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

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

To invent the perpendicular asymptotes we set the denominator resembling to cipher.

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

Look at the graph under.

Sometimes, the denominator is resembling to cipher at the selfselfsame x-value that makes the numerator cipher. This conciliate effect a instead of a perpendicular asymptote.

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