1. THE MODULUS FUNCTION

2. The graph of y = | x | can be found by starting with y = x:
The modulus of x, written | x | means the size of x ( ignoring the sign).
Hence, | – 3 | = 3,
| 7.1 | = 7.1
The graph of y = | x | can be found by starting with y = x: x y
y = x
The modulus of any negative
values of y, will be positive x y
y = | x |
For the graph of y = | f(x) |, sketch
y = f(x) and reflect the negative
values of f(x) in the x-axis.

3. Also consider the graph of y = f ( | x | ).
If x is negative, then | x | will be positive
i.e. f ( | – a | ) = f ( | a | )
So a graph of y = f ( | x | ) will be symmetrical about the y-axis.
For the graph of y = f ( | x | ), sketch
y = f(x) for x ≥ 0 and reflect the graph
in the y-axis.

4. Example 1: Draw a sketch of y = | 2x + 4 |, and hence solve the equation
Firstly, sketch y = 2x + 4
x = 0, y = 4
y = 0, x = –2
For the modulus graph, reflect the negative part in the x-axis
The line segment for x ≥ –2 has equation:
y = 2x + 4
The line segment for x ≤ –2 has equation:
y = –2x – 4
Add the line y = 6: y x 4 –2
For A:
–2x – 4 = 6
– 2x = 10 A B
y = 6
x = –5
For B:
2x + 4 = 6
2x = 2
x = 1

5. Example 2: Shown here is a graph of y = f(x). The graph crosses the
y-axis at the point ( 0, 7 ) and has maximum points at the
points ( 3, 10 ) and ( –4, 5 ). Sketch the graph of y = f ( | x | ).
( 3, 10 ) 7 x y
( –4, 5 )
y = f(x)
The part of the graph for negative
values of x disappears:
( 3, 10 ) 7 x y
( –3, 10 )
The part of the graph for positive
values of x is reflected in the y axis:
Hence, there is a maximum
point at ( –3, 10 ).

6. Hence solve the equation | 2x – 8 | = x + 2.
Example 3: Draw a sketch of y = | 2x – 8 |, and on the same axes sketch
the graph of y = x + 2.
Hence solve the equation | 2x – 8 | = x + 2.
Firstly, sketch y = 2x – 8 –8
x = 0, y =
For the modulus graph, reflect the negative part in the x-axis
y = 0, x = 4
The line segment for x ≥ 4 has equation:
y = 2x – 8
y = 2x – 8 x y
The line segment for x ≤ 4 has equation:
y = –2x + 8
Add the line y = x + 2: x y 4 8 2 B A
y = x + 2
For A:
– 2x + 8 = x + 2
6 = 3x
x = 2
For B:
2x – 8 = x + 2
x = 10

7. a) Find the co-ordinates of the points P and Q and R.
Example 4: Shown here is a sketch of y = f(x), where f(x) = | x + 1 | – 2
a) Find the co-ordinates of the points P and Q and R.
b) Sketch the graph of y = | f(x) | c) Solve | f(x) | = 2x. x y P R Q
a) For P and R, y = 0
| x + 1 | – 2 = 0
| x + 1 | = 2
x = – 3 or 1
i.e. P is ( –3, 0 ), and R is ( 1, 0 )
For Q, the smallest value of | x + 1 | is
x = –1
i.e. Q is ( –1, –2 ) x y –3 1
b) For y = | f(x) |, we have:
( –1, 2 )
Q will move to the point:

8. For the point of intersection: 1 – x = 2x 1 = 3xy
( 1, 0 )
( –3, 0 )
( –1, 2 )
c) We have, y = | f(x) |:
Where f(x) = | x + 1 | – 2 A
y = 2x
Add the line y = 2x
For A, we need the equation of the line through ( –1, 2 ) and ( 1, 0 ).
2 – 0
–1 – 1
The gradient, m = = –1
y – 0 = –1( x – 1 )
y = 1 – x
x = 1 3
For the point of intersection:
1 – x = 2x
1 = 3x

9. For the graph of y = | f(x) |, reflect the
Summary of key points:
For the graph of y = | f(x) |, reflect the
negative values of f(x) in the x-axis.
The part of the graph of y = f(x) which is positive and has
therefore not been reflected in the x-axis, has equation y = f(x).
The part of the graph of y = f(x) which is negative and has
been reflected in the x-axis has equation y = – f(x).
For the graph of y = f ( | x | ), sketch
y = f(x) for x ≥ 0 and reflect the graph
in the y-axis.
This PowerPoint produced by R.Collins ; Updated Feb. 2009