Question

How do you find the vertical, horizontal or slant asymptotes for `(2x^2+x-7)/(x^2-1)`?

Answer 1

The vertical asymptotes are `x=-1` and `x=1`
The horizontal asymptote is `y=2`

Explanation:

The denominator `(x^2-1)=(x+1)(x-1)`
As we cannot divide by zero, the vertical asymptotes are `x=-1` and `x=1`
As the degree of the numerator and denominator are the same, we don't have slant asymptotes
limit `(2x^2+x-7)/(x^2-1)=(2x^2)/x^2=2`
`x->+-oo`
So `y=2` is a horizontal asymptote
graph{(2x^2+x-7)/(x^2-1) [-10, 10, -5, 5]}