# What are the asymptote(s) and hole(s), if any, of  f(x) =(3x^2)/(x^2-x-1)?

Jun 12, 2017

$\text{vertical asymptotes at "x~~-0.62" and } x \approx 1.62$
$\text{horizontal asymptote at } y = 3$

#### 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.

$\text{solve } {x}^{2} - x - 1 = 0$

$\text{here " a=1,b-1" and } c = - 1$

$\text{solve using the "color(blue)"quadratic formula}$

$x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

$\Rightarrow x \approx 1.62 , x \approx - 0.62 \text{ are the asymptotes}$

$\text{Horizontal asymptotes occur as}$

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

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

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

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

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

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(3x^2)/(x^2-x-1) [-10, 10, -5, 5]}