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

1 Answer
Apr 26, 2016

Answer:

vertical asymptotes #x=-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 #

#rArr x = -1/2" and " x = 3" are the asymptotes "#

Horizontal asymptotes occur as # lim_(x to +- oo) , f(x) 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]}