is identical to that of y 3 x. When x 0, y 3. For x 0 , the graph is identical to that of y 3 (x) 3 x. Hence we can sketch the graph of y 3 |x|. Alternatively, the graph of y 3 | x | can be obtained from the graph of y | x | by a reflection in the x-axis followed 0 by a translation of . 3 y 3 3 O 3 x (b) It is necessary to consider three separate intervals in order to sketch y |x 1| |x 2|. Firstly, when x 1, |x 1| (x 1) x 1 and also |x 2| (x 2) x 2. Therefore y |x 1| |x 2| x 1 (x 2) 3 2x. The graph is identical to y 3 2x for x 1. Next, when 1 x 2, |x 1| x 1. But |x 2| (x 2) x 2. Therefore y |x 1| |x 2| x 1 (x 2) 1. The graph is identical to y 1 for 1 x 2. Finally, when x 2 |x 1| x 1 and also |x 2| x 2. Therefore y |x 1| |x 2| x 1 x 2 2x 3. The graph is identical to y 2x 3 for x 2. The graph of y |x 1| |x 2| can now be sketched. The end points of these intervals were not included but serve as useful check points. y 3 When x 1, y 0 1 1. 2 When x 2, y 1 0 1. 1 1 O 1 2 3 x B Chap02 023-041.qxd 14/2/05 1:04 pm Page 33 C3: Transformations of graphs and the modulus function 33 EXERCISE 2C Sketch the graph of each of the following, showing the values of any intercepts on the axes. 1 y |3x| 2 y |x 4| 3 y |3x 5| 4 y |5 2x| 5 y |x| 5 6 y 4 |x| 2 7 y |x 1| 8 y |(x 5)(x 2)| 9 y |x| |x 3| 10 y |x| |x 3| 11 y |(x 1)(x 2)(x 3)| 12 y |x2 5| 4 13 y |x| |x 1| |x 2| 14 y sin|x|, 2 x 2 15 y |cos x|, 2 x 2 16 y |tan x|, 0 x 17 y |cos 3x|, x 18 y 1 |sin 2x|, x 2 2.4 Equations involving modulus functions Often the graphical approach is best since you can see the approximate solutions and how many solutions to expect. Worked example 2.8 Solve the equation |3 2x| x 1. Solution y Method 1 Firstly, sketch the graph of y |3 2x|. This is the V-shaped graph. The sections have gradients 2. 5 Now add the straight line y x 1, which has gradient 1. Hence there are two points of intersection. 3 This means the equation |3 2x| x 1 will have two solutions, and from the graph one of these is less than 1.5 and the other is greater than 1.5. 3 When x , |3 2x| 3 2x. 2 One solution is given by 3 2x x 1 4 ⇒ 3 1 2x x ⇒ x . 3 3 Similarly, when x , |3 2x| (3 2x). 2 The second solution is given by (3 2x) x 1 ⇒ 3 2x x 1 ⇒ 2x x 3 1 ⇒ x 2. 4 The two solutions are x 2, . 3 4 2 1 1 O 1 2 3 4 5 6 x 1 3 This solution is valid since x . 2 3 This solution is valid since x . 2 B Chap02 023-041.qxd 34 14/2/05 1:04 pm Page 34 C3: Transformations of graphs and the modulus function Method 2 Square both sides of the equation |3 2x| x 1. (|3 2x|)2 (x 1)2 ⇒ 9 12x 4x2 x2 2x 1. ⇒ 3x2 10x 8 0 ⇒ (3x 4)(x 2) 0. 4 The two solutions are x 2, . 3 It is always a little dangerous when you square both sides of an equation, since you may produce spurious answers. You should therefore check that these solutions satisfy the original equation. In this case the solutions are valid. Worked example 2.9 Solve the equation |x 2| 3. Solution An approach involving squaring both sides yields (|x 2|)2 (3)2 ⇒ x2 4x 4 9 ⇒ x2 4x 5 0. ⇒ (x 5)(x 1) 0 ⇒ x 5, x 1. Checking each of these values in the original equation shows that neither of the answers are correct solutions. When x 5, |x 2| |5 2| 3. When x 1, |x 2| |(1)2| |3| 3. y 4 3 2 1 2 1 O 1 2 3 4 5 x 1 2 3 The graph of y |x 2| is never negative and so the equation |x 2| 3 has no solutions. Can you spot the flaw in the following argument? For solutions to |x 2| 3, solve x 2 3 ⇒ x 2 3 1. Also solve (x 2) 3 ⇒ x 2 3 ⇒ x 5. Hence the solutions are x 1 and x 5. A sketch can easily reveal when no solutions actually exist since the graphs of y | x 2| and y 3 do not intersect. B Chap02 023-041.qxd 14/2/05 1:04 pm Page 35 C3: Transformations of graphs and the modulus function Worked example 2.10 35 y Solve the equation |x 1| 6x. 2 Solution A sketch showing the graphs of y |x2 1| and y 6x enables you to see that there are two solutions, one in the interval 0 x 1 and the other satisfying x 1. For 0 x 1, you can write |x2 1| (x2 1) x2 1 Hence, solving 36 4 6 x2 1 6x ⇒ 0 x2 6x 1 ⇒ x 2 40 ⇒ x 3 3 10 . 