let $f(x)=\dfrac{n(x)}{d(x)}$ where $n(x)=x-4$ and $d(x)=x+8$

There are 3 regions of note

i) $n(x), d(x)<0$

ii) $n(x)<0, 0 <d(x)$

iii) $0<n(x), 0<d(x)$

$n(x) < 0 \Rightarrow x<4$

$d(x)<0 \Rightarrow x<-8$

so

i) is $x < -8$

ii) is $-8 < x < 4$

iii) is $4 < x$

In region 1 the negative signs cancel each other so we are left with the original expression.

In region 2 we have the negative of the original expression

In region 3 we have no negative signs so we have the original expression.

So we have the original expression on $(x<-8) \cup (4<x)$

and the negative of the original expression on $-8 < x < 4$

Finally note that $x \neq -8$ as this would lead to division by zero, but x=4 is allowed. So we refine our intervals as

Original expression on $(-\infty, -8) \cup (4, \infty)$

Negative expression on $(-8, 4]$

so we have finally

$f(x) = \begin{cases}

\dfrac{x-4}{x+8} &(-\infty, -8)\cup (4,\infty)\\

\dfrac{4-x}{x+8} &(-8,4]\end{cases}$

which is answer #1