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Math Help - graph asymptote question

  1. #1
    Senior Member
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    graph asymptote question

    hi, im having a problem with this question. Im not sure how even go about getting one of these multiple choices.

    The graph of f(x) = \frac{1}{x^2-mx -n}, where m and n are real constants, has no vertical asymptotes if

    A: m^2 < 4n
    B: m^2 > 4n
    C: m^2 = -4n
    D: m^2 < -4n
    E: m^2 > -4n

    help appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by scorpion007 View Post
    hi, im having a problem with this question. Im not sure how even go about getting one of these multiple choices.

    The graph of f(x) = \frac{1}{x^2-mx -n}, where m and n are real constants, has no vertical asymptotes if

    A: m^2 < 4n
    B: m^2 > 4n
    C: m^2 = -4n
    D: m^2 < -4n
    E: m^2 > -4n

    help appreciated.
    <obligatory moan about LaTeX not being working yet>

    For f to have a vertical asymtote requires that x^2-mx-n have at least
    one real root. This is equivalent to the discriminanat being >=0.

    The discriminant is the term under the square root sign in the quadratic
    formula, so in this case the disciminant is: m^2+4n, and so f has vetical
    asymtots if (and only if):

    m^2+4n >=0,

    or:

    m^2 >= -4n.

    RonL
    Last edited by CaptainBlack; September 25th 2006 at 04:33 AM. Reason: added "only if"
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  3. #3
    Senior Member
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    hah, sorry about the latex, i wont do it next time.

    Also -- i found the answer myself to the problem shortly after posting, lol.
    BTW, the question asks for the opposite of what you found. i.e. when the graph has *no* vert asymptotes. But all you have to do is change the >= to <.

    Thanks anyway
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