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

The vertical asymptotes of ##y=(x^2+1)/(3x-2x^2)## are ##x=0## and ##x=3/2##.

To find the vertical asymptotes we set the denominator equal to zero.

##3x-2x^2=x(3-2x)=0rArrx=0 or x=3/2##

Look at the graph below.

Sometimes, the denominator is equal to zero at the same x-value that makes the numerator zero. This will produce a instead of a vertical asymptote.

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