# What are the asymptotes of y=(2x^2 +1)/( 3x -2x^2)?

Dec 23, 2015

Vertical Asymptotes:
$x = 0 \wedge x = - \frac{3}{2}$

Horizontal Asymptote:

$y = - 1$

#### Explanation:

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

1. Verical Asymptotes
Since denominator could not be 0
we find the possible values of x that would make the equation in the denominator 0

$x \left(2 x + 3\right) = 0$

Therefore

$x = 0$

$\left(2 x + 3\right) = 0 \implies x = - \frac{3}{2}$

are vertical asymptotes.

1. Horizontal asymptotes

Since the degree of numerator and denominator is the same, we have an horizontal asymptotes

$y \approx - \frac{2 {x}^{2}}{2 {x}^{2}} = - 1$

$\therefore y = - 1$ is a horizontal asymptotes for $x \rightarrow \pm \infty$

graph{-(2x^2+1)/(x(2x+3)) [-25.66, 25.65, -12.83, 12.82]}