How do you find the vertical, horizontal or slant asymptotes for #(2x+1)/(x^2-5)#?

1 Answer
Mar 25, 2016

Answer:

Vertical asymptotes at #x=-sqrt5,sqrt5#; horizontal asymptote at #y=0#

Explanation:

Finding vertical asymptotes:

In a rational function such as this one, vertical asymptotes will occur when the denominator of the function equals #0#.

#x^2-5=0" "=>" "x^2=5" "=>" "x=+-sqrt5#

Thus, there are two vertical asymptotes: one at #x=-sqrt5# and the other at #x=sqrt5#.

Finding horizontal asymptotes:

Again, for a rational function, when the degree of the denominator is larger than the degree of the numerator, the function's horizontal asymptote is at #y=0#.

We can graph the function to see if we're on the right track:

graph{(2x+1)/(x^2-5) [-18.09, 17.95, -8.5, 9.52]}

The vertical asymptotes seem to be around or approaching #pmsqrt5approxpm2.24#. The horizontal asymptote is at #y=0#. Since we have a horizontal asymptote, the function does not have a slant asymptote.