Absolute Value Functions

Lesson

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`.

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

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

Using the same function as before, $|4x-12|$|4`x`−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 $|ax+b|$|`a``x`+`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.

Consider the function $y=\left|x\right|$`y`=|`x`|.

Complete the table.

$x$ `x`$-2$−2 $-1$−1 $0$0 $1$1 $2$2 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Hence sketch a graph of the function.

Loading Graph...State the equation of the axis of symmetry.

State the coordinates of the vertex.

Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

Write the equation and gradient for the two lines that make up the graph of the function.

Equation Gradient $x<0$ `x`<0$y$ `y`$=$=$\editable{}$$\editable{}$ $x>0$

`x`>0$y$ `y`$=$=$\editable{}$$\editable{}$

Consider the function $y=\left|4x-12\right|$`y`=|4`x`−12|.

Determine the coordinates of the $y$

`y`-intercept.Intercept $=$=$\left(\editable{},\editable{}\right)$(,)

State the coordinates of the vertex.

Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

Graph the function.

Loading Graph...

Michael is training for the land speed record and takes his new car for a test drive by driving straight down a closed highway and back. His distance $y$`y` in miles from the end of the highway $x$`x` minutes after he takes off is given by the function $y=\left|5x-40\right|$`y`=|5`x`−40|.

Loading Graph...

How far does he drive in total?

How long does Michael take to reach the end of the highway?

The next day Michael goes for another drive down the same route and his distance from the end of highway is given by $y=\left|4x-40\right|$

`y`=|4`x`−40|. Which of the following statements is true?Loading Graph...Michael is driving at the same speed the next day

AMichael is driving faster the next day

BMichael is driving slower the next day

CMichael is driving at the same speed the next day

AMichael is driving faster the next day

BMichael is driving slower the next day

C

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

Apply graphical methods in solving problems