Let \(x\) represent the number of miles on the highway and \(f (x)\) the number of miles from the mall. For \(x\)-values less than \(15\text{,}\) the graph is a straight line with slope \(-1\) and \(y\)-intercept at \((0, 15)\text{,}\) so its equation is \(y = -x + 15\text{.}\) Thus,
\begin{equation*}
f (x) = -x + 15~~~ \text{ when } ~~~0 \le x \lt 15
\end{equation*}
On the other hand, when \(x \ge 15\text{,}\) the graph of \(f\) is a straight line with slope \(1\) that passes through the point \((15, 0)\text{.}\) The point-slope form of this line is
\begin{equation*}
y = 0 + 1(x - 15)
\end{equation*}
so \(y = x - 15\text{.}\) Thus,
\begin{equation*}
f (x) = x - 15~~~ \text{ when }~~~ x \ge 15
\end{equation*}
Combining the two pieces, we obtain
\begin{equation*}
f (x) =
\begin{cases}
-x + 15~~~ \amp \text{when }~~~ 0\le x\lt 15\\
x - 15~~~ \amp \text{when }~~~ x\ge 15
\end{cases}
\end{equation*}
The graph of \(f (x)\) is a part of the graph of \(y = \abs{x - 15}\text{.}\) If we think of the highway as a portion of the real number line, with Marlene’s on-ramp located at the origin, then the outlet mall is located at \(15\text{.}\) Marlene’s coordinate as she drives along the highway is \(x\text{,}\) and the distance from Marlene to the mall is given by \(f (x) = \abs{x - 15}\text{.}\)