Introduction This chapter focuses on multiple transformations of

Introduction • This chapter focuses on multiple transformations of graphs • It will introduce a new concept, ‘modulus’ • We will also look at how to solve equations involving this…

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: Sketch the graph of: y=x 1) Sketch the graph ignoring the modulus So you can effectively think of a modulus as swapping all negative values to positive ones… y = |x| 2) Reflect the negative part in the x-axis 5 A

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: Sketch the graph of: y = 3 x - 2 1) Sketch the graph ignoring the modulus -2 2/ So you can effectively think of a modulus as swapping all negative values to positive ones… 3 y = |3 x - 2| 2) Reflect the negative part in the x-axis When the modulus is applied to the whole equation, we are changing the output (y-values), hence the reflection in the x-axis… 2 2/ 3 5 A

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: Sketch the graph of: y = x 2 – 3 x - 10 1) Sketch the graph ignoring the modulus -2 5 -10 So you can effectively think of a modulus as swapping all negative values to positive ones… y = |x 2 – 3 x - 10| 2) Reflect the negative part in the x-axis 10 -2 5 5 A

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: Sketch the graph of: 1) Sketch the graph ignoring the modulus 0 y = sinx π/ 2 π So you can effectively think of a modulus as swapping all negative values to positive ones… 3π/ 2 2π y = |sinx| 2) Reflect the negative part in the x-axis 0 π/ 2 π 3π/ 2 2π 5 A

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = f(|x|) The difference here is that we are changing the value of the inputs (x) We are not going to put any negative values into the function Sketch the graph of: y = |x| - 2 1) Sketch the graph for x ≥ 0, ignoring the modulus… 2) Reflect the graph in the yaxis -2 2 -2 The result is that the value we get at x = -3 will be the same as at x = 3 The graph will be reflected in the y-axis… 5 B

Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = f(|x|) The difference here is that we are changing the value of the inputs (x) We are not going to put any negative values into the function Sketch the graph of: 1) Sketch the graph for x ≥ 0, ignoring the modulus… -2 2 2) Reflect the graph in the yaxis y =4|x| - |x|3 The result is that the value we get at x = -3 will be the same as at x = 3 The graph will be reflected in the y-axis… 5 B

Transforming Graphs of Functions Solve the Equation: You need to be able to solve equations involving a modulus Solutions to these equations are the places where the two graphs cross (if each side of the equation is plotted as a graph) 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3 y=3 3/ 4 -3/2 You must pay careful attention to where they cross, on the original graph or the reflected part Try to keep sketches reasonably accurate by working out key points… y = 2 x – 3/2 y = |2 x – 3/2| 2) Alter the graphs to take into account any modulus effects 3 y=3 3/ 2 3/ 4 5 C

Transforming Graphs of Functions Solve the Equation: You need to be able to solve equations involving a modulus Solutions to these equations are the places where the two graphs cross (if each side of the equation is plotted as a graph) You must pay careful attention to where they cross, on the original graph or the reflected part 3) If a solution is on the reflected part, use –f(x) For example point A is on the original blue line, but the reflected red line… Solution A y = |2 x – 3/2| A B 3 y=3 3/ 2 3/ 4 Solution B Try to keep sketches reasonably accurate by working out key points… Using –f(x) for the equation of the red line Solution B is on both original curves, so no modification needed… 5 C

Transforming Graphs of Functions y = |5 x – 2| y = |2 x| You need to be able to solve equations involving a modulus Solve the Equation: 2) Alter the graphs to take into account any modulus effects 2 0 2/ 5 A B Solution A (Reflected Red, Original Blue) y = 5 x – 2 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 0 2/ 5 -2 y = 2 x 3) If a solution is on the reflected part, use –f(x) Solution B (Original Red, Original Blue) 5 C

Transforming Graphs of Functions y = |x 2 – 2 x| You need to be able to solve equations involving a modulus Solve the Equation: 2) Alter the graphs to take into account any modulus effects 0 1/ 8 A 2 y = 1/4 - 2 x Solution A (Original Red, Original Blue) y = x 2 – 2 x 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 1/ 8 0 2 3) If a solution is on the reflected part, use –f(x) or y = 1/4 - 2 x x < 0 at point A so the second solution is the correct one 5 C

