Question

Write a rational function satisfying the following criteria. vertical Asymptote: x=-1, slant asymptote: y=x+2, zero of the function: x=3
I had f(x)=x^2+3x+2/x+1,
that only works for the asymptotes and not the zero can someone please help me figure how to satisfy all three requirements

Answer 1

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

does not have a vertical asymptote. It just has a hole at x = -1

We want something with (x-3) on top, and (x+1) in the bottom, so

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

has the zero and the vertical asymptote. Note that for large x, f(x) -> 1. So, we can just multiply that by (x+2) to make the slant asymptote:

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

Now, that has two zeros.

If you play around some, you can get rid of the extra zero.

http://www.wolframalpha.com/input/?i=%28x-3%29%28x%2B2%29%2F%28x%2B1%29