# Rational Functions Questions

1. Mar 15, 2009

### TrueStar

1. The problem statement, all variables and given/known data

Find an equation of a rational function f that satisfies the given conditions:
vertical asymptotes: x=-3, x=1
horizontal asymptote: y=0
x-intercept: -1; f(0)= -2
hole at x=2

2. Relevant equations

3. The attempt at a solution

Vertical Asymptotes are (x+3) and (x-1). Also (x-2) because of the hole

(x+1) would go in the numerator because it corresponds with the x-intercept. I'm not sure where f(0)= -2 fits in this. (x+2) goes in the numerator because of the hole.

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

The answer looks like 6x^2-6x-12/x^3-7x+6 so I'm obviously missing something, but I don't know what.

Last edited: Mar 15, 2009
2. Mar 15, 2009

### CompuChip

You're almost there. The graph of the formula you wrote down has all the required asymptotes. If you multiply the whole thing by a constant, that doesn't change. So instead of
f(x) = (x+1)(x-2)/(x+3)(x-1)(x-2)
you might equally well take
f(x) = A (x+1)(x-2)/(x+3)(x-1)(x-2).
Plug in x = 0. What should A be to satisfy the last requirement?

When you are satisfied with your answer, multiply out the brackets in the denominator and the numerator. You'll see that it agrees with the answer given.

3. Mar 15, 2009

### TrueStar

I cleaned up my original post, sorry about that.

I figured out what piece I was missing. The horizontal asymptote is zero because there is a higher power in the denominator. The constant in the numerator doesn't change this fact....I incorrectly though the 6 would affect both.

Thank you for pointing that out!