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lilkrazy1@rocketmail.com
lilkrazy1@rocketmail.com asked in Science & MathematicsMathematics · 1 decade ago

finding the vertical asymptote of function (x^2+1)/ (3x-2x^2)?

steps in finding the vertical asymptote?

thanks

3 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    3x-2x^2 = 2x^2 + 3x -0

    Put the bottom of the fraction equal to zero, and your values are the asymptotes

    x(2x+3)

    so x=0 and x=-3/2 are the asyptotes

  • nazgul855
    1 decade ago

    Vertical asymptotes happen at values of x where the function is undefined. Here this happens when the denominator is 0 and only then.

    3x-2x^2=0=x[3-2x]

    So there are vertical asymptotes at x=0 and x=1.5 only.

  • math393
    1 decade ago

    This function will have an asymptote where its value is undefined, or where the denominator is equal to zero.

    Thus, 3x-2x^2=0

    x(3-2x)=0

    x=0, x=3/2

    Asymptotes will occur at x=0 and x=3/2

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