By 'identifying characteristics of absolute value functions', we mean recognising where critical points occur at which the gradient changes abruptly, and distinguishing regions where the function is increasing or decreasing.

A function of the form $y(x)=a+|x-b|$y(x)=a+|x−b| must have a critical point at $x=b$x=b. This is the point at which the expression $x-b$x−b changes in value from negative to positive.

When $xx<b, $x-b$x−b is negative and so the function in this region can be written $y(x)=a-x+b$y(x)=a−x+b, or simply $y(x)=-x+a+b$y(x)=−x+a+b. This is a linear function with gradient $-1$−1.

On the other hand, when $x>b$x>b, $x-b$x−b is positive and in this part of the domain the function can be written $y(x)=x+a-b$y(x)=x+a−b. This is a line with gradient $1$1.

These two lines meet at a vertex, the critical point $x=b$x=b. By substituting this value of $x$x into either of the linear expressions, we see that this corresponds to the function value $y(b)=a$y(b)=a. That is, the vertex is at the point $(b,a)$(b,a).

Example 1

Locate and describe the critical point for the function given by $y(x)=2+|3x-1|$y(x)=2+|3x−1|.

The critical point is at $x=\frac{1}{3}$x=13 since this is where $|3x-1|=0$|3x−1|=0 and $3x-1$3x−1 changes sign there. Now, $y\left(\frac{1}{3}\right)=2$y(13)=2. So, the vertex is at the point $\left(\frac{1}{3},2\right)$(13,2).

To the left of $x=\frac{1}{3}$x=13, the function can be written $y(x)=3-3x$y(x)=3−3x. So, the gradient in this part of the domain is $-3$−3. To the right of $x=\frac{1}{3}$x=13, the function has the form $y(x)=3x+1$y(x)=3x+1 so that the gradient there is $3$3.

Example 2

Locate and describe the critical point for the function given by $y(x)=3+|-2x+1|$y(x)=3+|−2x+1|.

This time, the coefficient of $x$x inside the absolute value bracket is negative. We apply the same reasoning as before.

The critical value is at $x=\frac{1}{2}$x=12. To the left, the function can be written $y(x)=4-2x$y(x)=4−2x, and to the right of $x=\frac{1}{2}$x=12, the function is $y(x)=2x+2$y(x)=2x+2. Thus, the gradient to the left of the critical point is $-2$−2 and to the right it is $2$2. Since $y\left(\frac{1}{2}\right)=3$y(12)=3, we see that the vertex is at the point $\left(\frac{1}{2},3\right)$(12,3).

Comparison of Examples $1$1 and $2$2 shows that changing the sign of the coefficient of $x$x inside the absolute value bracket causes the shape of the graph to be inverted. If the coefficient is positive, the gradients go from negative to positive. If the coefficient is negative, the gradients are arranged in the opposite way.

The following applet allows exploration of the effects of changing various coefficients in the linear, quadratic and cubic cases.

Created with Geogebra

What general statement can you make about the connection between the original curve and the absolute value function associated with it?

How would you describe them?

How can you use the characteristics of the original function, to help you construct the graph of it's absolute value?

Worked examples

Question 1

Consider the function $f\left(x\right)=\left|x\right|$f(x)=|x| that has been graphed. Notice that it opens upwards.

Loading Graph...

What is the gradient of the function for $x>0$x>0?
What is the gradient of the function for $x<0$x<0?
The graph below shows the graph of $y$y that results from reflecting $f\left(x\right)$f(x) across the $x$x-axis. State the equation of $y$y.

Loading Graph...
Select all the correct statements.
A downward absolute value function goes from decreasing to increasing.
A
An upward absolute value function goes from decreasing to increasing.
B
An upward absolute value function goes from increasing to decreasing.
C
A downward absolute value function goes from increasing to decreasing.
D

Question 2

Consider the function $y=\left|x\right|-5$y=|x|−5.

Does the graph of the function open upwards or downwards?
$Downwards$ .
A
$Upwards$ .
B
State the coordinate of the vertex.
State the equation of the line of symmetry.
Which of the following functions would have narrower graphs than $y=\left|x\right|-5$y=|x|−5?
$y=\frac{\left|x\right|}{2}-5$y=|x|2−5
A
$y=\left|x-5\right|$y=|x−5|
B
$y=\left|x\right|-5$y=|x|−5
C
$y=-4\left|x\right|-5$y=−4|x|−5
D

Question 3

Consider the function $y=\left|6-2x\right|-5$y=|6−2x|−5.

Does the function have a minimum $y$y value or a maximum $y$y value?
Maximum $y$y value.
A
Minimum $y$y value.
B
State the coordinates of the vertex.
Which of the following functions have a vertex that lies above the vertex of $y=\left|6-2x\right|-5$y=|6−2x|−5? Select all the correct options.
$y=\left|x-2\right|$y=|x−2|
A
$y=\left|x+5\right|+6$y=|x+5|+6
B
$y=-5\left|6-2x\right|-5$y=−5|6−2x|−5
C
$y=-4-\left|x-6\right|$y=−4−|x−6|
D

Outcomes

M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

91257

Apply graphical methods in solving problems

Identify Characteristics of Absolute Value Functions II