### 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: A rational function has a vertical asymptote at any value of $x$ that makes the denominator equal to zero.
To find the possible values of $x$ that do this, I need to factor the denominator and set it equal to zero.
Factoring $x^2+x-20$, I get $(x+5)(x-4)$.
Setting this equal to zero, I get $x+5=0$ or $x-4=0$, which means $x=-5$ or $x=4$.
These are the two values of $x$ that could potentially be vertical asymptotes, but I need to check if they are also zeros of the numerator.
If they are, then they would be holes in the graph, not asymptotes.
To check this, I need to plug in $x=-5$ and $x=4$ into the numerator and see if it equals zero.
Plugging in $x=-5$, I get $(-5)^2-(-5)+c = 25+5+c = 30+c$.
This is zero only when $c=-30$.
Plugging in $x=4$, I get $(4)^2-4+c = 16-4+c = 12+c$.
This is zero only when $c=-12$.
Therefore, the graph of $f(x)$ has exactly one vertical asymptote when either $c=-30$ or $c=-12$, because in either case, one of the potential asymptotes becomes a hole and the other remains an asymptote.