jmc

There are factors of $x+1$ and $x+2$ in both the numerator and denominator, and the factors in the denominator cancel the factors in the numerator, so the graph has a hole at $x=-1$ and $x=-2.$

There are two factors of $x$ in the denominator, so there is a vertical asymptote at $x=0.$ There are five factors of $x+3$ in the numerator, and 11 factors of $x+3$ in the denominator, so there is a vertical asymptote at $x=-3.$

Hence, there are $\boxed{2}$ vertical asymptotes.