# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(3x^2+2) / (x^2 -1)?

Sep 1, 2016

vertical asymptotes at x = ± 1
horizontal asymptote at y = 3

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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=0rArrx^2=1rArrx=±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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2} = \frac{3 + \frac{2}{x} ^ 2}{1 - \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3 + 0}{1 - 0}$

$\Rightarrow y = 3 \text{ is the asymptote}$

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{(3x^2+2)/(x^2-1) [-20, 20, -10, 10]}