# How do you find vertical, horizontal and oblique asymptotes for (2x-2) /( 2x+2 )?

May 8, 2016

vertical asymptote x = -1
horizontal asymptote y = 1

#### Explanation:

Begin by factorising numerator/denominator

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

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve: x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

divide terms on numerator/denominator by x

$\frac{\frac{x}{x} - \frac{1}{x}}{\frac{x}{x} + \frac{1}{x}} = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}$

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

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

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) hence there are no oblique asymptotes.
graph{(x-1)/(x+1) [-10, 10, -5, 5]}