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Math Help - Power Function

  1. #1
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    Power Function

    Let f(x)= [((3x^5) - 2x + 3) / (2x -5)]. Find a power function end behavior model for f.[/FONT]
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    Quote Originally Posted by h4hv4hd4si4n View Post
    Let f(x)= [((3x^5) - 2x + 3) / (2x -5)]. Find a power function end behavior model for f.[/font]
    When you say "end behaviour" do you mean as x \rightarrow \pm \infty?
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    Yes
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    Do you know how to divide polynominals by long division?
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    yes, but i'm confused by what the question is asking for.
    How do you find the power function and end behavior function?
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    Quote Originally Posted by h4hv4hd4si4n View Post
    Let f(x)= [((3x^5) - 2x + 3) / (2x -5)]. Find a power function end behavior model for f.[/font]
    Can you figure out what power functions can be the representatives for the polynomials for the numerator and denominator in the long run? Then the end behavior of a rational function can be found by looking at the ratio of the power functions you found.
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  7. #7
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    If you perform long division you get

     \frac{3x^5-2x^2+3}{2x^2-5} = \frac{3x^3}{2} + \frac{15 x}{4} - 1 + \frac{75x-8}{4(2x^2-5)}

    As x \rightarrow \pm \infty then the last term drops and the end behaviour looks like the function

     \frac{3x^5-2x^2+3}{2x^2-5} \approx \frac{3x^3}{2} + \frac{15 x}{4} - 1
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    Quote Originally Posted by danny arrigo View Post
    If you perform long division you get

     \frac{3x^5-2x^2+3}{2x^2-5} = \frac{3x^3}{2} + \frac{15 x}{4} - 1 + \frac{75x-8}{4(2x^2-5)}

    As x \rightarrow \pm \infty then the last term drops and the end behaviour looks like the function

     \frac{3x^5-2x^2+3}{2x^2-5} \approx \frac{3x^3}{2} + \frac{15 x}{4} - 1

    Your approach is right, however the problem calls for a power function.
    Actually y=\frac{3}{2}x^3 would be the answer. Another way to look at this is by looking at the leading terms of the polynomials on the numerator and denominator (they each describe the end behavior of the two polynomials), which in this case is 3x^5 and 2x^2 and by looking at its ratio, we get y=\frac{3}{2}x^3
    Last edited by chabmgph; January 15th 2009 at 03:47 PM. Reason: better phrasing?
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  9. #9
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    Quote Originally Posted by chabmgph View Post
    Your approach is right, however the problem calls for a power function.
    Actually y=\frac{3}{2}x^3 would be the answer. Another way to look at this is by looking at the leading terms of the polynomials on the numerator and denominator (they each describe the end behavior of the two polynomials), which in this case is 3x^5 and 2x^2 and by looking at its ratio, we get y=\frac{3}{2}x^3
    Yes - I was hoping that the poster would come to this conclusion. Thanks for the clarifaction though.
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