How do you find the asymptotes for #y = (2x^2 - 11)/( x^2 + 9)#?

1 Answer
Feb 18, 2016

Answer:

This function has only one horizontal asymptote #y=2#

Explanation:

To find vertical asymptotes of such rational function you have to check if its denominator has any zeroes. Since #x^2+9# has no real roots (it only has 2 complex conjugate roots) the vertical asymptotes do not exist.

To find horizontal asymptotes you have to calculate

#lim _{x->-oo}f(x)# and

#lim _{x->+oo}f(x)#

#lim _{x->-oo}f(x)=#

#lim_{x->-oo}(2x^2-11)/(x^2+9)=lim_{x->-oo}(2-11/x^2)/(1+9/x^2)=2#

The same calculations can be repeated for #lim _{x->+oo}f(x)#. So we can write that #y=2# is a horizontal asymptote for the function #f(x)#