### Instruction: Find $q(x)$ if the graph of $\frac{x^3-2x^2-5x+3}{q(x)}$ has vertical asymptotes at $2$ and $-2$, no horizontal asymptote, and $q(3) = 15$.### Response: The graph of a rational function $\frac{p(x)}{q(x)}$ has vertical asymptotes at the zeros of $q(x)$, as long as they are not also zeros of $p(x)$.
So, $q(x)$ must have factors of $(x-2)$ and $(x+2)$.
The graph of a rational function $\frac{p(x)}{q(x)}$ has no horizontal asymptote if the degree of $p(x)$ is greater than the degree of $q(x)$.
So, $q(x)$ must have a degree less than $3$.
Combining these two facts, I can conclude that $q(x)$ must have the form $q(x) = a(x-2)(x+2)$, where $a$ is some constant.
To find $a$, I can use the given information that $q(3) = 15$.
Plugging in $3$ for $x$, I get
\begin{align*}
15 &= a(3-2)(3+2)\\
15 &= 5a\\
a &= 3
\end{align*}
So, $q(x) = 3(x-2)(x+2)$.