Question

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

Answer 1

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