### 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: To find the vertical asymptotes of a rational function, I need to set the denominator equal to zero and solve for $x$.
So I have $x^2+x-20 = 0$, which I can factor as $(x+5)(x-4) = 0$.
This means the graph has vertical asymptotes at $x = -5$ and $x = 4$, unless the numerator also has a zero at the same value, in which case the asymptote is canceled out by a hole in the graph.
To check if the numerator has any zeros, I need to set it equal to zero and solve for $x$.
So I have $x^2-x+c = 0$, which is a quadratic equation that I can solve using the quadratic formula.
The quadratic formula gives me $x = \frac{1 \pm \sqrt{1-4c}}{2}$, which are the zeros of the numerator.
Now, I want to find the values of $c$ that make one of these zeros match one of the vertical asymptotes, which are $x = -5$ and $x = 4$.
So I have two equations to solve: $\frac{1 \pm \sqrt{1-4c}}{2} = -5$ and $\frac{1 \pm \sqrt{1-4c}}{2} = 4$.
I'll start with the first one.
Multiplying both sides by 2, I get $1 \pm \sqrt{1-4c} = -10$.
Subtracting 1 from both sides, I get $\pm \sqrt{1-4c} = -11$.
Squaring both sides, I get $1-4c = 121$.
Solving for $c$, I get $c = -30$.
This is one possible value of $c$ that makes the graph have exactly one vertical asymptote, because it makes the numerator have a zero at $x = -5$, which cancels out the asymptote at that value, but not at $x = 4$.
Now I'll try the second equation.
Multiplying both sides by 2, I get $1 \pm \sqrt{1-4c} = 8$.
Subtracting 1 from both sides, I get $\pm \sqrt{1-4c} = 7$.
Squaring both sides, I get $1-4c = 49$.
Solving for $c$, I get $c = -12$.
This is another possible value of $c$ that makes the graph have exactly one vertical asymptote, because it makes the numerator have a zero at $x = 4$, which cancels out the asymptote at that value, but not at $x = -5$.