### Instruction: For what values of the constant $c$ does the graph of $f(x) = \frac{x^2-x+c}{x^2+x-20}$ have exactly one vertical asymptote?

Enter all possible values, separated by commas.### Response: I need to find the values of $c$ that make the denominator of the function $f(x)$ equal to zero for exactly one value of $x$.
To do that, I will factor the denominator and look for its roots.
The denominator is a quadratic expression, so I can use the difference of squares method to factor it: $x^2+x-20 = (x+5)(x-4)$.
Now I can see that the denominator is zero when $x=-5$ or $x=4$, so these are the possible vertical asymptotes of the function.
But I want to have exactly one vertical asymptote, not two, so I need to make one of these roots also a root of the numerator.
That means I need to find a value of $c$ that satisfies either $(-5)^2-(-5)+c=0$ or $(4)^2-4+c=0$.
Simplifying these equations, I get $c=-30$ or $c=-12$.
So these are the two possible values of $c$ that make the graph of $f(x)$ have exactly one vertical asymptote.