1. Transforming Graphs of Functions

2. 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
This means if a y value is calculated as negative you take the positive value instead. So anything below the x axis is reflected upwards 5A

3. 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
When the modulus is applied to the whole equation, we are changing the output (y-values), hence the reflection in the x-axis…
2/3
The red part of the graph is from the equation: y = 3x - 2 5A
The blue part of the graph is from the equation: y = -(3x - 2)

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 = 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
The red part of the graph is from the equation: y = x2 – 3x - 10
The blue part of the graph is from the equation: y = -(x2 – 3x – 10) 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:
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
The red part of the graph is from the equation: y = sinx
The blue part of the graph is from the equation: y = -(sinx) 5A

6. Try exercise 5A

7. 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

8. 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

9. Try Exercise 5B

10. 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

11. 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

12. 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

13. 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

14. 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

15. Exercise 5C