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C3 Chapter 5
TRANSFORMING GRAPHS OF FUNCTIONS
THE MODULUS FUNCTION y = |f(x)|
The modulus of a number a, written as |a|, is its positive
numerical value. It is also known as its absolute value.
The modulus of a function f(x) is:
When f(x)  0 , f(x) = f(x)
When f(x) < 0 , f(x) = – f(x)
The general method is to draw the graph of the function
without the modulus sign, then for the part of the graph below
the x-axis, reflect in the x-axis.
You should know the general shapes of this sort of graph:
Linear Graphs:
y
y
x
x
y = 5 – 2x
y = 2x + 3
Quadratic Graphs:
y
y
x
2
y = x – 6
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x
2
y = 10 – x
1
C3 Chapter 5
Reciprocal Graphs:
y
y
x
x
y = -2/x
y = 2/x
Cubic Graphs:
y
y
x
x
3
3
y = – x
y = x
y
y
x
3
2
y = x – 2x + 8x
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x
3
2
y = – x + 2x – 8x
2
C3 Chapter 5
Example 1
Sketch the graph of y = |x – 1|
First sketch the graph of y = x – 1
y
4
2
-4
-2
0
2
4
x
-2
-4
Now for the part of the graph below the x-axis, reflect in the
x-axis.
y
4
2
-4
-2
0
2
4
x
-2
-4
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3
C3 Chapter 5
Example 2
2
Sketch the graph of y = x + x – 12 
2
y = x + x – 12  = (x + 4)(x – 3) 
First draw the graph of y = (x + 4)(x – 3)
y
12
8
4
-6
-4
-2
0
-4
2
4
x
-8
-12
Now reflect the part of the graph below the x-axis in
the x-axis.
y
12
8
4
-6
-4
-2
0
-4
2
4
x
-8
-12
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4
C3 Chapter 5
Example 3
Sketch the graph of y = |cos x| for x between 0º and 720º
First draw the graph of y = cos x
y
2
0
180
360
540
720 x
-2
Now reflect the parts of the graph below the x-axis in
the x-axis.
y
2
0
180
360
540
720 x
-2
Ex 5A p 57
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5
C3 Chapter 5
THE FUNCTION y = f(|x|)
Since the value of y at x = a is the same as the value of y
at x = -a, it follows that the y-axis will be a line of symmetry.
Sketch the graph of y = f(x) for x ≥ 0 and then reflect in the
y-axis.
Example
Sketch the graph of y = 2|x| - 1
First draw the graph of y = 2x - 1 for x ≥ 0
y
6
4
2
-4
-2
0
-2
2
4
x
2
4
x
Now reflect this in the y-axis
y
6
4
2
-4
-2
0
-2
Ex 5B p 59
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6
C3 Chapter 5
SOLVING EQUATIONS INVOLVING A MODULUS
Use a sketch to locate the roots and then solve algebraically
using – f(x) for reflected parts of functions.
Example
Solve the equation |4x – 10| = |x|
Sketch the graphs of the two functions:
y
4
A
B
2
-4
0
-2
2
4
x
The intersection at A is on the original part of both graphs
At A,
4x – 10 = x
3x = 10
x =
10
3
 y =
10
3
The intersection at B is on the original graph of y = 2x but on
the reflected graph of y = 4x – 10
At B,
– (4x – 10) = x
– 4x + 10 = x
5x = 10  x = 2  y = 2
Therefore, solutions are:
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x = 2 or
10
3
Ex 5C p 62
7
C3 Chapter 5
COMBINATIONS OF TRANSFORMATIONS
TRANSFORMING GRAPHS SUMMARY
Starting with the graph of y = f(x)
y=f(x)+a
y=f(x+a)
y =af(x)
y = f(ax)
a
Translation a in y
direction
a
Translation –a in x
direction
a
a
Scaling, factor a in y
direction
(stretching)
Scaling, factor 1/a
in x direction
(compressing)
y = -f(x)
Reflection in x axis
y= f(-x)
Reflection in y axis
y = f-1(x)
Reflection in y = x
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8
C3 Chapter 5
Example
Sketch the graph of y = 2|x + 1| - 4
First sketch y = |x|
y
4
2
-4
-2
0
2
4
x
4
x
Now sketch y = |x + 1| by translating 1 unit ←
y
4
2
-4
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-2
0
2
9
C3 Chapter 5
Now sketch y = 2|x + 1| by stretching vertically factor 2
y
4
2
-4
-2
0
2
4
x
Now sketch y = 2|x + 1| - 4 by translating 4 units ↓
y
4
2
-4
-2
0
2
4
x
-2
-4
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10
C3 Chapter 5
Sketch the graph of y = |ln x| + 1
Example
First sketch the graph of y = ln x
y
2
0
-2
2
4
x
4
5 x
-2
-4
Now sketch the graph of y = |ln x|
y
4
2
-2
-1
0
1
2
3
Now sketch the graph of y = |ln x| + 1 by translating 1 unit ↑
y
4
2
Ex 5D p 67
-2
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-1
0
1
2
3
4
5 x
11
C3 Chapter 5
FURTHER SKETCHING OF TRANSFORMATIONS
Build up “one step at a time”
Example
The diagram shows the graph of y = f(x). The graph goes
through O and the points A(2, -2) & B(4, 12). Sketch the graph
of y = 3f(½x + 1) and find the coordinates of the images of O,
y
A and B.
15
B(4,12)
10
5
-4
0
-2
2
6
x
A(2,2)
-5
The graph of f(½x):
4
y
15
B'(8,12)
10
5
-4
-2
0
-5
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2
4
6
8
x
A'(4,2)
12
C3 Chapter 5
The graph of y = f(½x + 1) moves 1 unit ←
y
15
B''(7,12)
10
5
O''(1,0)
-4
-2
0
2
4
6
8
x
A''(3,2)
-5
The graph of y = 3f(½x + 1) stretches 3 times vertically:
y
36
B'''(7,36)
32
28
24
20
16
12
8
O'''(1,0)
-4
-2
4
0
-4
-8
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2
A'''(3,6)
4
6
8
x
Ex 5E p 69
Mixed Ex 5F p 70
13
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