Graphing Linear Absolute Value Functions I

A linear absolute value function is really a composite of two functions. It is defined by a rule like the following, where the coefficient $a$a can be assumed to be a positive real number.

Particular functions are obtained by substituting appropriate values for the coefficients $a$a and $b$b.

Such a function can be graphed if we know the position of the vertex and the gradients of the two parts of the function. It was shown in another lesson that the shape of the graph is a 'V' with its vertex lying on the horizontal axis. 

Alternatively, we can graph the two linear functions separately and then take just the positive parts of each of them to be the graph of the composite function.

The vertex of the graph is the unique point where the function value is zero. We can find the value of $x$x where the vertex is located by setting either of the separate linear functions in the composite to zero and then solving for $x$x.

Worked Examples

Question 1

Locate the vertex of the function given by $|4x-12|$|4x−12|.

One of the component linear functions of this absolute value function is given by $4x-12$4x−12. We set this to zero. Thus, $4x-12=0$4x−12=0 and therefore $x=3$x=3.

Note that we get the same result if the other linear component is used:

$-(4x-12)$−(4x−12)	$=$=	$0$0
$-4x+12$−4x+12	$=$=	$0$0
$12$12	$=$=	$4x$4x
$3$3	$=$=	$x$x

Having found the position of the vertex, we need to find a point on the graph to the right of the vertex and also a point on the graph to the left. The three points are sufficient for determining the two rays that form the graph.

 

Question 2

Using the same function as before, $|4x-12|$|4x−12|, locate points to the right and left of the vertex.

The vertex is at $x=3$x=3. To the right, we could choose $x=5$x=5 where the function value is $4\times5-12=8$4×5−12=8.

To the left, we could choose $x=-1$x=−1 where the function value is $-(4\times(-1)-12)=16$−(4×(−1)−12)=16.

Now we draw rays from the point $(3,0)$(3,0) passing through $(5,8)$(5,8) and $(-1,16)$(−1,16). Check that the sketch below has the correct features.

 

Another way to obtain a sketch of the two parts of the graph is to note that for $|ax+b|$|ax+b| the gradient of the part to the right of the vertex is $a$a and the gradient of the part to the left of the vertex is $-a$−a. (We are assuming that $a$a is a positive number.)

You should check that the graph above has gradient $4$4 to the right of $x=3$x=3 and $-4$−4 to the left of $x=3$x=3.

 

Question 3

Question 4

Question 5