1. Transforming Graphs of Functions

2. 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…

3. Teachings for Exercise 5A

4. 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:
So you can effectively think of a modulus as swapping all negative values to positive ones…
Sketch the graph of:
y = x
1) Sketch the graph ignoring the modulus
y = |x|
2) Reflect the negative part in the x-axis 5A

5. 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:
So you can effectively think of a modulus as swapping all negative values to positive ones…
Sketch the graph of:
y = 3x - 2
1) Sketch the graph ignoring the modulus
2/3 -2
y = |3x - 2|
2) Reflect the negative part in the x-axis 2
2/3
When the modulus is applied to the whole equation, we are changing the output (y-values), hence the reflection in the x-axis… 5A

6. 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:
So you can effectively think of a modulus as swapping all negative values to positive ones…
Sketch the graph of:
y = x2 – 3x - 10
1) Sketch the graph ignoring the modulus -2 5
-10
y = |x2 – 3x - 10| 10
2) Reflect the negative part in the x-axis -2 5 5A

7. 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:
So you can effectively think of a modulus as swapping all negative values to positive ones…
Sketch the graph of:
1) Sketch the graph ignoring the modulus
y = sinx
π/2 π
3π/2 2π
y = |sinx|
2) Reflect the negative part in the x-axis
π/2 π
3π/2 2π 5A

8. Teachings for Exercise 5B

9. Transforming Graphs of Functions
Sketch the graph of:
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
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…
y = |x| - 2
1) Sketch the graph for x ≥ 0, ignoring the modulus… -2 2
2) Reflect the graph in the y-axis -2 5B

10. Transforming Graphs of Functions
Sketch the graph of:
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
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…
1) Sketch the graph for x ≥ 0, ignoring the modulus… -2 2
2) Reflect the graph in the y-axis
y =4|x| - |x|3 5B

11. Teachings for Exercise 5C

12. 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
Try to keep sketches reasonably accurate by working out key points…
y = 2x – 3/2
1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3
y = 3
3/4
-3/2
y = |2x – 3/2|
2) Alter the graphs to take into account any modulus effects 3
y = 3
3/2
3/4 5C

13. 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
Try to keep sketches reasonably accurate by working out key points…
y = |2x – 3/2|
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… A B 3
y = 3
3/2
3/4
Solution A
Solution B
Solution B is on both original curves, so no modification needed…
Using –f(x) for the equation of the red line 5C

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

15. Transforming Graphs of Functions
y = |x2 – 2x|
You need to be able to solve equations involving a modulus
2) Alter the graphs to take into account any modulus effects
1/8 2
Solve the Equation: A
y = 1/4 - 2x
Solution A
(Original Red, Original Blue)
y = x2 – 2x
1) Draw both graphs, ignoring the modulus part. Work out key points if possible
3) If a solution is on the reflected part, use –f(x)
1/8 2 or
y = 1/4 - 2x
x < 0 at point A so the second solution is the correct one 5C

16. Transforming Graphs of Functions
y = |x2 – 2x|
You need to be able to solve equations involving a modulus
2) Alter the graphs to take into account any modulus effects B
1/8 2
Solve the Equation:
y = 1/4 - 2x
Solution B
(Reflected Red, Original Blue)
y = x2 – 2x
1) Draw both graphs, ignoring the modulus part. Work out key points if possible
3) If a solution is on the reflected part, use –f(x)
1/8 2
y = 1/4 - 2x or
x < 2 at point B so the second solution is the correct one 5C

17. Teachings for Exercise 5D

18. 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 5D

19. Transforming Graphs of Functions
y = (x – 2)2 + 3
y = (x – 2)2
y = x2
You need to be able to apply multiple transformations to the same curve
Sketch the graph of:
 Build the equation up from y = x2 7
Horizontal Translation, 2 units right
y-intercept where x = 0
Vertical Translation, 3 units up 5D

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

21. Transforming Graphs of Functions
You need to be able to apply multiple transformations to the same curve
Sketch the graph of:
 Build the equation up from y = cosx 2
y = cos2x 1
y = cosx 90
180
270
360 -1 -2
y = cos2x - 1
Horizontal ‘stretch’, scale factor 1/2
Vertical translation 1 unit down 5D

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

23. Teachings for Exercise 5E

24. 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)
Sketch the graph of:
y = 2f(x) – 1
and state the new coordinates of O, A and B…
B(6,4) O
O(0,-1)
A(2,-1)
y = f(x)
A(2,-2)
A(2,-3)
y = 2f(x)
y = 2f(x) - 1
Vertical Stretch, scale factor 2
 y-values double
Vertical translation 1 unit down
 y-values reduced by 1 5E

25. Transforming Graphs of Functions
y = f(x + 2) + 2
B(4,6)
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)
Sketch the graph of:
y = f(x + 2) + 2
and state the new coordinates of O, A and B…
B(6,4)
B(4,4)
A(0,1)
O(-2,2)
O(-2,0) O
A(0,-1)
A(2,-1)
y = f(x)
y = f(x + 2)
Horizontal translation 2 units left
 x-values reduced by 2
Vertical translation 2 units up
 y-values increased by 2 5E

26. Transforming Graphs of Functions
y = 1/4f(2x)
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)
Sketch the graph of:
y = 1/4f(2x)
and state the new coordinates of O, A and B…
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(2x)
Horizontal stretch, scale factor 1/2
 x-values divided by 2
Vertical stretch, scale factor 1/4
 y-values divided by 4 5E

27. 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
To the right is the graph of y = f(x)
Sketch the graph of:
y = -f(x - 1)
and state the new coordinates of O, A and B…
B(6,4)
B(7,4)
A(3,1)
O(1,0) O
A(2,-1)
A(3,-1)
y = f(x)
B(7,-4)
y = -f(x - 1)
Horizontal translation 1 unit right
 x-values increase by 1
Reflection in the x-axis
 y-values ‘swap sign’ (times -1) 5E

28. 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 co-ordinates