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

Jan 20, 2016

There exists a vertical asymptote at $x = - \frac{1}{3}$ and a horizontal asymptote at $y = - \frac{2}{3}$.

#### Explanation:

Vertical asymptotes occur at points that lead to division by zero or negative even square roots.

In this case, division by zero results if the denominator of the rational function is zero, ie if $3 x + 1 = 0$, which implies if $x = - \frac{1}{3}$.

Horizontal asymptotes occur at ${\lim}_{x \to {+}_{\infty}} g \left(x\right) = - \frac{2}{3}$.

The graph of the function verifies this.

graph{(-2x+3)/(3x+1) [-11.25, 11.245, -5.63, 5.62]}