# How do you find the asymptotes for  (1-x)/(2x^2-5x-3)?

Apr 26, 2016

vertical asymptotes $x = - \frac{1}{2} , x = 3$
horizontal asymptote y = 0

#### 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 :  2x^2-5x-3 = 0 → (2x+1)(x-3) = 0

$\Rightarrow x = - \frac{1}{2} \text{ and " x = 3" are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

If the degree of the numerator < degree of the denominator,as is the case here then the equation is always y = 0
graph{(1-x)/(2x^2-5x-3) [-10, 10, -5, 5]}