Find the vertical asymptotes of f(x)= [x(x^2+1)]/(2x^2-3x-2)

Question

How do you work this?

Answer 1

Hi Mark;

The vertical asymptotes are the value(s) of x which enable the denominator to equal zero.  Because the denominator cannot equal zero, the lines which these equations create will never cross these asymptotes.

f(x)= [x(x^2+1)]/(2x^2-3x-2)

We will focus on the denominator...

0=(2x²-3x-2)

For the FOIL...

FIRST must be (2x)(x)=2x²

OUTER and INNER must add-up to -3x.

LAST must be (2)(1) or (1)(2) and only one number must be negative to render the product of -2.

0=(2x+1)(x-2)

Let's FOIL...

FIRST...(2x)(x)=2x²

OUTER...(2x)(-2)=-4x

INNER...(1)(x)=1x=x

LAST...(1)(-2)=-2

0=2x²-4x+x-2

0=2x²-3x-2

0=(2x+1)(x-2)

0=2x+1.....0=x-2

-1=2x...........2=x

-1/2=x

 

Vertical asymptotes are

x=-1/2

x=2