jmc

By factoring the denominator, the equation becomes $\frac{x-1}{(x-1)(x+7)}$. So the denominator equals $0$ when $x=1$ and $x=-7$. However, since the term $x-1$ also exists in the numerator and is of the same degree as in the denominator, $x=1$ is not a vertical asymptote. Therefore, the equation has only $\boxed{1}$ vertical asymptote at $x=-7$.