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

##### 1 Answer
Jul 10, 2016

vertical asymptotes $x = - \frac{1}{2} , x = 2$
horizontal asymptote $y = \frac{1}{2}$

#### Explanation:

The denominator of this rational function cannot equal zero as this would give division by zero which is undefined. Setting the denominator equal to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: $2 {x}^{2} - 3 x - 2 = 0 \Rightarrow \left(2 x + 1\right) \left(x - 2\right) = 0$

$\Rightarrow x = - \frac{1}{2} \text{ and " x=2" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest exponent of x, that is ${x}^{2}$

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

as $x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{2 - 0 - 0}$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2) Hence there are no slant asymptotes.
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}