### 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 look at where the denominator is equal to zero, because that's when the function becomes undefined.
So I want to solve the equation $x^2+x-20=0$ for $x$.
This looks like a quadratic equation, so I can try to factor it or use the quadratic formula.
Factoring seems easier, so I look for two numbers that multiply to -20 and add to 1.
Those numbers are 5 and -4, so I can write $x^2+x-20=(x+5)(x-4)$.
Therefore, the denominator is zero when $x=-5$ or $x=4$.
These are the potential vertical asymptotes of the function.
However, the question asks for the values of $c$ that make the graph have exactly one vertical asymptote, not two.
This means that for one of these values of $x$, the numerator must also be zero, so that the function has a hole instead of an asymptote.
So I want to find the values of $c$ that make $x^2-x+c=0$ for either $x=-5$ or $x=4$.
I can plug in each value of $x$ and solve for $c$ separately.
Let me start with $x=-5$.
If $x=-5$, then I get $(-5)^2-(-5)+c=0$.
Simplifying, this becomes $25+5+c=0$, or $c=-30$.
So if $c=-30$, then the function has a hole at $x=-5$ and a vertical asymptote at $x=4$.
Next, I try $x=4$.
Then I get $(4)^2-(4)+c=0$.
Simplifying, this becomes $16-4+c=0$, or $c=-12$.
So if $c=-12$, then the function has a hole at $x=4$ and a vertical asymptote at $x=-5$.
These are the only two possible values of $c$ that make the function have exactly one vertical asymptote, because any other value of $c$ would either make the numerator never zero, or make it zero for both values of $x$.