How do you identify all asymptotes for #f(x)=(1-x^2)/x#?

1 Answer
Nov 4, 2016

Answer:

The only vertical asymptote is at #x=0#. The only slant asymptote is at #y=-x#. There are no horizontal asymptotes.

Explanation:

An asymptote is a line that approaches the curve but does not meet it at any finite distance. To find vertical asymptotes, find where the denominator is equal to zero (this is where the function divides by zero, and the function does not exist at this point).
Set denominator equal to zero:
#f(x)=(-x^2+1)/x#
#x=0# is the only vertical asymptote.

In this function, the degree in the numerator is greater than the degree in the denominator, so there is a slant asymptote, and not a horizontal asymptote. To find the slant asymptote, we have to divide each term by x.
#f(x)=(-x^2+1)/x#
#f(x)=-x+1/x#
This gives us that the slant asymptote is at y = -x.