# How do you identify all asymptotes or holes for f(x)=(x^2-x)/(2x^2+4x-6)?

Sep 9, 2016

hole at $\left(1 , \frac{1}{8}\right)$
vertical asymptote at x = 3
horizontal asymptote at $y = \frac{1}{2}$

#### Explanation:

Holes occur when a duplicate factor is cancelled in the numerator/denominator of f(x).
The first step is therefore to factorise and simplify f(x).

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

set $x - 1 = 0 \Rightarrow x = 1$

substitute x = 1 into 'simplified' version of f(x)

$\Rightarrow f \left(1\right) = \frac{1}{2 \left(1 + 3\right)} = \frac{1}{8}$

$\Rightarrow \text{there is a hole at} \left(1 , \frac{1}{8}\right)$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $2 \left(x + 3\right) = 0 \rightarrow x = - 3 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as $x \to \pm \infty , f \left(x\right) \to \frac{1}{2 + 0}$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$
graph{(x^2-x)/(2x^2+4x-6) [-10, 10, -5, 5]}