Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)=(6x^2+2x-1 )/ (x^2-1)`?

Answer 1

The vertical asymptotes are `x=1` and `x=-1`
The horizontal asymptote is `y=6`
No oblique asymptote

Explanation:

The denominator is `(x^2-1)=(x-1)(x+1)`

As we cannot divde by `0`, `x!=+-1`

Therefore the vertical asymptotes are `x=1` and `x=-1`

As the degree of the numerator `=` degree of the denominator, we don't expect a slant asymptote.

For the limit, we take the term with the highest coefficient

`lim_(x->+-oo)f(x)=lim_(x->+-oo)(6x^2)/x^2=6`

The horizontal asymptote is `y=6`

graph{(y-(6x^2+2x-1)/(x^2-1))(y-6)=0 [-12.83, 12.49, -1.84, 10.83]}