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

1 Answer
Jan 2, 2016

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]}