Question

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

Answer 1

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)`