# How do you find the asymptotes for (3x-2) / (x+1)?

Sep 4, 2016

vertical asymptote at x = - 1
horizontal asymptote at y = 3

#### Explanation:

The denominator of the function cannot equal zero as this would make the function 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 : $x + 1 = 0 \Rightarrow x = - 1 \text{ is the asymptote}$

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

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

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

$\Rightarrow y = 3 \text{ is the asymptote}$
graph{(3x-2)/(x+1) [-20, 20, -10, 10]}