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Hi,

I was wondering if you could give me step by step instructions to write a rational function with a slant/oblique asymptote given the vertical and slant asymptotes and x and y intercepts.

For example, vertical asymptote of 3, slant asymptote of y=x+1, and x intercept at 2.

Thanks so much!

First, let's get all our information out:

f(x) = p(x) / q(x)

This is a rational function, so p(x) and q(x) are polynomials. The conditions are:

1. Vertical asymptote at 3 means q(3)=0 and p(q)≠0.

2. Slant asymptote at x+1 means p(x)/q(x) = x + 1 + p'(x)/q(x) where p' has smaller degree than q. This means you could to divide out p(x)/q(x) to get x+1 plus some remainder (which is p').

3. The x-intercept at 2 means f(2)=0. For this, you need p(2)=0 and q(2)≠0.

So how can you do all of this?

First, conditions 1 & 3 are easy to get with the first guess:

f(x) = (x-2) / (x-3)

That is almost right -- but it does not have the slant asymptote you want. How can you "fix" this so that it does?

Well, what happens if you divide this out? You get:

(x-2) / (x-3) = 1 + 1/(x-3)

It would be better if you had:

f(x) = x + 1 + 1/(x-3)

That doesn't quite work either. Now the asymptotes are right, but f(2) = 2.

The catch is then, you know you want it to look like this:

f(x) = x + 1 + A/(x-3)

where A could be any number. You just need f(2)=0. But:

f(2) = 2 + 1 + A/(2-3)= 3-A

So A=3.

Thus f(x) = x + 1 + 3/(x-3) = (x^2-2x) / (x-3)

Generally speaking, if you want an asymptote at x=K, slant line mx+b, and f(t)=0, then you would have to get:

f(x) = mx + b + A/(x-K)

f(t) = mt + b + A/(t-K)

Solving for A gives: A = (k - t) (b + m t)

So in the end, you have:

f(x) = mx + b + (k-t)(b+mt)/(x-K)

Which simplifies as:

f(x) = (x-t)( m(x+t-k) + b ) / (x-K)

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