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Math Help - Finding Vertical/Horizontal/Slant Asymptotes

  1. #1
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    Finding Vertical/Horizontal/Slant Asymptotes

    Hi.

    I have a couple of really confusing (well, at least to me ) problems that I really need help with. Actually, I have about 4 of them.

    I can find the asymptote(s) and other stuff in problems like this...


    ...but I can't do ones like this:

    ^^ Those are the four I need help with. I don't want the answers directly (although that would be nice), but rather a little walkthrough. (I made that image in PowerPoint).

    Thanks.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by BeSweeet View Post
    Hi.

    I have a couple of really confusing (well, at least to me ) problems that I really need help with. Actually, I have about 4 of them.

    I can find the asymptote(s) and other stuff in problems like this...


    ...but I can't do ones like this:

    ^^ Those are the four I need help with. I don't want the answers directly (although that would be nice), but rather a little walkthrough. (I made that image in PowerPoint).

    Thanks.
    Hi BeSweeet,

    Speaking of PowerPoint, let me suggest that you check this out: Asymptote Tutorial

    [1] f(x)=x+\frac{4}{x^2+1}

    f(x)=\frac{x(x^2+1)+4}{x^2+1}

    f(x)=\frac{x^3+x+4}{x^2+1}

    (a) No vertical asymptote because any real number is valid for x in the denominator.
    (b) No horizontal asymptote because the degree of the numerator is not less than or equal to the degree of the denominator.
    (c) Has a slant asymptote since the degree of the numerator is exactly one degree bigger than the denominator. Use long division to divide the denominator into the numerator and discard the remainder. The slant asymptote is at y=x-1

    [2] f(x)=x+\frac{32}{x^2}Try to follow [1] to answer this one

    [3] f(x)=\frac{3x^4-5x+3}{x^4+1}

    (a) No vertical asymptote (see reason for [1] above)
    (b) Has a horizontal asymptote since the degree of the numerator is equal to the degree of the denominator. The horizontal asymptote is y = a/b where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator. Horizontal asymptote is y = 3
    (c) No slant asymptote (see reason for [1] above)

    [4] f(x)=\frac{x-1}{1+3x^2}

    (a) No vertical asymptote
    (b) Has horizontal asymptote since degree of numerator is less than the degree of the denominator. Horizontal asymptote in this case is y = 0
    (c) No slant asymptote.
    Last edited by masters; April 9th 2009 at 11:26 AM.
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  3. #3
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    Haha, I actually looked at that PowerPoint before I made this thread.

    Thanks a bunch for your reply.

    When doing long division for the first problem, I came up with this:


    So, I guess that whenever you only have one 'divisor' (or whatever), you just add +1 (or something like that)?

    If I plug the problem into my graphing calculator, it looks like this:


    That doesn't look anything like how I would have sketched it.

    For the second problem, the graph on the graphing calculator looked like this:


    Again, I have no idea how I would go about manually sketching it.

    Then for the last two, I have no idea how to find out anything...

    The way my teacher explained it is extremely weird...

    She said that, to find the horizontal asymptote, we put the problem into our graphing calculator, go to 2nd>Graph [table], and put in some sample numbers, like -1,000,000, -10,000,000, 1,000,000, and 10,000,000, to see if we would get anything. If the answers are near each other, then that answer would be the horizontal asymptote.
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  4. #4
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    Quote Originally Posted by BeSweeet View Post
    Haha, I actually looked at that PowerPoint before I made this thread.

    Thanks a bunch for your reply.

    When doing long division for the first problem, I came up with this:


    So, I guess that whenever you only have one 'divisor' (or whatever), you just add +1 (or something like that)?
    Sorry, BeSweeet, but your screenshots are too dark for me to see. And I made a mistake on the first slant asymptote. It should've been y=x-1. I've edited it so that it is correct now.


    The way my teacher explained it is extremely weird...

    She said that, to find the horizontal asymptote, we put the problem into our graphing calculator, go to 2nd>Graph [table], and put in some sample numbers, like -1,000,000, -10,000,000, 1,000,000, and 10,000,000, to see if we would get anything. If the answers are near each other, then that answer would be the horizontal asymptote.
    I suggest you follow the procedure from the PP presentation. If your teacher wants you to "eyeball" the graph and analyze the table values, then I guess I would follow her lead. But remember, the calculator doesn't make you smart.

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  5. #5
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    I figured it out. Thanks for the help!
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