3.06 Review: Graphs and characteristics of absolute value functions

The characteristics of the parent absolute value function

In our lesson on transformations, we were briefly introduced to the parent absolute value function, $f(x)=\left|x\right|$f(x)=|x|. Below is a bit more detail about this parent function.

Table of values:

$x$x	$-2$−2	$-1$−1	$0$0	$1$1	$2$2
$f(x)=\left|x\right|$f(x)=|x|	$2$2	$1$1	$0$0	$1$1	$2$2

Graph:

Key characteristics:

$y$y-intercept: $\left(0,0\right)$(0,0)

$x$x-intercept: $\left(0,0\right)$(0,0)

Vertex: $\left(0,0\right)$(0,0)

Line of symmetry: $x=0$x=0

Slope for $x<0$x<0 (left of vertex): $-1$−1

Slope for $x>0$x>0 (right of vertex): $1$1

Domain: $x$x is any real number, $x\in(-\infty,\infty)$x∈(−∞,∞)

Range: $y\ge0$y≥0, $y\in[0,\infty)$y∈[0,∞)

 

Practice question

Question 1

 

 

Transformations of $f(x)=\left|x\right|$f(x)=|x|

Recall from our lesson on transformations that we can have dilations, reflections and translations done to our parent function.

We will look at transformation of $f(x)=\left|x\right|$f(x)=|x| to give us $g(x)=a\left|b\left(x-h\right)\right|+k$g(x)=a|b(x−h)|+k.

Recall the transformations resulting from each coefficient or constant:

$a$a: vertical dilation (stretch if $a>1$a>1 and compression if $00<a<1) and vertical reflection over $x$x-axis if $a<0$a<0

$b$b: horizontal dilation (compression if $b>1$b>1 and stretch if $00<b<1) and horizontal reflection over $y$y-axis if $b$b<0

$h$h: horizontal translation (right if $h>0$h>0 and left if $h<0$h<0)

$k$k: vertical translation (up if $k>0$k>0 and down if $k<0$k<0)

 

Worked examples

Question 2

Let $f(x)=|x|$f(x)=|x| be defined on the interval $[-3,3]$[−3,3]. Construct a new function $f_1(x)=f(x)+1$f1​(x)=f(x)+1. That is, $f_1(x)=|x|+1$f1​(x)=|x|+1. Plot on the same set of axes and state the domain and range of both $f$f and $f_1(x)$f1​(x).

Think: We are translating $f(x)$f(x) up $1$1 unit to get $f_1(x)$f1​(x). How will this change the key features of the graph?

Do: 

Key Feature	$f(x)$f(x)	$f_1(x)$f1​(x)
$x$x-intercept	$\left(0,0\right)$(0,0)	none
$y$y-intercept	$\left(0,0\right)$(0,0)	$\left(0,1\right)$(0,1)
Vertex	$\left(0,0\right)$(0,0)	$\left(0,1\right)$(0,1)
Domain	$[-3,3]$[−3,3]	$[-3,3]$[−3,3]
Range	$[0,3]$[0,3]	$[1,4]$[1,4]

 

Question 3

Let $f(x)=|x|$f(x)=|x| be defined on the interval $[-3,3]$[−3,3]. A new function $f_2$f2​ is defined by $f_2(x)=-2|x|$f2​(x)=−2|x|. Plot on the same set of axes and state the domain and range of both $f$f and $f_2(x)$f2​(x).

Think: We are vertically dilating $f(x)$f(x) by a factor of $2$2 and reflecting in the x-axis to get $f_2(x)$f2​(x). How will this change the key features of the graph?

Do: 

Key Feature	$f(x)$f(x)	$f_2(x)$f2​(x)
Vertex	$\left(0,0\right)$(0,0)	$\left(0,0\right)$(0,0)
Opening	Upwards	Downwards
Slope for $x<0$x<0	$-1$−1	$2$2
Slope for $x>0$x>0	$1$1	$-2$−2
Domain	$[-3,3]$[−3,3]	$[-3,3]$[−3,3]
Range	$[0,3]$[0,3]	$[-6,0]$[−6,0]

 

Question 4

How would the graphs of $f(x)=2|x|$f(x)=2|x|,  $f_1(x)=|2x|$f1​(x)=|2x| and $f_2(x)=|-2x|$f2​(x)=|−2x| compare?

Think: We can manipulate these algebraically to compare. 

Do: 

$f_1(x)$f1​(x)	$=$=	$\left|2x\right|$|2x|
	$=$=	$\left|2\right|\left|x\right|$|2||x|
	$=$=	$2\left|x\right|$2|x|
	$=$=	$f(x)$f(x)

$f_2(x)$f2​(x)	$=$=	$\left|-2x\right|$|−2x|
	$=$=	$\left|-2\right|\left|x\right|$|−2||x|
	$=$=	$2\left|x\right|$2|x|
	$=$=	$f(x)$f(x)

Reflect: These three functions will all actually have the same graph.

 

Question 5

Graph $g(x)=\left|-3x-2\right|$g(x)=|−3x−2|. State the key features including the domain and range using interval notation.

Think: We can think about this two ways. We can think of this as the graph of $y=-3x-2$y=−3x−2 with all values below the $x$x-axis reflected to become positive. Or we can think $g(x)$g(x) as a transformation of the form $g(x)=a\left|b\left(x-h\right)\right|+k$g(x)=a|b(x−h)|+k. Let's use the second approach.

Do: We need to factor the inside of the absolute value to get it to this form. From there, we can find the key features and graph.

$g(x)$g(x)	$=$=	$\left|-3x-2\right|$|−3x−2|
	$=$=	$\left|-3\left(x+\frac{2}{3}\right)\right|$|−3(x+23​)|
	$=$=	$\left|-3\right|\left|x+\frac{2}{3}\right|$|−3||x+23​|
	$=$=	$3\left|x+\frac{2}{3}\right|$3|x+23​|

So this is the absolute value function by dilated vertically by a stretch factor of $3$3 and translated left $\frac{2}{3}$23​ of a unit.

Key Feature	$g(x)=3\left|x+\frac{2}{3}\right|$g(x)=3|x+23​|
Vertex	$\left(\frac{-2}{3},0\right)$(−23​,0)
Opening	Upwards
Slope for $x<\frac{-2}{3}$x<−23​	$-3$−3
Slope for $x>\frac{-2}{3}$x>−23​	$3$3
Domain	$(-\infty,\infty)$(−∞,∞)
Range	$[0,\infty)$[0,∞)

Reflect: If we had graphed $y=-3x-2$y=−3x−2 and reflected all negative values in the $x$x-axis we would have had a $y$y-intercept of $y=-2$y=−2 become a $y$y-intercept of $y=2$y=2 and a graph of slope $-3$−3 which becomes $-3$−3 and $3$3 respectively on either side of the vertex.

 

Practice questions

Question 6

Question 7

Question 8

Question 9