A zero at x = 3
A hole when x = 5
A vertical asymptote at x = -1
A horizontal asymptote at y = 3
A y-intercept at y = -2
What I have so far is $$ y=\frac{3(x-3)(x-5)}{(x+1)(x-5)} $$
I figured the y intercept right now is -9 but no matter how I transform it to try and get a y intercept of -2, it just messes up the whole equation. Is there a way to do a vertical shift without changing the entire equation? Thank you!
In order for a rational function to have a zero at $x=3$, a hole at $x=5$, and a vertical asymptote at $x=-1$, it must be of the form
$$(x-3)(x-5)P(x)\over(x+1)(x-5)Q(x)$$
If there is a nonzero horizontal asymptote, $P$ and $Q$ must have the same degree. This still allows a lot of freedom to get a $y$-intercept of $-2$ and a horizontal asymptote of $y=3$. Can you take it from here? (If you're willing to allow additional zeroes and vertical asymptotes, you can let $P$ and $Q$ be linear; if you want to avoid them, you can take $P$ and $Q$ to be quadratics like $x^2+1$.)