How do you find the Vertical, Horizontal, and Oblique Asymptote given $y = \frac{8 x^{2} + x - 2}{x^{2} + x - 72}$ $y = (8 x^2 + x - 2)/(x^2 + x - 72)$ ?

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Answer 1 · 2016-11-28T07:37:06

The vertical asymptotes are $x = - 8$ $x=-8$ and $x = 9$ $x=9$
No oblique asymptote
The horizontal asymptote is $y = 8$ $y=8$

Let's factorise the denominator

$x^{2} + x - 72 = (x + 8) (x - 9)$ $x^2+x-72=(x+8)(x-9)$

So,

$y = \frac{8 x^{2} + x - 2}{x^{2} + x - 72} = \frac{8 x^{2} + x - 2}{(x + 8) (x - 9)}$ $y=(8x^2+x-2)/(x^2+x-72)=(8x^2+x-2)/((x+8)(x-9))$

The domain of $y$ $y$ is $D_y=RR-{-8,+9}$

As we cannot divide by $0$ , $x!=-8$ and $x!=9$

The vertical asymptotes are $x=-8$ and $x=9$

As the degree of the numerator is $=$ to the degree of the denominator, there are no oblique asymptotes.

To calculate the limits of $x->+-oo$ , we take the terms of highest degree in the numerator and the denominator.

$lim_(x->+-oo)y=lim_(x->+-oo)(8x^2)/x^2=8$

So, the horizontal asymptote is $y=8$

graph{(y-(8x^2+x-2)/(x^2+x-72))(y-8)=0 [-58.5, 58.55, -29.3, 29.23]}

How do you find the Vertical, Horizontal, and Oblique Asymptote given y = (8 x^2 + x - 2)/(x^2 + x - 72)y=8x2+x−2x2+x−72y = (8 x^2 + x - 2)/(x^2 + x - 72)?