# How do you identify all asymptotes or holes for f(x)=(x^2-1)/(-3x^2+12)?

Aug 23, 2016

vertical asymptotes at x = ± 2
horizontal asymptote at $y = - \frac{1}{3}$

#### 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: $- 3 {x}^{2} + 12 = 0 \Rightarrow 3 {x}^{2} = 12 \Rightarrow {x}^{2} = 4 \Rightarrow x = \pm 2$

$\Rightarrow x = - 2 \text{ and " x=2" are the asymptotes}$

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 ${x}^{2}$

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2}{\frac{- 3 {x}^{2}}{x} ^ 2 + \frac{12}{x} ^ 2} = \frac{1 - \frac{1}{x} ^ 2}{- 3 + \frac{12}{x} ^ 2}$

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

$\Rightarrow y = - \frac{1}{3} \text{ is the asymptote}$

Holes occur when there are duplicate factors on the numerator/denominator. There are no duplicate factors here hence there are no holes.
graph{(x^2-1)/(-3x^2+12) [-10, 10, -5, 5]}