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Finding horizontal and vertical asymptotes

  1. Oct 26, 2006 #1


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    the equation is:

    x^2 + xy + y^2 = 25

    my method was to implicitely differentiate and then set the numerator = 0 to get points of horizontal asymptotes, and then set the denominator = 0 to get the points for the veritcal asymptotes.

    when i use implicit differentiation i get:

    y' = (-2x - y)/(x + 2y)

    am i correct so far? now im stuck because if i set the top equal to 0 i just get y = -2x. what can i do from here to solve for the points where there are both horizontal and vertical asymptotes? thanks.
  2. jcsd
  3. Oct 26, 2006 #2


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    Do you have any reason to think the graph of that expression has horizontal and vertical asymptotes? I get that the only asymptotes are the oblique asymptotes y= x and y= -3x.
  4. Oct 28, 2006 #3


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    how can i know if it does have any to begin with?

    and can you explain how you got those oblique asymptotes?
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