How do you find the asymptotes for #y=(2x^2 + 3)/(x^2 - 6)#?

1 Answer
Jan 2, 2016

Answer:

VA: #x=+-sqrt6#
HA: #x=2#

Explanation:

Vertical Asymptotes

These will occur when the denominator equals #0#.

#x^2-6=0#
#x^2=6#
#x=+-sqrt6#

Thus, there are vertical asymptotes at #x=-sqrt6# and #x=sqrt6#.

Horizontal Asymptotes

When the degree of the numerator and denominator are equal, like they are here, divide the terms with the greatest degree. In this case, those are #2x^2# and #x^2#.

#(2x^2)/x^2=2#

There is a horizontal asymptote at #x=2#.

graph{(2x^2+3)/(x^2-6) [-19.62, 20.93, -8.83, 11.45]}