CHAPTER 1 The Modulus Function
Example 7 Sketch the graph of y = | 2x β 3 |. Hence, determine the values of x for which | 2x β 3 | = x. Solution 3 . 2 3 β(2x β 3), if x < . 2
| 2x β 3 | =
W
(2x β 3), if x β₯
The values of x which satisfy the equation | 2x β 3 | = x are the x-coordinates of the points of intersection between the graphs of
IE
y = | 2x β 3 | and y = x.
V
y
The gradient of y = x is less than the gradient of y = 2x β 3. Thus, the line
E R
P
A 0
At point A, y = x and y = β(2x β 3). x = β(2x β 3) =1 At point B, y = x and y = 2x β 3. x = 2x β 3 =3
6
B
3
l
y = x will intersect the line segment y = 2x β 3.
y = |2x β 3| y=x
x 3 2