Just build it a piece at a time.
Here's a rational function: \(\displaystyle y = \frac{\;\;\;\;\;}{\;\;\;\;\;}\)
"with a hole at (1,2)"
\(\displaystyle y = \frac{x-1}{x-1}\)
We'll worry about the 2 later. Just do the eyeball pieces first.
"x intercept at (-4,0). "
\(\displaystyle y = \frac{(x-1)(x+4)}{x-1}\)
Now, we just need to tune it up a bit. We have two points to find, (1,2) and (0,3/2). But first, since there are quite a few possible solutions, you should pick one. Plot your three points and decidee on a shape. Personally, I like a double vertical asymptote at x = -2 and another zero at (-1,0).
The double asymptote
\(\displaystyle y = \frac{(x-1)(x+4)}{(x-1)(x+2)^2}\)
The second zero
\(\displaystyle y = \frac{(x-1)(x+4)(x+2)}{(x-1)(x+2)^2}\)
The other zero
\(\displaystyle y = \frac{(x-1)(x+4)(x+1)}{(x-1)(x+2)^2}\)
Now, if we define f(x) = y and g(x) = y with (x-1) eliminated from numerator and denominator, we have two more criteria. g(1) = 2 and f(0) = 3/2
Well, we can't just multiply by a constant and get both. We might get lucky and do that. Let's add two parameters and see what happens.
\(\displaystyle y = a\cdot \frac{(x-1)(x+4)(x+1)(x-b)}{(x-1)(x+2)^2}\)
Using the two relations, g(1) = 2 and f(0) = 3/2
This leads to a = 3/10 and b = -5
\(\displaystyle y = \frac{3\cdot(x-1)(x+4)(x+1)(x+5)}{10\cdot(x-1)(x+2)^2}\)
Of course, I added another zero at (-5,0) and gained an oblique asymptote at the same time.
Really, you don't have to make it anywhere near this complicated. I just wanted to show you that you can do ANYTHING you want. It can be fun!!
