Answer
See graph.
Work Step by Step
Given $f(x)=\frac{x^2-4}{x^2+3x+2}=\frac{(x+2)(x-2)}{(x+1)(x+2)}=\frac{x-2}{x+1}, (x\ne-2)$, we can identify: V.A. $x=-1$, H.A. $y=1$, O.A. $none$, hole $(-2,4)$. See graph.
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