### Instruction: Find $q(x)$ if the graph of $\frac{3x^3-x^2-10x}{q(x)}$ has a hole at $x=2$, a vertical asymptote at $x=-1$, no horizontal asymptote, and $q(1) = -6$.### Response: I want to find a polynomial $q(x)$ that makes the given rational function have the specified features.
A hole at $x=2$ means that $q(2) = 0$ and the factor $(x-2)$ cancels out in the numerator and denominator.
So I can write $q(x) = (x-2) r(x)$ for some polynomial $r(x)$.
A vertical asymptote at $x=-1$ means that $q(-1) = 0$ and the factor $(x+1)$ does not cancel out in the numerator and denominator.
So I can write $q(x) = (x-2) (x+1) s(x)$ for some polynomial $s(x)$.
No horizontal asymptote means that the degree of the numerator is greater than the degree of the denominator.
Since the degree of the numerator is 3, the degree of the denominator must be less than 3.
So $s(x)$ must be a constant, say $s(x) = k$.
So far, I have $q(x) = (x-2) (x+1) k$.
To find $k$, I can use the given value of $q(1) = -6$.
Plugging in $x=1$, I get
\begin{align*}
-6 &= q(1) \\
-6 &= (1-2) (1+1) k \\
-6 &= -2 k \\
k &= 3
\end{align*}
Therefore, $q(x) = (x-2) (x+1) 3$.