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Thread: Vertical Asymptotes

  1. #1
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    Vertical Asymptotes

    f(x) = 4 / (x^2 - 2)^(1/2)

    Is zero a vertical Asymptote in this function?
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  2. #2
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    Re: Vertical Asymptotes

    No, it is not! Why would you think it was? (This function is only defined for x> \sqrt{2} or x< -\sqrt{2}.)
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  3. #3
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    Re: Vertical Asymptotes

    $x=0$ is not in the domain of this function. Domain is $(-\infty,-\sqrt{2}) \cup (\sqrt{2}, \infty)$
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  4. #4
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    Re: Vertical Asymptotes

    When I plug zero inside the square root, I get a negative value which is undefined. I thought any undefined value in the denominator will be a vertical asymptote for the function.

    Then, if it is not a vertical asymptote, what is the correct way to recognize the values of the vertical asymptote?
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  5. #5
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    Re: Vertical Asymptotes

    vertical asymptotes occur when a factor in the denominator equals 0 with no common factor in the numerator

    for example ...

    $y = \dfrac{(x+1)(x-2)}{(x-3)}$ has a vertical asymptote at $x=3$

    $y = \dfrac{(x+1)(x-3)}{(x-3)}$ has a point discontinuity (a hole) at $x=3$


    the function you posted

    f(x) = 4 / (x^2 - 2)^(1/2)
    has vertical asymptotes at $x=\sqrt{2}$ and $x=-\sqrt{2}$
    Attached Thumbnails Attached Thumbnails Vertical Asymptotes-vert_asym.png  
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  6. #6
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    Re: Vertical Asymptotes

    Thanks a lot. I have got the idea.
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