# For (x^2+1)/(2x^2-3x-2), how do you find horizontal and vertical asymptotes?

Apr 12, 2018

Horizontal asymptote is $y = \frac{1}{2}$
Vertical asymptotes are: $x = - \frac{1}{2}$ and $x = 2$

#### Explanation:

For the Horizontal asymptote, you look at the degrees of the numerator and denominator. Since the degree of the numerator and denominator are the same, we use a ratio of the leading coefficients.

$y = \text{lead coef."/"lead coef.} = \frac{1}{2}$

So the Horizontal asymptote is $y = \frac{1}{2}$

For the Vertical asymptote, we look at the zeros of the denominator. Set the denominator equal to zero and solve for x.

$2 {x}^{2} - 3 x - 2 = 0$

$\left(2 x + 1\right) \left(x - 2\right) = 0$

$2 x + 1 = 0$ and $x - 2 = 0$

$x = - \frac{1}{2}$ and $x = 2$

$x = - \frac{1}{2}$ and $x = 2$