What are the asymptotes of #y=(2x^2 +1)/( 3x -2x^2)#?

1 Answer

Vertical Asymptotes:
#x= 0^^x=-3/2#

Horizontal Asymptote:

#y=-1#

Explanation:

#y=(2x^2+1)/(3x-2x^2)=-(2x^2+1)/(2x^2+3x)=-(2x^2+1)/(x(2x+3))#

  1. Verical Asymptotes
    Since denominator could not be 0
    we find the possible values of x that would make the equation in the denominator 0

#x(2x+3)=0#

Therefore

#x=0#

#(2x+3)=0=>x=-3/2#

are vertical asymptotes.

  1. Horizontal asymptotes

Since the degree of numerator and denominator is the same, we have an horizontal asymptotes

#y~~-(2x^2)/(2x^2)=-1#

#:.y=-1# is a horizontal asymptotes for #xrarr+-oo#

graph{-(2x^2+1)/(x(2x+3)) [-25.66, 25.65, -12.83, 12.82]}