Answer
See graph.
Work Step by Step
Given $f(x)=\frac{(x^2-9)(x+2)}{(x^2-4)(x+3)}=\frac{(x+3)(x-3)(x+2)}{(x+2)(x-2)(x+3)}=\frac{x-3}{x-2}, (x\ne-3,-2)$, we can identify: V.A. $x=2$, H.A. $y=1$, O.A. $none$, holes $(-3,\frac{6}{5}),(-2,\frac{5}{4})$. See graph.