# How to find the asymptotes of f(x)=(x^2+1)/(2x^2-3x-2)?

Mar 29, 2016

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

#### Explanation:

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

solve : 2x^2-3x-2 = 0 → (2x+1)(x-2) = 0

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

Horizontal asymptotes occur as lim_(xtooo) f(x) → 0

If the degree of the numerator/denominator are equal , as is the case here , both of degree 2. Then the equation is the ratio of the coefficients of the highest powers.

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

Here is the graph of f(x).
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}