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Math Help - Asymptote

  1. #1
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    Asymptote

    Determine function y=\frac{2x^3+6x-1}{4-x^2} graphic's asymptotes.
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  2. #2
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    Quote Originally Posted by Bernice View Post
    Determine function y=\frac{2x^3+6x-1}{4-x^2} graphic's asymptotes.

    The verical asymptotes occuer at the zero's of the denominator

    4-x^2=0 \iff (2-x)(2+x)=0

    so we get x= \pm 2

    It does not have any horizontal asymptotes because the degree of the numerator is larger than the denominator.

    It does have a slant asymptote( the degree of the numerator is one larger than the denominator) This can be calculated by dividing the denominator into the numerator. (long division)

    Or by this trick

    \frac{2x^3+6x-1}{4-x^2}=\frac{2x^3-8x+14x-1}{4-x^2}=\frac{-2x(4-x^2)+14x-1}{4-x^2}=

    \frac{-2x(4-x^2)}{4-x^2}+\frac{14x-1}{4-x^2}=-2x+\frac{14x-1}{4-x^2}

    So the slant asymptotes is y=-2x
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  3. #3
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    Hello,

    You has just to make a division :

    2x^3+6x-1=-2x(4-x^2)+14x-1

    Hence \color{blue}\boxed{y+2x=\frac{14x-1}{4-x^2}\underset{x\to \pm \infty}{\longrightarrow} 0}

    So \color{red} x\mapsto -2x is the asymptote.

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  4. #4
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    Reply

    Quote Originally Posted by Bernice View Post
    Determine function y=\frac{2x^3+6x-1}{4-x^2} graphic's asymptotes.
    For vertical Asymptote, Denominator = 0

    4 - x^2 =0

    \Rightarrow x=-2,2

    Vertical asymptote (VA) are x = -2, and x = 2

    The degree of Numerator is one more than Denominator, so, it will also have linear-oblique asymptote.

    Divide Numerator with Denominator, using long division.

    long divide polynomials

    Since y = - 2x is answer, so the linear-oblique asymptote (LOA) is y=-2x

    See graph attached. The green lines show asymptote.
    Attached Thumbnails Attached Thumbnails Asymptote-graph33.jpg  
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  5. #5
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    Thank you all for explanations. They are very useful!
    Last edited by Bernice; May 9th 2009 at 01:40 PM.
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  6. #6
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    TheEmptySet > Is it possible to calculate the integral of your signature without the residue theorem ?
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  7. #7
    Junior Member Infophile's Avatar
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    I found the answer.
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