Find the horizontal asympote, if any, of the rational function.

f(x)=5/x^2+1

Question

Find the horizontal asympote, if any, of the rational function.

f(x)=5/x^2+1

myininaya · Answer

y=0 is a horizontal asy

anonymous · Answer

y=0 is a vertical asymptote.  There's a horizontal one at y=1

myininaya · Answer

just divide x^2 on both top and bottom
5/x^2 goes to 0 as x goes to infinity

myininaya · Answer

$$y=\frac{5}{x^2+1}?$$

anonymous · Answer

$$5/x ^{2}+1$$ is what I read, which explains the discrepancy

myininaya · Answer

did  you mean  y=5/(x^2+1) or y=5/x^2+1
i just assumed he doesnt write parenthesis like the last problem

myininaya · Answer

you could be right

anonymous · Answer

If it reads as myin read it, there are no asymptotes.  If it reads as I read it, there's a vert at x=0 and a horizont at y=1

anonymous · Answer

the original problem doesn't have parenthesis
its written f(x)=5/x^2+1

anonymous · Answer

myininaya wrote it correct

anonymous · Answer

so the answer is "none"

myininaya · Answer

$$y=\frac{5}{x^2+1}$$
you can write that as 5/(x^2+1) 
but if you write it as y=5/x^2+1, you mean$$y=\frac{5}{x^2}+1$$
so we have $$y=\frac{5}{x^2+1}$$ we have y=0 is a horizontal asymptote

anonymous · Answer

yes, the way that you wrote it the first time is the correct way to write the problem.

myininaya · Answer

lol