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

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Answer 1 · 2016-11-10T13:07:56

Only a horizontal asymptote $y=2$ .

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]}

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