Question

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

Answer 1

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

Explanation:

Let's factorise the denominator

`x^2+x-72=(x+8)(x-9)`

So,

`y=(8x^2+x-2)/(x^2+x-72)=(8x^2+x-2)/((x+8)(x-9))`

The domain of `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]}