Answer
The vertical asymptote will be $x=4$; there will be a hole at $x=-6$.
Work Step by Step
Write $h\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& h\left( x \right)=\frac{x+6}{{{x}^{2}}+2x-24} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
So, $p\left( x \right)=x+6$ and $q\left( x \right)={{x}^{2}}+2x-24$.
For the vertical asymptotes, put $q\left( x \right)=0$:
$\begin{align}
& q\left( x \right)={{x}^{2}}+2x-24=0 \\
& {{x}^{2}}+2x-24=0 \\
& \left( x+6 \right)\left( x-4 \right)=0
\end{align}$
Therefore, the vertical asymptotes are $x=-6,\text{ or }x=4$.
But, after rearranging, $x+6$ from the denominator will be eliminated.
So, the vertical asymptote will be $x=-4$.
To find the holes, observe the expression:
$\begin{align}
& h\left( x \right)=\frac{x+6}{{{x}^{2}}+2x-24} \\
& =\frac{x+6}{\left( x+6 \right)\left( x-4 \right)} \\
& =\frac{1}{x-4}
\end{align}$
Here, $x+6$ will be eliminated; when x approaches $-6$ , the polynomial approaches to $-\frac{1}{10}$ but at the point $x=-6$ the polynomial will be discontinuous. So, there will be a hole at $x=-6$
Thus, the vertical asymptote will be $x=4$; there will be a hole at $x=-6$.
