### Instruction: For what values of the constant $c$ does the graph of $f(x) = \frac{x^2-x+c}{x^2+x-6}$ have exactly one vertical asymptote?

Enter the possible values of $c,$ separated by commas.### Response: To find the vertical asymptotes of a rational function, I need to look at the values of $x$ that make the denominator equal to zero.
So, I want to factor the denominator and set it equal to zero: $x^2+x-6 = (x+3)(x-2) = 0.$
This means the vertical asymptotes are at $x=-3$ and $x=2.$
However, the problem asks for the values of $c$ that make the graph have exactly one vertical asymptote, not two.
This means I need to somehow eliminate one of the asymptotes by making the numerator also equal to zero at that value of $x.$
For example, if I want to eliminate the asymptote at $x=-3,$ I need to find a value of $c$ such that $f(-3) = 0.$
Plugging in $x=-3$ into the numerator, I get $(-3)^2 - (-3) + c = 9 + 3 + c = 12 + c.$
So, I need to choose $c = -12$ to make $f(-3) = 0.$
Similarly, if I want to eliminate the asymptote at $x=2,$ I need to find a value of $c$ such that $f(2) = 0.$
Plugging in $x=2$ into the numerator, I get $(2)^2 - (2) + c = 4 - 2 + c = 2 + c.$
So, I need to choose $c = -2$ to make $f(2) = 0.$