# How do you find the vertical, horizontal or slant asymptotes for (2x^2-1) / (3x^3-2x+1)?

Jan 8, 2017

Horizontal : $\leftarrow y = 0 \rightarrow$
Vertical : $\uparrow x = - 2 \downarrow$.

#### Explanation:

Let $y = \frac{2 {x}^{2} - 1}{3 {x}^{3} - 2 x + 1}$

There is no quotient in this division. So, existence of slant asymptote

is ruled out..

The y-intercept ( x = 0 ) is $- 1$.

x-intercepts ( y = 0 ) are $x = \pm \frac{1}{\sqrt{2}}$.

Cross multiplying,

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

If one factor in LHS is 0, the other has to be infinite, when the

product $\in \left(0 , \infty\right)$.

Now, y = 0 gives horizontal asymptote.

The other factor has a factor (x +1 ) and its other factor has complex

So, x + 2 = 0 gives the vertical asymptote.

graph{ y(y (3x^3-2x+1) - 2x^2+1)=0 [-8, 8 -3.26, 3.276]} #

See the Socratic graph.