How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (2x^2 - 11)/( x^2 + 9)#?

1 Answer
Nov 10, 2016

Answer:

Only a horizontal asymptote #y=2#.

Explanation:

As #y=(2x^2-11)/(x^2+9)=(2-11/x^2)/(1+9/x^2)#

#Lt_(x->+-oo)(2-11/x^2)/(1+9/x^2)=2#

Also observe that the denominator #x^2+9>0#, as #x^2# is always positive.

Hence we have only a horizontal asymptote #y=2#.
graph{(y-(2x^2-11)/(x^2+9))(y-2)=0 [-10, 10, -5, 5]}