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# 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, $f(x)=|x|.$ Since the absolute value of an expression is never negative, the graph of the function $f(x)=|x|$ will always lie on or above the $x$-axis. Note that $|\text{-} 5| = 5 \quad \text{and} \quad |5|=5.$ This means that the points $(\text{-} 5,5)$ and $(5,5)$ both lie on $f(x)=|x|.$ 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 $y=a|x-h|+k$

where $a,$ $h,$ and $k$ are real numbers and $a\neq0.$
fullscreen
Exercise

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

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

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 $f(x)=|x+3|-1.$

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 $x$- and $y$-intercepts are the points where the graph intersects with the $x$-axis and $y$-axis, respectively. It can be seen that $f(x)$ intersects the $x$-axis at two different points, $(\text{-} 4,0)$ and $(\text{-} 2,0).$ The function intersects the $y$-axis at $(0,2).$ Therefore, the intercepts are $(\text{-} 4,0), (\text{-} 2, 0), \text{and } (0,2).$ Since $f$ 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 $x=\text{-} 3,$ $f(x)$ decreases. Additionally, from $x=\text{-} 3$ to the right side of the graph, $f(x)$ increases. Thus, we can express the increasing and decreasing intervals of $f$ as follows. \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 $f(x)$ can be written as follows. \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 $(\text{-} 3, \text{-} 1).$ Let us now show these features on the graph.