2 But the only valid solution in the interval 0 x 1 is 0.162… . You must reject the solution that is x 3 10 negative. For x 1, you can write |x2 1| x2 1 4 6 36 Hence, solving x2 1 6x ⇒ 0 x2 6x 1 ⇒ x 2 40 ⇒ x 3 3 10 . Once again, rejecting the negative 2 6.162… . solution gives the solution x 3 10 The two solutions are x 3 10 0.162… 6.162… . x 3 10 EXERCISE 2D Solve the following equations. In some cases there are no solutions. 1 |x 3| 2 2 |4 x| 5 3 |x 2| 7 4 |x 1| x 5 |2 x| x 1 6 |2x 3| x 1 7 |2x 3| 2 8 |4 3x| x 9 |x2 2| 3x 10 |2x2 3x| 1 11 |4 3x2| x 12 |x2 2x| 3x 2 13 |x2 x| 2 0 14 |4x2 x| 3 15 |4x2 3| x 2.5 Inequalities involving modulus functions It is advisable to draw a sketch and then to find the critical points when trying to solve inequalities involving modulus functions. 2 1 1 O 1 2 Using the quadratic equation b ac b2 4 formula x . 2a x B Chap02 023-041.qxd 36 14/2/05 1:04 pm Page 36 C3: Transformations of graphs and the modulus function Worked example 2.11 Solve the inequality |x 2| 2x 3. y Solution The graphs y |x 2| and y 2x 3 are drawn opposite and there is a single point of intersection when x is negative. 3 2 Since x 2 for this point of intersection, you can write 1 |x 2| (x 2) 2 x. 2 Solving 2 x 2x 3 gives 1 3x. Hence x 13. This is the critical point and by looking at the graph, the y-value for the graph of y |x 2| is less than the y-value on the graph of y 2x 3 whenever you are to the right-hand side of this critical point. 1 O 1 2 x You can check this solution by taking a test value. For example when x 0, | x 2 | 2 and 2x 3 3. Since 2 3, it gives a check on the solution. The solution is therefore x 13. Worked example 2.12 Solve the inequality |x2 3x| 2. Solution The graph of y x2 3x x(x 3) is a parabola cutting the x-axis at (0, 0) and (3, 0). The graph of y |x2 3x| is sketched opposite together with the straight line y 2. y 4 There are four points of intersection and hence four critical points. 2 Two of these are given by solving x2 3x 2 ⇒ x2 3x 2 0. This equation must be solved using the formula 3 1 1 O b 2 4 8 b ac 3 9 x . Hence x . 2a 2 The approximate values of x are 3.56 … and 0.56… . The other two critical points are given by solving (x2 3x) 2 ⇒ x2 3x 2 0 ⇒ (x 1)(x 2) 0. Therefore x 1, x 2. By considering the graph, the intervals satisfying |x2 3x| 2 are 3 17 3 17 x , 1 x 2, x . 2 2 1 2 3 4 5 The four critical points in ascending order are 3 17 , 2 1, 2, 3 17 . 2 x B Chap02 023-041.qxd 14/2/05 1:04 pm Page 37 C3: Transformations of graphs and the modulus function 37 EXERCISE 2E Solve the following inequalities: 1 |x 5| 2 2 |4 x| 1 3 |x 3| 7 4 |x 2| x 5 |3 x| x 4 6 |2x 3| x 2 7 |2x 5| 2 x 8 |2 5x| x 9 |x2 4| 5x 10 |2x2 x| 1 2 11 |7 3x2| 2x 1 12 |x2 2x| 7x 6 Worked examination question The function f is defined for all real values of x by f(x) |x| |x 3|. (a) For values of x such that x 0 , show that f(x) 3 2x. (b) Write down expressions for f(x) in a form not involving modulus signs for each of the intervals: (i) x 3; (ii) 0 x 3. (c) Sketch the graph of f and write down the equation of its line of symmetry. (d) State the range of f. (e) Solve the equation f(x) 4. (f) Explain whether it is possible to find an inverse of the function f. [A] Solution (a) When x 0 , |x| x and |x 3| (x 3) 3 x. Hence f(x) x 3 x 3 2x. (b) (i) When x 3, |x| x and |x 3| x 3. Hence f(x) x x 3 2x 3. (ii) When 0 x 3, |x| x and |x 3| (x 3) 3 x. Hence f(x) x 3 x 3. (c) The graph consists of the three sections defined above and is sketched opposite. 3 The line of symmetry has equation x . 2 (d) Since the function has least value equal to 3, the range is given by f(x) 3. (e) From the graph, the equation f(x) 4 has two solutions. One of these is when x 0 ⇒ 3 2x 4 ⇒ x 12. The other is when x 3 ⇒ 2x 3 4 ⇒ x 312. (f) Since the graph of f is many-one, it does NOT have an inverse. y 5 4 3 2 1 1 O 1 2 3 4 5 x B Chap02 023-041.qxd 38 14/2/05 1:04 pm Page 38 C3: Transformations of graphs and the modulus function MIXED EXERCISE 1 The function f is defined for all real values of x by f(x) |2x 3| 1. (a) Sketch the graph of y f(x). Indicate the coordinates of the points where the graph crosses the coordinate axes. (b) State the range of y f(x). (c) Find the values of x for which f(x) x. [A] 2 Sketch the graphs of y |2x 3|and y |2x 5| on the same axes. Hence solve the inequality |2x 3| |2x 5|. 