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Math Help - Discontinuity and Vertical Asymptotes

  1. #1
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    Discontinuity and Vertical Asymptotes

    So I have a problem that I have all parts figured out except for the final part of the question.

    The function is f(x)=(x^2 - 1) / (2x^2 - 2x).

    The function is discontinuous at 0 and 1.

    x=0 is a vertical asymptote.

    The horizontal asymptote is y=1/2

    Now the final part of the question is: Since there is only one vertical asymptote, indicate on the graph what occurs at the other point of discontinuity. Be specific by giving numerical values.

    I am really unsure what happens here. Does the point not exist? I'm confused Any help would be appreciated
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  2. #2
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    f(x) = \frac{(x^2 - 1)}{(2x^2 - 2x)}

    Expand it & Simplify:

    f(x) = \frac{(x - 1)(x + 1)}{2x(x - 1)}

    So we know the degree of the numerator is the same as the degree of the denominator

    There is a vertical asymptote at x = 0 because 2x = 0 and therefore x = 0

    Now we have an (x - 1) in the numerator and the denominator
    This means we can cancel these out giving us a point of discontinuity at x = 1
    Now sub it back into the equation with the (x - 1) removed

    f(x) = \frac{(x+1)}{2x}
    f(x) = 1

    Therefore there is a point of discontinuity at the point (1,1)

    Hope this helps : )
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  3. #3
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    thanks man appreciate it
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