• November 30th 2008, 10:13 AM
live_laugh_luv27
If I have the fraction (x+2)/(x^2-9), where do I find the zeros -- the numerator or the denominator?

Would the zeros of the equation be just -2, or would they be 3 and -3 (x+3, x-3 from the denominator)?

• November 30th 2008, 10:30 AM
masters
Quote:

Originally Posted by live_laugh_luv27
If I have the fraction (x+2)/(x^2-9), where do I find the zeros -- the numerator or the denominator?

Would the zeros of the equation be just -2, or would they be 3 and -3 (x+3, x-3 from the denominator)?

I assume you have a rational function here.

$f(x)=\frac{x+2}{(x+3)(x-3)}$

The zero in this case is -2. You have two vertical asymptotes at x=3 and x=-3
• November 30th 2008, 10:39 AM
live_laugh_luv27
Quote:

Originally Posted by masters
I assume you have a rational function here.

$f(x)=\frac{x+2}{(x+3)(x-3)}$

The zero in this case is -2. You have two vertical asymptotes at x=3 and x=-3

That's what I thought...thanks for confirming :)