Calculating the number of Asymptotes?

[NS10]

Ok, so I can never seem to learn how to find the number of asymptotes in an equation. For example, number 74 on TMSCA 13 says: The graph of y=(x^2-2x-1)/(2x-2) has how many asymptotes? For some reason, I thought the answer should be 3, but the answer is actually 2. Obviously, I don't know my asymptote rules well enough. I was thinking there would be one slant, one horizontal, and one vertical? Anybody know what I'm doing wrong? Thanks

[darksaber21]

A graph usually never has both a horizontal asymptote and a slant asymptote. Typically, it is one or the other. Since we can see that there is a slant asymptote from the equation, then we know there isn't a horizontal asymptote. You can prove the fact that there isn't a horizontal asymptote by taking the limit of the function as x approaches infinity and negative infinity, which doesn't exist (as the result is infinity or negative infinity).

[NS10]

Okay, I think I understand that. Can it only have a slant asymptote if the degree of the numerator is 1 greater than the degree of the numerator? For example, what if the degree of the numerator is 2 greater? Or 3? Does it have neither a slant nor a horizontal? Thanks again guys

[darksaber21]

It would only have a slant asymptote if the denominator is one degree less than the numerator. If the degrees differ any further, then there would be the potential for a polynomial asymptote, (quadratic if the denominator is two degree less than the nunerator, and vice-versa), but those types are not usually tested. All you would need to do is divide the equation out, and typically the equation left (if there is a remains a remainder of course) is the equation for the horizontal asymptote.

It is totally possible to have neither. Take the function [unparseable or potentially dangerous latex formula]. Because the equation simplifies, there is a hole at x=1, but not an asymptote.

[NS10]

Thanks, I meant to say if the degree of the numerator is one greater than the degree of the denominator, but looks like you answered my question anyways.