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

##### 2 Answers
May 21, 2018

When $x = - 1 \implies - \frac{5}{0}$ so there is an asymptote when $x = - 1$

May 21, 2018

#### Answer:

$\text{vertical asymptote at } x = - 1$
$\text{horizontal asymptote at } y = 3$

#### Explanation:

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

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.

$\text{solve "x+1=0rArrx=-1" is the asymptote}$

$\text{horizontal asymptotes occur as}$

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

$\text{divide terms on numerator/denominator by } x$

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

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

$\Rightarrow y = 3 \text{ is the asymptote}$
graph{(3x-2)/(x+1) [-10, 10, -5, 5]}