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Math Help - Oblique Asymptotes

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    Senior Member sakonpure6's Avatar
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    Oblique Asymptotes

    hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to f(x) = (x^3-16x)/(-4x^2+4x+24) how come?

    I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: y= 1/4 x + 1/4.
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    Re: Oblique Asymptotes

    Quote Originally Posted by sakonpure6 View Post
    hi , I learned in class that if an Oblique asymptote is present, then the degree of the numerator should be exactly 1 more than the denominator's. However this does not apply to f(x) = (x^3-16x)/(-4x^2+4x+24) how come?

    I mean I looked at its appropriate graph, and I could not see an OA, According to my calculations it is: y= 1/4 x + 1/4.
    You are off by a - sign. See below.

    -Dan
    Oblique Asymptotes-oblique.jpg
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    Senior Member sakonpure6's Avatar
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    Re: Oblique Asymptotes

    Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.
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    Re: Oblique Asymptotes

    Hello, sakonpure6!

    \text{Find the oblique asymptote: }\:f(x) \:=\: \frac{x^3-16x}{-4x^2+4x+24}

    \text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}


    \text{Divide numerator and denominator by }x^2\!:

    . . f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}


    \text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}


    \text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x
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    Forum Admin topsquark's Avatar
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    Re: Oblique Asymptotes

    Quote Originally Posted by sakonpure6 View Post
    Oh okay! By the way on the right hand side of the graph, does the function approach the OA or get farther from it? Because to me it looks like it is getting farther.
    It lines up. Graph it on a calculator and zoom out. (Or you could do it for real and find the limit. )

    -Dan
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    Re: Oblique Asymptotes

    The oblique asymptote is y=-x/4 -1/4
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    Re: Oblique Asymptotes

    Quote Originally Posted by Soroban View Post
    Hello, sakonpure6!


    \text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}


    \text{Divide numerator and denominator by }x^2\!:

    . . f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}


    \text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}


    \text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x
    You are looking at the behavior of the function in the asymtotic region. You suppressed what is happening near the origin. Just divide the polynomials and you will get -x/4 - 1/4. The complete graph is attached.
    Oblique Asymptotes-untitled2.gif
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    Re: Oblique Asymptotes

    Quote Originally Posted by topsquark View Post
    You are off by a - sign. See below.

    -Dan
    Click image for larger version. 

Name:	oblique.jpg 
Views:	8 
Size:	23.7 KB 
ID:	29548
    What graphing utility you are using
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    Forum Admin topsquark's Avatar
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    Re: Oblique Asymptotes

    Quote Originally Posted by votan View Post
    What graphing utility you are using
    Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.

    -Dan
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    Re: Oblique Asymptotes

    Quote Originally Posted by topsquark View Post
    Graph. It's freeware and it's pretty awesome, really. No pesky ads or anything.

    -Dan
    I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?
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    Re: Oblique Asymptotes

    Quote Originally Posted by votan View Post
    I am using dplot95, freeware from uncle sam. It is superior. The reason I asked is because your graph seems to cross the asymptote near about x=6, suggesting there is a point of inflection somewhere. Is there one?
    I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.

    I'll look into dplot, thanks.

    -Dan
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    Re: Oblique Asymptotes

    Quote Originally Posted by Soroban View Post
    Hello, sakonpure6!


    \text{We have: }\:f(x) \;=\;\frac{x^3-6x}{\text{-}4x^2 + 4x + 24}


    \text{Divide numerator and denominator by }x^2\!:

    . . f(x) \;=\;\dfrac{\frac{x^3}{x^2} - \frac{6x}{x^2}}{\frac{\text{-}4x^2}{x^2} + \frac{4x}{x^2} + \frac{24}{x^2}} \;=\;\frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}}


    \text{Hence: }\:\lim_{x\to\inty}f(x) \;=\;\lim_{x\to\infty} \frac{x - \frac{16}{x}}{\text{-}4 + \frac{4}{x} + \frac{24}{x^2}} \;=\;\frac{x-0}{\text{-}4+0+0} \;=\;\frac{x}{\text{-}4}


    \text{The oblique asymptote is: }\:y \:=\:\text{-}\tfrac{1}{4}x
    Another way to see this is to use "long division": \frac{x^3- 16x}{-4x^2+ 4x+ 24}= -\frac{1}{4}x- \frac{1}{4} - \frac{9x- 6}{-4x^2+ 4x+ 24}.
    Obviously as x goes to infinity (or negative infinity) the remaining fraction goes to 0 so the graph approaches -\frac{1}{4}x- \frac{1}{4}.
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    Re: Oblique Asymptotes

    Quote Originally Posted by topsquark View Post
    I don't think there is anything to be read into the asymptote cutting across the function. It's about the limiting properties of the function at large |x|, not in the region "near" the origin.

    I'll look into dplot, thanks.

    -Dan
    Can we insert a power point presentation here?
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