Transforming Graphs of Functions y = |x 2 – 2 x| You need to be able to solve equations involving a modulus Solve the Equation: 2) Alter the graphs to take into account any modulus effects B 0 1/ 8 2 y = 1/4 - 2 x Solution B (Reflected Red, Original Blue) y = x 2 – 2 x 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 1/ 8 0 2 y = 1/4 - 2 x 3) If a solution is on the reflected part, use –f(x) or x < 2 at point B so the second solution is the correct one 5 C

Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve f(x + a) is a horizontal translation of –a units f(x) + a is a vertical translation of a units f(ax) is a horizontal stretch of scale factor 1/a af(x) is a vertical stretch of scale factor a -f(x) is a reflection in the x-axis f(-x) is a reflection in the y-axis 5 D

Transforming Graphs of Functions y = (x – 2)2 + 3 2 y = x 2 y = (x – 2) You need to be able to apply multiple transformations to the same curve Sketch the graph of: 7 Build the equation up from y = x 2 Horizontal Translation, 2 units right Vertical Translation, 3 units up y-intercept where x=0 5 D

Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve y = 2/x + 5 y = 1/x 2/ 5 Sketch the graph of: Build the equation up from y = 1/x Horizontal Translation, 5 units left Vertical stretch, scale factor 2 At the yintercept, x = 0 5 D

Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve 2 Sketch the graph of: 1 0 Build the equation up from y = cosx 90 180 270 360 -1 -2 Horizontal ‘stretch’, scale factor 1/2 y = cos 2 x - 1 Vertical translation 1 unit down 5 D

Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve y = 3|x – 1| y=x-1 3 Sketch the graph of: y = |x – 1| 1 1/ 3 Build the equation up from y = x - 1 11 -1 5/ 3 y = 3|x – 1| - 2 Reflect negative values in the x-axis Vertical stretch, scale factor 3 Vertical translation 2 units down You will need to do more than one sketch for these – do not do lots on the same diagram! 5 D

Transforming Graphs of Functions B(6, 8) B(6, 7) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) B(6, 4) O O(0, -1) A(2, -2) A(2, -3) Sketch the graph of: y = 2 f(x) – 1 and state the new coordinates of O, A and B… y = f(x) y = 2 f(x) - 1 Vertical Stretch, scale factor 2 y-values double Vertical translation 1 unit down y-values reduced by 1 5 E

Transforming Graphs of Functions y = f(x + 2) + 2 When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates B(4, 6) B(4, 4) A(0, 1) O(-2, 2) O(-2, 0) O A(0, -1) To the right is the graph of y = f(x) B(6, 4) A(2, -1) y = f(x + 2) Sketch the graph of: y = f(x + 2) + 2 and state the new coordinates of O, A and B… Horizontal translation 2 units left x-values reduced by 2 Vertical translation 2 units up y-values increased by 2 5 E

Transforming Graphs of Functions When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) y = 1/4 f(2 x) B(3, 4) B(6, 4) B(3, 1) A(1, -0. 25) O A(1, -1) A(2, -1) y = f(x) y = f(2 x) Sketch the graph of: y = 1/4 f(2 x) and state the new coordinates of O, A and B… Horizontal stretch, scale factor 1/2 x-values divided by 2 Vertical stretch, scale factor 1/4 y-values divided by 4 5 E

Transforming Graphs of Functions y = f(x - 1) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates B(6, 4) O(1, 0) B(7, 4) A(3, 1) O A(2, -1) A(3, -1) To the right is the graph of y = f(x) B(7, -4) Sketch the graph of: y = -f(x - 1) and state the new coordinates of O, A and B… Horizontal translation 1 unit right x-values increase by 1 Reflection in the x-axis y-values ‘swap sign’ (times -1) 5 E

Summary • We have learnt about modulus graphs • We have seen how sketches help us solve equations involving a modulus • We have also practised multiple transformations and tracked given coordinates