vertical asymptotes??

Question

hey everyone
I completely forget how to find the horizontal asymptotes for a function.
the specific function I need asymptotes for is:

9x2 – 36 / x2 - 9
(the x's are squared, it wouldn't let me type it with exponents)
I know how to get vertical asymptotes...but if anyone wants to do a little math or at least remind me how to get horizontal asymptotes, that would be great
thanks!

Answer 1

Remember polynomial long division?
Divide 9x² - 36 by x² - 9, getting 9 + 45/(x²-9). limit as x→∞ is 9 + 45/∞ = 9 + 0 = 9, so y = 9 is the horizontal asymptote.

Answer 2

On a polynomial function, where you have something divided by something else, the vertical asymptotes are where the denominator is zero (after you've simplified the function), making the entire function undefined, and the quotient of the leading terms is the horizontal asymptote. In other words, the horizontal asymptote of (2x^2-1)/(3x^2+5x) is just 2/3, you can ignore the rest of the terms. If you know calculus, it's just the limit as x approaches infinity and negative infinity.

Answer 3

when the leading terms are to the same degree you simply divide the leading coefficients to get the horizontal asymptotes = 9 in this case

Answer 4

for horizontal asymptotes take the limit as x--> infinity, which is infinity in this case, so this equation as none