Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Graphing Absolute Value Functions

Just as the shape of the graph of a linear function is unique to that family, absolute value functions have distinct graphs. Below, these graphs and their characteristics will be explored.
Concept

Absolute Value Function

An absolute value function is any function that contains the absolute value of a variable expression. These functions can also be described as any function that is a transformation of the absolute value parent function, f(x)=x. f(x)=|x|. Since the absolute value of an expression is never negative, the graph of the function f(x)=xf(x)=|x| will always lie on or above the xx-axis. Note that -5=5and5=5. |\text{-} 5| = 5 \quad \text{and} \quad |5|=5. This means that the points (-5,5)(\text{-} 5,5) and (5,5)(5,5) both lie on f(x)=x.f(x)=|x|. As it turns out, these points lie directly across from each other. In fact, this symmetry exists for all inverse input values. Thus, absolute value graphs have a distinct V-shape.

Any function belonging to the absolute value function family can be written using the equation y=axh+k y=a|x-h|+k

where a,a, h,h, and kk are real numbers and a0.a\neq0.
fullscreen
Exercise

Graph the absolute value function f(x)=x23f(x)=|x-2|-3 using a table of values.

Show Solution
Solution
To graph the given function using a table of values, we can substitute various xx-values into the rule and solve for the corresponding yy-values. The absolute value of a number is always the positive value of that number. For instance, -2=2.|\text{-} 2| = 2. Let's first calculate the yy-value that corresponds with x=-3.x=\text{-} 3.
f(x)=x23f(x)=|x-2|-3
f(-3)=-323f({\color{#0000FF}{\text{-} 3}})=|{\color{#0000FF}{\text{-} 3}}-2|-3
f(-3)=-53f(\text{-} 3)=|\text{-} 5|-3
f(-3)=53f(\text{-} 3)= 5-3
f(-3)=2f(\text{-} 3)= 2
We have found that f(-3)=2.f(\text{-} 3)= 2. Thus, the point (-3,2)(\text{-} 3,2) lies on the graph. We can find other points on the graph in the same way.
xx x23|x-2|-3 f(x)f(x)
-2{\color{#0000FF}{\text{-} 2}} -223|{\color{#0000FF}{\text{-} 2}}-2|-3 11
-1{\color{#0000FF}{\text{-} 1}} -123|{\color{#0000FF}{\text{-} 1}}-2|-3 00
0{\color{#0000FF}{0}} 023|{\color{#0000FF}{0}}-2|-3 -1\text{-} 1
1{\color{#0000FF}{1}} 123|{\color{#0000FF}{1}}-2|-3 -2\text{-} 2
2{\color{#0000FF}{2}} 223|{\color{#0000FF}{2}}-2|-3 -3\text{-} 3
3{\color{#0000FF}{3}} 323|{\color{#0000FF}{3}}-2|-3 -2\text{-} 2
4{\color{#0000FF}{4}} 423|{\color{#0000FF}{4}}-2|-3 -1\text{-} 1
5{\color{#0000FF}{5}} 523|{\color{#0000FF}{5}}-2|-3 00
6{\color{#0000FF}{6}} 623|{\color{#0000FF}{6}}-2|-3 11
7{\color{#0000FF}{7}} 723|{\color{#0000FF}{7}}-2|-3 22

To draw the graph, we can plot these points, then connect them with a V-shaped curve.

Theory

Describing Key Features of Absolute Value Functions

Key features are certain characteristics of the graphs of functions that are noteworthy. For absolute value functions, the graph's intercepts, the intervals in which the function is increasing or decreasing, the minimum and maximum values, and the end behavior are important.
fullscreen
Exercise

The graph shows the function f(x)=x+31.f(x)=|x+3|-1.

Describe the function's key features including intercepts, the intervals for which it increases and decreases, its minimum and maximum values, and its end behavior. Then, show the features on the graph.

Show Solution
Solution

To begin, we'll describe each of the features. A graph's xx- and yy-intercepts are the points where the graph intersects with the xx-axis and yy-axis, respectively. It can be seen that f(x)f(x) intersects the xx-axis at two different points, (-4,0)(\text{-} 4,0) and (-2,0).(\text{-} 2,0). The function intersects the yy-axis at (0,2).(0,2). Therefore, the intercepts are (-4,0),(-2,0),and (0,2). (\text{-} 4,0), (\text{-} 2, 0), \text{and } (0,2). Since ff is an absolute value function, it has a V-shaped graph. This means the function will have both an increasing and a decreasing interval. Looking from left to right on the graph, it can be seen that from the left side of the graph until x=-3,x=\text{-} 3, f(x)f(x) decreases. Additionally, from x=-3x=\text{-} 3 to the right side of the graph, f(x)f(x) increases. Thus, we can express the increasing and decreasing intervals of ff as follows. Decreasing Interval:-<x<-3Increasing Interval:--3<x<+\begin{aligned} \textbf{Decreasing Interval} &: \quad \text{-} \infty < x < \text{-} 3 \\ \textbf{Increasing Interval} &: \quad \phantom{\text{-}} \text{-} 3 < x < + \infty \end{aligned} Looking at the graph, we can see that both the left end and the right end extend upward. Thus, the end behavior of f(x)f(x) can be written as follows. As x-, f(x)+As x+, f(x)+\begin{aligned} \text{As}\ x \rightarrow \text{-} \infty , && \ f(x) \rightarrow +\infty \\ \text{As}\ x \rightarrow +\infty , && \ f(x) \rightarrow +\infty \end{aligned} Since the function is decreasing in one interval and increasing in another, the function has a minimum. From the graph, we can tell that the minimum is the point (-3,-1).(\text{-} 3, \text{-} 1). Let us now show these features on the graph.

{{ 'mldesktop-placeholder-grade-tab' | message }}
{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!
{{ grade.displayTitle }}
{{ exercise.headTitle }}
{{ 'ml-tooltip-premium-exercise' | message }}
{{ 'ml-tooltip-programming-exercise' | message }} {{ 'course' | message }} {{ exercise.course }}
Test
{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}
{{ 'ml-btn-previous-exercise' | message }} arrow_back {{ 'ml-btn-next-exercise' | message }} arrow_forward