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

Mar 2, 2016

There is a horizontal asymptote: $y = 3$

There is a vertical asymptote: $x = - 1$

Explanation:

You can rewrite the expression.

$\frac{3 x - 2}{x + 1} = \frac{3 x + 3 - 5}{x + 1}$

$= \frac{3 x + 3}{x + 1} + \frac{- 5}{x + 1}$

$= 3 - \frac{5}{x + 1}$

From this, you can see that

${\lim}_{x \to \infty} \frac{3 x - 2}{x + 1} = {\lim}_{x \to \infty} \left(3 - \frac{5}{x + 1}\right) = 3$

Similarly,

${\lim}_{x \to - \infty} \frac{3 x - 2}{x + 1} = {\lim}_{x \to - \infty} \left(3 - \frac{5}{x + 1}\right) = 3$

There is a horizontal asymptote: $y = 3$

You can also see that $x = - 1$ results in division by zero. When $x$ approaches $- 1$ from the left, the denominator will become infinisimally less than zero, while the numerator is negative. So,

${\lim}_{x \to - {1}^{-}} \frac{3 x - 2}{x + 1} = \infty$

Similarly,

${\lim}_{x \to - {1}^{+}} \frac{3 x - 2}{x + 1} = - \infty$

There is a vertical asymptote: $x = - 1$

Below is a graph for your reference.
graph{(3x-2)/(x + 1) [-20, 20, -10, 10]}