Answer
$V.A.$ $x=-3$,
$H.A$ $none$,
$O.A.$ $y=x-3$.
Work Step by Step
Given $f(x)=\frac{x^2-1}{x+3}=x-3+\frac{8}{x+3}$, we can identify the following:
vertical asymptote $V.A.$ $x=-3$,
horizontal asymptote $H.A$ $none$,
oblique asymptote $O.A.$ $y=x-3$.
