How to find the numerator of a rational function given slant asymptote and denominator

I am trying to create a function with slant asymptote y = (7/4)x + 2, vertical asymptotes at x = ±6, and a hole at x = 3. So far what I have managed to create is this:

(x-3)
-----------------
(x-6)(x+6)(x-3)

However, when I try to give it a slant asymptote of (7/4)x + 2, I multiply the fraction by the SA, but I end up with a graph showing me a line. What am I doing wrong, and how can I do this right?

You've got the basics. Now what you need is something that is a cubic which gives the slant asymptote. You need a cubic because you need something greater than the quadratic on the bottom. We can go ahead and disregard the (x - 3)'s and just work with:

p(x) / (x² - 36)

This is actually opposite of what you normally do--normally you are given a rational function and have to use polynomial long division to find the slant asymptote...

The procedure is the same--if you understand polynomial long division--which you probably don't!

Here is what we have:

p(x) / (x² - 36) = ax + (bx² + cx + d)/(x² - 36)
--> we know that a = 7/4 and b = 2

That's all that is important. We can find MANY different c's and d's that will satisfy this. We can assume ANY value for c and d (that is consistent)...let's choose c = d = 0, so that we have:

p(x) / (x² - 36) = 7x/4 + 2x/(x² - 36)
--> now just simplify the left side:

( 7x(x² - 36) - 2x(4) ) / (x² - 36)
-->

(7x³ - 36x - 8x) / (x² - 36) = (7x³ - 44x)/(x² - 36)
-->

Now this alone will give A solution:

(x - 3)(7x³ - 44x) / { (x + 6)(x - 6)(x - 3) }

The only problem is that this will have a zero at x = 0--which is not mentioned.