TITLE: How to determine parameters $a, b,$ and $d$ so that a rational function models a given graph?
QUESTION [5 upvotes]: I'm trying to solve a function problem. It states:
Determine the values of $a$, $b$, and $d$ so that the rational function $$f(x) = \frac{(x+a)(x-1)(x-b)}{(x-c)(x+d)(x-3)}$$
correctly models this graph:

I've been looking at this for a while and I just can't figure out how I'm supposed to approach it. I think it has something to do with limits and I know they're all going to be integer numbers. Any ideas beyond randomly plugging numbers into Desmos until I get the right graph?

REPLY [3 votes]: For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc. 
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$\frac{(0+2)(0-1)(0-b)}{(0-c)(0+d)(0-3)}.$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.