Answer
Domain: $x \ne \pm 3$
Vertical asymptotes: $x=3$
Horizontal asymptotes: $y=0$
Work Step by Step
Domain:
$x^2 -9 \ne 0$
$x^2 \ne 9$
$x \ne \pm3$
Identify vertical asymptotes:
$\frac{x+3}{x^2 -9}=\frac{x+3}{(x-3)(x+3)}= \frac {1}{x-3}$
$x=3$
Identify any horizontal asymptotes:
$f(x) \rightarrow 0$ as $x\rightarrow \infty$
$y=0$
