Question

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

Answer 1

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]}