### 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 for the values of $x$ that make the denominator zero, since then the function will be undefined.
So, I want to factor the denominator and see what its roots are: $x^2 + x - 6 = (x+3)(x-2)$.
This means that the function has vertical asymptotes at $x = -3$ and $x = 2$, unless the numerator also has a root at those values, in which case the function has a hole there instead.
So, I need to check if the numerator can be zero when $x = -3$ or $x = 2$.
I can do this by plugging in those values and seeing what value of $c$ would make the numerator zero.
If $x = -3$, then the numerator is $(-3)^2 - (-3) + c = 12 + c$.
This is zero when $c = -12$.
If $x = 2$, then the numerator is $(2)^2 - (2) + c = 2 + c$.
This is zero when $c = -2$.
Therefore, if $c = -12$ or $c = -2$, the function has only one vertical asymptote, since the other one is canceled by a hole.
If $c$ is any other value, the function has two vertical asymptotes.