How do you find the vertical, horizontal or slant asymptotes for (6x^2+2x-1) /( x^2-1)?

Sep 15, 2016

vertical asymptotes at $x = \pm 1$
horizontal asymptote at y = 6

Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: ${x}^{2} - 1 = 0 \Rightarrow x = \pm 1$

$\Rightarrow x = - 1 \text{ and " x=1" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x that is ${x}^{2}$

$f \left(x\right) = \frac{\frac{6 {x}^{2}}{x} ^ 2 + \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2} = \frac{6 + \frac{2}{x} - \frac{1}{x} ^ 2}{1 - \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{6 + 0 - 0}{1 - 0}$

$\Rightarrow y = 6 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(6x^2+2x-1)/(x^2-1) [-20, 20, -10, 10]}