College Algebra, Chapter 2, 2.2, Section 2.2, Problem 32

Make a table of values and sketch the graph of the equation $y = |4 - x|$. Find the $x$ and $y$ intercepts.

By using the Property of Absolute Value,

$y = |4 - x| \to y = \left\{ \begin{array}{cc}
4 - x & \text{for } x > 0 \\
-(4 - x) & \text{for } x < 0
\end{array} \right.$

$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\text{Let} x & -2 & -1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\
\hline\\
f(x) & -6 & -5 & 3 & 2 & 1 & 0 & 1 & 2 & 3 & 4\\
\hline
\end{array}
$




To solve for $x$ intercept, we set $y = 0$


$
\begin{equation}
\begin{aligned}

0=& 4 - x
\\
\\
x =& 4

\end{aligned}
\end{equation}
$


The $x$ intercept is at $(4,0)$

To solve for the $y$ intercept, we set $x = 0$


$
\begin{equation}
\begin{aligned}

y =& 4 - 0
\\
\\
y =& 4

\end{aligned}
\end{equation}
$


The $y$ intercept is at $(0, 4)$