jmc

Factoring the denominator $x^2+7x-30$, we get $$y=\frac{x-3}{x^2+7x-30}=\frac{x-3}{(x-3)(x+10)}.$$The graph has a vertical asymptote at $x=-10$. There is no vertical asymptote at $x=3$, because the factors of $x-3$ in the numerator and denominator cancel. Therefore, the graph has $\boxed{1}$ vertical asymptote.