Answer
Vertical asymptote: $x=4$
Hole at $x=-6$
Work Step by Step
Locating Vertical Asymptotes, (page 395) tells us that for $f(x)=\displaystyle \frac{p(x)}{q(x)}$,
if p(x) and q(x) have NO common factors,
and a is a zero of the denominator $q(x)$ ,
then the line $x=a$ is a vertical asymptote.
BUT, if they DO have common factors,
after REDUCING the form of the function's equation,
the number a may not cause the denominator to be zero any more.
in which case there will be a hole at x=a.
The point is that both p(x) and q(x) have to be factored,
and if we can, then we reduce the expression for f(x).
------------------
$h(x)=\displaystyle \frac{x+6}{x^{2}+2x-24}$
$p(x)=x+6\qquad $... fully factored
$q(x)=x^{2}+2x-24$
(find two factors of $-24$ whose sum is +$2$ ...
...found $+6$ and $-4$ )
$q(x)=(x+6)(x-4)$
We can reduce the expression for $g(x):$
$h(x)=\displaystyle \frac{x+6}{x^{2}+2x-24}$
$=\displaystyle \frac{(x+6)}{ (x+6)(x-4)}=\frac{1}{x-4},\quad x\neq-6$
Vertical asymptote: $x=4$
Hole at $x=-6$
