# Thread: Find Asymptotes of Rational Function

1. ## Find Asymptotes of Rational Function

Find the vertical, horizontal and obligue asymptotes, if any, of the rational function below.

F(x) = (x -1)/(x - x^3)

2. (x-1)/(x (1-x^2) = (x-1)/(x (1-x)(1+x) ) = -1/(x (1+x) )

Take note that since we canceled 1-x, then there is a hole at x = 1. That is, f(x) is not defined at x = 1.

There is no oblique asymptotes.

The vertical asymptote is at x = 0 and x = -1 (These are the points that makes the function undefined).

The horizontal asymptote is zero since the limit as x goes to ± infinity equals 0. In Algebra, you would probably know it by this way:

- Degree of Numerator > Degree of Denominator, then no horizontal asymptote exist.
- Degree of Numerator = Degree of Denominator, then horizontal asymptotes is the y = a/b, where a and b is the coefficients for the leading terms in numerator and denominator, respectively.
- Degree of Numerator < Degree of Denominator, then horizontal asymptote is y = 0

3. Originally Posted by Chop Suey
(x-1)/(x (1-x^2) = (x-1)/(x (1-x)(1+x) ) = -1/(x (1+x) )

Take note that since we canceled 1-x, then there is a hole at x = 1. That is, f(x) is not defined at x = 1.

There is no oblique asymptotes.

The vertical asymptote is at x = 0 and x = -1 (These are the points that makes the function undefined).

The horizontal asymptote is zero since the limit as x goes to ± infinity equals 0. In Algebra, you would probably know it by this way:

- Degree of Numerator > Degree of Denominator, then no horizontal asymptote exist.
- Degree of Numerator = Degree of Denominator, then horizontal asymptotes is the y = a/b, where a and b is the coefficients for the leading terms in numerator and denominator, respectively.
- Degree of Numerator < Degree of Denominator, then horizontal asymptote is y = 0
The graph verifies what you have done! (Good work! )

--Chris

4. ## It's clear....

Both replies are clear.

Thanks a million!