Answer
See graph.
Work Step by Step
Given $f(x)=\frac{(x^2-16)(x+3)}{(x^2-9)(x+4)}=\frac{(x+4)(x-4)(x+3)}{(x+3)(x-3)(x+4)}=\frac{x-4}{x-3}, (x\ne-4,-3)$, we can identify: V.A. $x=3$, H.A. $y=1$, O.A. $none$, holes $(-4,\frac{8}{7}),(-3,\frac{7}{6})$. See graph.