Answer
The vertical asymptote will be $x=3$; there will be a hole at $x=-7$.
Work Step by Step
$\begin{align}
& h\left( x \right)=\frac{x+7}{{{x}^{2}}+4x-21} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
So, $p\left( x \right)=x+7$ and $q\left( x \right)={{x}^{2}}+4x-21$.
For the vertical asymptotes, put $q\left( x \right)=0$
$\begin{align}
& q\left( x \right)={{x}^{2}}+4x-21=0 \\
& {{x}^{2}}+4x-21=0 \\
& {{x}^{2}}+7x-3x-21=0 \\
& x\left( x+7 \right)-3\left( x+7 \right)=0 \\
& \left( x+7 \right)\left( x-3 \right)=0
\end{align}$
Therefore, the vertical asymptotes are:
$x=-7,\text{ or }x=3$.
But, after rearranging, $x+7$ from the denominator will be eliminated.
So, the vertical asymptote will be $x=3$.
Now let us find the holes.
$\begin{align}
& h\left( x \right)=\frac{x+7}{{{x}^{2}}+4x-21} \\
& =\frac{x+7}{\left( x+7 \right)\left( x-3 \right)} \\
& =\frac{1}{x-3}
\end{align}$
Here, $x+7$ will be eliminated; when x approaches -7, the polynomial approaches to $-\frac{1}{10}$ but at the point $x=-7$ the polynomial will be discontinuous. So, there will be a hole at $x=-7$
Therefore, the vertical asymptote will be $x=3$; there will be a hole at $x=-7$.
