Answer
$$
y =\frac{x^{2}+7 x+12}{x+4}
$$
There are no asymptotes, but there is a hole at $x=-4$.
$x$-intercept: $-3$ the value when $y= 0$.
$y$-intercept: $3$ the value when $x= 0$.
Work Step by Step
$$
y =\frac{x^{2}+7 x+12}{x+4}
$$
To find a asymptote let
$$
\begin{aligned} y &=\frac{x^{2}+7 x+12}{x+4} \\ &=\frac{(x+3)(x+4)}{x+4} \\ &=x+3, x \neq-4 \end{aligned}
$$
There are no asymptotes, but there is a hole at $x=-4$.
$x$-intercept: $-3$ the value when $y= 0$.
$y$-intercept: $3$ the value when $x= 0$.
