Graphing Absolute Value Functions

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 $x=\text{-}3.$ 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.

$x$	$|x-2|-3$	$f(x)$
$\text{-}2$	$|\text{-}2-2|-3$	$1$
$\text{-}1$	$|\text{-}1-2|-3$	$0$
$0$	$|0-2|-3$	$\text{-}1$
$1$	$|1-2|-3$	$\text{-}2$
$2$	$|2-2|-3$	$\text{-}3$
$3$	$|3-2|-3$	$\text{-}2$
$4$	$|4-2|-3$	$\text{-}1$
$5$	$|5-2|-3$	$0$
$6$	$|6-2|-3$	$1$
$7$	$|7-2|-3$	$2$

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

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. $$\begin{array}{rl}\text{Decreasing Interval}& :\text{-}\infty <x<\text{-}3\\ \text{Increasing Interval}& :\text{-}3<x<+\infty \end{array}$$ 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. $$\begin{array}{rlr}\text{As}x\to \text{-}\infty ,& & f(x)\to +\infty \\ \text{As}x\to +\infty ,& & f(x)\to +\infty \end{array}$$ 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.