# How do you find the Vertical, Horizontal, and Oblique Asymptote given (x^2+1)/(2x^2-3x-2) ?

##### 1 Answer
Jun 7, 2016

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

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero, To find the equation/s set the denominator equal to zero.

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

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

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by ${x}^{2}$

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

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

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

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