How do you find the asymptotes of f(x)=x/(x^2+1)? Do they exist?
I already know how to graph it.

Question

How do you find the asymptotes of f(x)=x/(x^2+1)? Do they exist?
I already know how to graph it.

sailorsoju · Answer

i'm kinda shaky on this too, but from my understanding:

to find the horizontal asymptote, if the degree of the denominator is greater than the degree of the numerator, it would be y=0

sailorsoju · Answer

at least that's per my notes... let me find an example

sailorsoju · Answer

$$\frac{ x-3 }{ x^2 -4 }$$ = horizontal asymptote is y=0

sailorsoju · Answer

to find the vertical asymptote, set the denominator = 0... so in my example, (x+2)(x-2) = 0, so x= 2, -2

algtrigcalc · Answer

@sailorsoju what would be the asymptote of y=1/x + x^2?

sailorsoju · Answer

vertical asymptote would be to set the denominator = 0, so in that case that would be x=0... i'm not sure if there's a horizontal asymptote