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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#

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]}

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Answer 2

Isaiah Allen

To find the vertical asymptotes of the function y = (x^2+1)/(3x-2x^2), we need to determine the values of x for which the denominator becomes zero.

Setting the denominator equal to zero, we solve the equation 3x - 2x^2 = 0.

Factoring out an x, we get x(3 - 2x) = 0.

This equation is satisfied when x = 0 or 3 - 2x = 0.

Solving 3 - 2x = 0, we find x = 3/2.

Therefore, the vertical asymptotes of the function y = (x^2+1)/(3x-2x^2) are x = 0 and x = 3/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.