Describing Transformations of Absolute Value Functions

The graphs of $f(x)=|x|$ and $g(x)=a|x|$ both have vertices located at the origin, so we know that there haven't been any translations. However, $g(x)$ is upside down and somewhat stretched. The fact that $g(x)$ is upside down means that it has been reflected in the $x$-axis. This means $a$ is a negative number. $$\begin{array}{c}a<0\end{array}$$ Let's assume, temporarily, that $a=\text{-}1.$ In this case, we would only change the sign of the graph's $y$-values, taking us from the graph of $f(x)=|x|$ to the graph of $y=\text{-}|x|,$ shown below.

To get from the graph of $y=\text{-}|x|$ to the graph of $g(x),$ we need to stretch $y=\text{-}|x|$ away from the $x$-axis. Let's look at a point on each graph with the same $x$-values and compare their $y$-values.

The graph of $y=\text{-}|x|$ passes through $(1,\text{-}1)$ and the graph of $g(x)$ passes through $(1,\text{-}4).$ This means that the $y$-values in $y=\text{-}|x|$ have been stretched by a factor of $4.$ $$\begin{array}{c}\underset{\stackrel{y\text{-value}}{y=\text{-}|x|}}{\underbrace{\text{-}1}}\times \underset{\stackrel{\text{stretch}}{\text{factor}}}{\underbrace{4}}=\underset{\stackrel{y\text{-value}}{p(x)=a|x|}}{\underbrace{\text{-}4}}\end{array}$$ Therefore, to get both a reflection in the $x$-axis and a stretch of $4,$ we must have that $a=\text{-}4.$