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Math Help - Which of these horizontal asymptote equations are wrong?

  1. #1
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    Which of these horizontal asymptote equations are wrong?

    Hi, I have to match 6 functions with the equation for its horizontal asymptote, but I can't get it right.

    I know at least 1 or more are wrong, can anyone tell me which, and how to solve it/them?

    1. f(x) = (6x^3 + 2x + 1) / (2x^3 - 10x^2 + 13x +100)
    y = 3

    2. f(x) = arccot(x^2 - x^4)
    y = pi

    3. f(x) = arccot(x)
    y = 0 and y = pi

    4. f(x) = e^x
    y = 0

    5. f(x) = arctan(x)
    y = pi/2 and y = -pi/2

    6. f(x) = (6x^2 + 2x + 1) / (2x^3 - 10x^2 +13x +100)
    none

    Thanks!
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  2. #2
    Super Member 11rdc11's Avatar
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    To find horizontal asymptote you take the limit to infinity.

    I'll the first 1 as an example

    \lim_{x \to \infty} \frac{6x^3 + 2x + 1}{2x^3 -10x^2 + 3x +100}

    now divide by the largest degree of the polynomial in the denominator

    \lim_{x \to \infty} \frac{6+ \frac{2}{x^2} +\frac{1}{x^3}}{2 - \frac{10}{x} + \frac{3}{x^2} + \frac{100}{x^3}}

    so you end up with 3 when taking the limit to infinity.

    You could also use L'Hopital rule
    Last edited by 11rdc11; September 22nd 2008 at 06:05 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    To find horizontal asymptote you take the limit to infinity.
    and negative infinity, as there can be as many as two horizontal asymptotes
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  4. #4
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    Yes I know that, but my question is which one of those functions doesn't match the equation for its horizontal asymptote i've given?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coldfire View Post
    Hi, I have to match 6 functions with the equation for its horizontal asymptote, but I can't get it right.

    I know at least 1 or more are wrong, can anyone tell me which, and how to solve it/them?

    1. f(x) = (6x^3 + 2x + 1) / (2x^3 - 10x^2 + 13x +100)
    y = 3
    correct

    2. f(x) = arccot(x^2 - x^4)
    y = pi
    incorrect

    3. f(x) = arccot(x)
    y = 0 and y = pi
    incorrect

    4. f(x) = e^x
    y = 0
    correct

    5. f(x) = arctan(x)
    y = pi/2 and y = -pi/2
    correct

    6. f(x) = (6x^2 + 2x + 1) / (2x^3 - 10x^2 +13x +100)
    none
    incorrect


    we already told you how to solve them
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  6. #6
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    Thanks for the reply.

    Can you show me a way to solve the ones I got wrong (specifically)?
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    and negative infinity, as there can be as many as two horizontal asymptotes
    Thanks I knew I forgot something lol
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  8. #8
    Super Member 11rdc11's Avatar
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    f(x) = \frac{6x^2 + 2x + 1} {2x^3 - 10x^2 +13x +100}

    Ok using L'hopital rule

    \lim_{x \to \infty} \frac{6x^2 + 2x + 1} {2x^3 - 10x^2 +13x +100}

    \lim_{x \to \infty} \frac{12x+2}{6x^2 -20x + 13}

    \lim_{x \to \infty} \frac{12}{12x -20}

    so you have an asymptote at 0

    Edit: o yea remember to test negative infinity aswell
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  9. #9
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    thanks guys!
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