# Vertical and horizontal Asymptotes.

• Dec 29th 2009, 08:36 PM
integral
Vertical and horizontal Asymptotes.
How would you find the asymptotes of:

$\displaystyle y=\frac{1}{x^2-1}$

So far all my attempts have come up with something dealing with n/0

(I know how to do it graphically.)
• Dec 29th 2009, 09:22 PM
bigwave
Quote:

Originally Posted by integral
How would you find the asymptotes of:

$\displaystyle y=\frac{1}{x^2-1}$

So far all my attempts have come up with something dealing with n/0

(I know how to do it graphically.)

when the rational function becomes undefined in this case when $\displaystyle x^2-1$becomes $\displaystyle 0$ then you have a vertical asymptote. so there is vertical asymptotes at $\displaystyle \pm1$

also, since f(x)=0 has no solutions, the graph a horizonal asymptote at y=0
if the degree of the denominator is greater that the degree of the numerator, the x-axis is the horzontal asymptote

also, what appears sometimes to be asymptotes on a graphing calculator is really just a line going from point to point where there really is a hole in the graph
• Dec 29th 2009, 10:05 PM
integral
Ah, I see.

So if an equation can not be set to zero then the point were it is undefined, and y=0 (if the denominator is larger than the numerator) are the asymptotes?

And just so I do not spam with more then one thread.
Could you please explain How to find points of inflections without using a graph? :D
Such as:

$\displaystyle \sqrt[3]{x}$ were the point of inflection would be 0
• Dec 30th 2009, 09:04 AM
bigwave
basically you find inflection points by taking the $\displaystyle f''(x)$ and setting x = 0

in this case $\displaystyle f''(x) = \left(-\frac{2}{9x^{\frac{5}{3}}}\right)$

and is undefined at x=0