### 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 where the denominator is equal to zero.
So I want to factor the denominator of $f(x)$ and see what values of $x$ make it zero.
The denominator $x^2+x-6$ can be factored as $(x+3)(x-2),$ so the possible vertical asymptotes are $x=-3$ and $x=2.$
But the problem asks for the values of $c$ that make the graph have exactly one vertical asymptote, not two.
That means I need to eliminate one of the possible asymptotes by making the numerator also zero at that value of $x.$
If I plug in $x=-3$ into the numerator, I get $(-3)^2-(-3)+c = 9+3+c = 12+c.$
If I plug in $x=2$ into the numerator, I get $(2)^2-(2)+c = 4-2+c = 2+c.$
So to make the numerator zero at $x=-3,$ I need to have $c=-12,$ and to make the numerator zero at $x=2,$ I need to have $c=-2.$
If $c=-12,$ then the graph has a vertical asymptote at $x=2$ and a hole at $x=-3.$
If $c=-2,$ then the graph has a vertical asymptote at $x=-3$ and a hole at $x=2.$
In either case, the graph has exactly one vertical asymptote.
Therefore, the possible values of $c$ are $-12$ and $-2.$