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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!

A vertical assyptote at x=3 means the equation has /(x-3) in it.

A x intercept of 2 means the equation has a +2 at the end and the rest of the equation is *x.

If we look at y = x/(x-3) + 2, at x=0, y=2 and at x=3, there is a vertical assymptote.

To have a slant assymtote of x+1, we need the equation of the form and keep the assymtote at x=3, the equation would be of the form y = x + 1 + 1/(x-3).

As x goes to ±infinity, this goes to the line x+1 since 1/(x-3) goes to 0,

so that condition is met.

When x=3, 1/(x-3) is undefined, so that condition is met.

To have an x intercept of 2 means when we set y to 0 we get 2 for x.

Note that we could also have this equation in the form y = x + 1 + a/(x-3) for some a

at the point (2,0). Putting this in gives 0 = 2 + 1 + a/-1. Subtracting 3 from both

sides gives -3 = -a. This means a=3.

This means the final equation should be y = x + 1 + 3/(x-3).

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