Constructing a Rational Function with Given Asymptotes and Intercepts

[TrueStar]

Homework Statement

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

Homework Equations

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.

 

[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.

 

[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!