A graph has a $y$-intercept at $-5$, no $x$-intercepts, and discontinuous points at $(-1,-5)$ and $(3, -5)$. I want to form an equation for this graph, but I don't know how the $y$-intercept relates to the graph of a rational function. Here's all I have so far:
$$y = \frac{(x+1)(x-3)}{(x+1)(x-3)}$$
I realize this isn't even close to being complete; I need some direction on how to finish the equation. Any help would be appreciated. Thanks.
You have handled the $x$-values of the "hole" requirements with placing the monic linear terms in both numerator and denominator. We can worry about the $y$-values later.
You get no $x$-intercepts by having either no factors after removing the duplicated monic linear terms, by having constant factors, or by having only irreducible quadratic factors (such as $x^2+1$). This prevents any value of zero being possible in the numerator (again, after removing the duplicated factors). It is simplest to add no more factors. Your current expression meets this requirement as well.
You get a $y$-intercept of $-5$ by making $y=-5$ when $x=0$. Your expression makes $y=1$. You can get your desired $-5$ by multiplying your expression by $-5$.
So, our final answer is
$$y=-\frac{5(x+1)(x-3)}{(x+1)(x-3)}$$
Graphing this shows a horizontal line with holes at exactly the right places, so we are done. If we needed other $y$-values at the holes we could have played with irreducible quadratic factors, but that is not needed here.