How do you find the vertical, horizontal or slant asymptotes for $y = \frac{8 x^{2} + x - 2}{x^{2} + x - 72}$ $y = (8 x^2 + x - 2)/(x^2 + x - 72)$ ?

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Answer 1 · 2016-10-23T18:20:06

The vertical asymptotes are $x = 8$ $x=8$ and $x = - 9$ $x=-9$
and the horizontal asymptote is $y = 8$ $y=8$

We start by factorising the denominator
$x^{2} + x - 72 = (x - 8) (x + 9)$ $x^2+x-72=(x-8)(x+9)$

As we cannot divide by so $x \neq 8$ $x!=8$ and $x \neq - 9$ $x!=-9$
So the vertical asymptotes are $x = 8$ $x=8$ and $x = - 9$ $x=-9$

As the degree of the polynomial of the numerator and denominator are the same, there is no slant asymptote.

To find the horizontal asymptote, we find the limit as $x \to \pm \infty$ $x->+-oo$
We take the highest order of the polynomials

$lim y=8/1$
$x->-oo$

And
$lim y=8/1$
$x->oo$
So the horizontal asymptote is $y=8$

How do you find the vertical, horizontal or slant asymptotes for 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)?