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

1 Answer
Feb 25, 2016

Answer:

Two vertical asymptotes are #2x+1=0# and #x-3=0#.

Explanation:

To find all the asymptotes for function #y=(1-x)/(2x^2-5x−3)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #(2x^2-5x−3)=0#.

Let us find factors of #(2x^2-5x−3)# by splitting middle term in #-6s# and #x# i.e. #(2x^2-6x+x−3=2x(x-3)+1(x-3)=(2x+1)(x-3)#.

As the factors of denominators are #(2x+1)# and #(x-3)#, two vertical asymptotes are #2x+1=0# and #x-3=0#.

As the highest degree of numerator is less than that of denominator, there is no horizontal or slanting asymptote.