3 The function f is defined by f(x) |x 3|, x . (a) Sketch the graph of y f(x). (b) Solve the inequality |x 3| 12x. [A] 4 The functions f and g are defined for all real values of x by f(x) x2 10 and g(x) |x 2|: (a) Show that ff(x) x4 20x2 90. Find all the values of x for which ff(x) 26. (b) Show that gf(x) |x2 12|. Sketch the graph of y gf(x). Hence or otherwise, solve the equation gf(x) x. [A] 5 (a) Determine the two values of x for which |2x 3| |5 x|. y (b) The function f is defined for all real values of x. The graph of y |f(x)| is sketched opposite. Sketch two possible graphs of y f(x) on separate axes. [A] 6 (a) Sketch the graphs of (i)i y x2 6x 5; (ii) y |x2 6x 5|. (b) Calculate the four roots of the equation |x2 6x 5| 3, expressing the irrational solutions in surd form. (c) Using this result and the sketch to (a) (ii), or otherwise, solve the inequality |x2 6x 5| 3. [A] O x B Chap02 023-041.qxd 14/2/05 1:04 pm Page 39 C3: Transformations of graphs and the modulus function 39 7 The diagram shows a sketch of the curve with equation y sin 2x for x . 2 2 y 2 1 π2 O π 2 x 1 (a) Draw on the same diagram sketches of the graphs with equations y |x| and y |sin 2x| for x . 2 2 (b) Hence state the number of times the graph of the curve with equation y |sin 2x| |x| intersects the x-axis in [A] the interval x . 2 2 8 (a) Sketch the graph of y |2x 4|. Indicate the coordinates of the points where the graph meets the coordinate axes. (b) (i) The line y x intersects the graph of y |2x 4| at two points P and Q. Find the x-coordinates of the points P and Q. (ii) Hence solve the inequality |2x 4| x. (c) The graph of y |2x 4| k touches the line y x at only one point. Find the value of the constant k. [A] 9 A function f is defined for all real values of x by f(x) 3 |2x 1|. (a) (i) Sketch the graph of y f(x). Indicate the coordinates of the points where the graph crosses the coordinate axes. (ii) Hence show that the equation f(x) 4 has no real roots. (b) State the range of f. (c) By finding the values of x for which f(x) x, solve the inequality f(x) x. [A] B Chap02 023-041.qxd 40 14/2/05 1:04 pm Page 40 C3: Transformations of graphs and the modulus function Key point summary 1 A translation of b transforms the graph of y f(x) a p23 into the graph of y f(x a) b. 2 The graph of y f(x) is transformed into the graph of y f(x) by a reflection in the line y 0 (the x-axis). p23 3 The graph of y f(x) is transformed into the graph of y f(x) by a reflection in the line x 0 (the y-axis). p23 4 The graph of y f(x) is transformed into the graph of p24 y d f(x) by a stretch of scale factor d in the y-direction. 5 The graph of y f(x) is transformed into the graph of p24 x y f by a stretch of scale factor c in the x-direction. c 6 The modulus function |x| is defined by x when x 0. |x| x when x 0. p30 Test yourself 1 Describe geometrically how the curve: What to review Section 2.1 (a) y x5 is transformed into y (x 1)5 3, (b) y tan x is transformed into y tan 4x. 2 Describe geometrically a sequence of transformations that transforms y x2 into y 3(x 1)2 4. Section 2.2 3 Describe geometrically a sequence of transformations that transforms y sin x into y 3 2 sin 4x. Section 2.2 4 The function f is defined for all values of x by Section 2.3 f(x) |x 5|. (a) Express f(x) in a form not involving modulus signs when x 5. (b) Sketch the graph of y f(x), indicating any values where the graph meets the axes. 5 (a) Sketch the graph of y 4 x2. Section 2.3 (b) Hence sketch the graph of y |4 x2|. (c) State the number of roots of the equation |4 x2| 1. 6 Solve the equation |3x 5| 2 x. Section 2.4 7 Solve the inequality|2x 7| 3 x. Section 2.5 2 1 (a) Translation through 13; (b) Stretch in x-direction with scale factor 41. 2 Translation of 01, followed by stretch in y-direction with scale factor 3, followed by translation of 40. 3 Stretch in x-direction with scale factor 41 and stretch in y-direction with scale factor 2, followed by translation of 30. 4 (a) f(x) 5 x, when x 5; (b) y 5 O 5 (a) 5 x (b) y 4 2 O 2 y 4 2 x O 2 x (c) 4 roots. 6 x 121, 143. 7 131 x 10. Test yourself ANSWERS C3: Transformations of graphs and the modulus function B Chap02 023-041.qxd 14/2/05 1:04 pm 41 Page 41

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