# How do you find the vertical asymptotes of the function y= (x^2+1)/(3x-2x^2) ?

Jun 18, 2018

$\text{vertical asymptotes at "x=0" and } x = \frac{3}{2}$

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve "3x-2x^2=0rArrx(3-2x)=0

$x = 0 \text{ and "x=3/2" are the asymptotes}$
graph{(x^2+1)/(3x-2x^2) [-10, 10, -5, 5]}