jmc

We can factor the denominator to get $$f(x)=\frac{x^2-x+c}{(x-4)(x+5)}.$$Hence, the graph of $f(x)$ has vertical asymptotes at $x=-5$ and $x=4$, unless there is a factor of $x-4$ or $x+5$ in the numerator that cancels out the corresponding factor in the denominator (in this case there will be a hole at that point rather than an asymptote). So, we need to find $c$ such that $x^2-x+c$ has a factor of $x-4$ or $x+5,$ but not both.

That is to say, we need $c$ such that either $4$ or $-5$ is a root. If $x=4$ is a root, we must have $(4{)}^2-4+c=0$ which gives us $c=-12.$ If $-5$ is a root, then we must have $(-5{)}^2-(-5)+c=0,$ or $c=-30.$

Thus, the values that work are $c=\boxed{-12\text{ or }-30}.$