Question

Find the vertical asymptotes, if any, of the graph of the rational function.

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

Answer 1

First, please clarify your function:
Do you mean f(x) = x/(x^2 + 1), or do you mean f(x) = x/(x^2) + 1? It helps to use parentheses, which remove ambiguities.

Answer 2

can't you just take the limit as x goes to infinity and - infinity to get te asymptote??\[\lim_{x \rightarrow \infty}\frac{ x }{ x^2+1 }\]
and
\[\lim_{x \rightarrow -\infty}\frac{ x }{ x^2+1 }\]

Answer 3

basically you are finding what the whole function approaches as x gets larger and larger

Answer 4

As you can see in the first limit as x gets larger and larger the denominator is always bigger than the numerator so it will become a smaller and smaller fraction which means the function approaches 0. This means y=0 is one of the asymptotes.

Answer 5

if you take the limit as x approaches -infinity you also will get zero so the only asympttote is y=0

Answer 6

but that asymptote is horizontal not vertical so there are no vertical asymptotes

Answer 7

Refer to the attached plots.

Answer 8

Spectacular graphs, Robert! Thanks. Neat that you also produced two different graphs, one for each possible interpretation of the given function.

It's clear from Robert's graph that there is a horizontal asymptote; it's the line y = 0.

It's moderately clear that the slope of the graph is never infinite, meaning that there is no vertical asymptote.

Those familiar with calculus could find the first derivative of the given function

This derivative is \[y' = \frac{ -x ^{2}+1 }{ (x ^{2}+1)^{2} }.\]

Case 1: horiz. asympt.: as x approaches infinity, y' approaches 0 (which fits the concept of horizontal asymptote).

Case 2: vertical asymptote: There is no real x value that makes the denominator of either the original function or the derivative zero. Thus, there is NO vertical asymptote.