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

Jun 18, 2018

$\text{vertical asymptotes at "x=0" and } x = \frac{3}{2}$
solve "3x-2x^2=0rArrx(3-2x)=0
$x = 0 \text{ and "x=3/2" are the asymptotes}$