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

Nov 16, 2016

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

#### Explanation:

The denominator is $\left({x}^{2} - 1\right) = \left(x - 1\right) \left(x + 1\right)$

As we cannot divde by $0$, $x \ne \pm 1$

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

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{6 {x}^{2}}{x} ^ 2 = 6$

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]}