SAT Question 128c75e2: Nonlinear Functions answer and explanation with Desmos solution

It's given that the function $g$ is defined by $g\left(x\right)=\frac{|x|}{a}-14$, where $a<0$. Substituting $15a$ for $x$ in function $g$ yields $g\left(15a\right)=\frac{|15a|}{a}-14$. This function can be rewritten as $g\left(15a\right)=\frac{15|a|}{a}-14$, or $g\left(15a\right)=15\left(\frac{|a|}{a}\right)-14$. Since $a<0$, it follows that $\frac{|a|}{a}=-1$. Substituting $-1$ for $\frac{|a|}{a}$ in $g\left(15a\right)=15\left(\frac{|a|}{a}\right)-14$ yields $g(15a)=15(-1)-14$, or $g(15a)=-29$. Similarly, substituting $7a$ for $x$ in function $g$ yields $g\left(7a\right)=\frac{|7a|}{a}-14$. This function can be rewritten as $g\left(7a\right)=\frac{7|a|}{a}-14$, or $g\left(7a\right)=7\left(\frac{|a|}{a}\right)-14$. Since $a<0$, it again follows that $\frac{|a|}{a}=-1$. Substituting $-1$ for $\frac{|a|}{a}$ in $g\left(7a\right)=7\left(\frac{|a|}{a}\right)-14$ yields $g(7a)=7(-1)-14$, or $g(7a)=-21$. Therefore, $g(15a)=-29$ and $g(7a)=-21$. Thus, the product of $g(15a)$ and $g(7a)$ is $(-29)(-21)$, or $609$.