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

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Answer 1 · 2015-12-23T07:06:31

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

Horizontal Asymptote:

$y=-1$

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

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.

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]}

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