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Math Help - [SOLVED] Find the first TWO non-negative asymptotes and the first negative asymptote

  1. #1
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    [SOLVED] Find the first TWO non-negative asymptotes and the first negative asymptote

    Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of in radians

    .

    smallest non-negative asymptote:


    second non-negative asymptote:
    first negative asymptote:


    okay, i know the first one is pi/5

    but the other two i do not know how to find. can some one please help me?

    but i keep thinkin the other two answers are x=+-(pi/10). (which is wrong)
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    Member sinewave85's Avatar
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    Quote Originally Posted by lsnyder View Post
    Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of in radians

    .

    smallest non-negative asymptote:


    second non-negative asymptote:
    first negative asymptote:


    okay, i know the first one is pi/5

    but the other two i do not know how to find. can some one please help me?

    but i keep thinkin the other two answers are x=+-(pi/10). (which is wrong)

    Hi, Isnyder. The trick to this is to realize that what you are doing is simply solving for those values of x that make cot undefined (they cause division by 0): -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi

    Let \cot\left(x - \frac{\pi}{5}\right) = \cot{\theta} so that x - \frac{\pi}{5} = \theta

    So the first non-negative asymptote is at \frac{\pi}{5} as you guessed:

    x - \frac{\pi}{5} = 0

    x = \frac{\pi}{5}

    And the second non-negative asymptote:

    x - \frac{\pi}{5} = \frac{\pi}{2}

    x = \frac{\pi}{2} + \frac{\pi}{5} = \frac{2\pi + 5\pi}{10} = \frac{7\pi}{10}

    Lastly, the first negative asymptote:

    x - \frac{\pi}{5} = -\frac{\pi}{2}

    x = -\frac{\pi}{2} + \frac{\pi}{5} = \frac{-3\pi}{10}
    Last edited by sinewave85; March 26th 2009 at 03:19 PM. Reason: I improved the format.
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    actually the last two answers are 6pi/5 and -4pi/5.

    because you add pi instead of pi/2. because the pi is smaller.

    but if it wasn't for you saying all that, I would of never have gotten the right answer. Thank you!
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    Member sinewave85's Avatar
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    Quote Originally Posted by lsnyder View Post
    actually the last two answers are 6pi/5 and -4pi/5.

    because you add pi instead of pi/2. because the pi is smaller.

    but if it wasn't for you saying all that, I would of never have gotten the right answer. Thank you!

    I am fairly sure that \pi > \frac{\pi}{2}.

    \pi \approx 3.14159

    \frac{\pi}{2} \approx 1.57080

    Your answers are the third non-negative and the second negative asymptotes.
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    Quote Originally Posted by sinewave85 View Post
    I am fairly sure that \pi > \frac{\pi}{2}.

    \pi \approx 3.14159

    \frac{\pi}{2} \approx 1.57080

    yes but think about it.

    (pi/5)+pi = 6pi/5 rather than (pi/5)+(pi/2)=7pi/10
    6pi/5 < 7pi/10

    same for the negative

    (pi/5)-pi =-4pi/5 rather than (pi/5)-(pi/2) = -3pi/10
    -4pi/5 <-3pi/10


    also, it came up right in my homework when i typed it in.
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    Member sinewave85's Avatar
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    Quote Originally Posted by lsnyder View Post
    yes but think about it.

    (pi/5)+pi = 6pi/5 rather than (pi/5)+(pi/2)=7pi/10
    6pi/5 < 7pi/10

    same for the negative

    (pi/5)-pi =-4pi/5 rather than (pi/5)-(pi/2) = -3pi/10
    -4pi/5 <-3pi/10


    also, it came up right in my homework when i typed it in.
    Your answers are correct in the sense that they are asyptotes, they just are not the ones your assignment asks for:

    \frac{6\pi}{5} \approx 3.7699 is actually greater than (>) \frac{7\pi}{10} \approx 2.19911

    and

    \frac{-4\pi}{5} \approx -2.51327 is less that \frac{-3\pi}{10} \approx-0.942478, but that means that it is further away from 0. The absolute value of the former is greater than the latter.
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