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Thread: Asymptote

  1. #1
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    Asymptote

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

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

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

    so we get $\displaystyle x= \pm 2$

    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

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

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

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

    You has just to make a division :

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

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

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

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

    $\displaystyle 4 - x^2 =0$

    $\displaystyle \Rightarrow x=-2,2$

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

    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
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    I found the answer.
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