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Math Help - Asymptote

  1. #1
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    Asymptote

    Is it correct to say that f(x) = \frac{4x^2+4x+1}{x^3-x} has an oblique asymptote f(x) = \frac{4}{x} and horizontal asymptotes x = 0, x = 1 and x = -1?
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  2. #2
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    The degree of the numerator must be one greater than the degree of the denominator in order to have a slant asymptote. In addition, 4/x is not a straight line. The horizontal asymptotes are all y = 0, because the degree of the denominator is greater than the degree of the numerator.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    The degree of the numerator must be one greater than the degree of the denominator in order to have a slant asymptote.
    Can you explain this condition bit more, please?
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  4. #4
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    As an example: the function

    \frac{x^{2}-1}{x}=x-\frac{1}{x}

    has a slant asymptote of x, because as x gets very large, the 1/x is very small in comparison to the x, and hence becomes negligible. See here for a more full explanation.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    As an example: the function

    \frac{x^{2}-1}{x}=x-\frac{1}{x}

    has a slant asymptote of x, because as x gets very large, the 1/x is very small in comparison to the x, and hence becomes negligible. See here for a more full explanation.
    okay, thank you. just one more question, so that I get this.


    what can we say about the asymptotes of f(x) = \frac{x^4+1}{x^3-1}? I see that it has vertical asymptote at x = 1, x = -1 and x = 0, but I'm not sure what else
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  6. #6
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    Why does it have a vertical asymptote at x = -1 and x = 0?
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  7. #7
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    Quote Originally Posted by pickslides View Post
    Why does it have a vertical asymptote at x = -1 and x = 0?
    Sorry, I meant

    f(x) = \frac{x^4+1}{x^3-x}
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  8. #8
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    Quote Originally Posted by Resilient View Post
    ...

    what can we say about the asymptotes of f(x) = \frac{x^4+1}{x^3-x}? I see that it has vertical asymptote at x = 1, x = -1 and x = 0, but I'm not sure what else
    Quote Originally Posted by Resilient View Post
    Sorry, I meant

    f(x) = \frac{x^4+1}{x^3-x}
    1. As you stated correctly the vertical asymptotes correspond to the zeros of the denominator.

    2. To get the horizontal or slanted asymptotes transform the term of the function into a complete polynomial and a fraction:
    Code:
                                           x^2 + 1
     (x^4        +  1)  (x^3 - x) = x + -----------
    -(x^4 - x^2)                           x^3 - x
    ------------
            x^2 + 1
    The none-fractional part of the term is the term of the slanted asymptote. That means in your case the slanted asymptote has the equation:

    y = x
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