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

1 Answer
Sep 26, 2016

Answer:

VA: 1,-1
HA: 0
SA: N/A

Explanation:

Vertical asymptote: Set denominator = 0 but factor first
#x^2-1#=#(x-1)(x+1)#
#(0-1)=-1#
#(0+1)=1#

#1# and #-1# are your vertical asymptotes

Horizontal asymptote: When the x with the biggest power in the numerator is less than the x with the biggest power in the numerator, the horizontal asymptote is #0#.

Since a horizontal asymptote exists, then a slanted asymptote does not exist. It would only exist if the x with the biggest power in the numerator is biggest than the x with the biggest power in the denominator.