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Math Help - Need help on finding the rule of an unknown function

  1. #1
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    Exclamation Need help on finding the rule of an unknown function

    Hey, could anyone tell me what is the rule of that function?
    Here are the numbers found in the table (numbers with periods are approximated by the original problem):
    (x.y)
    (0.100)
    (1.100)
    (2.90)
    (3.85,86)
    (4.82,68)
    (6.77,63)
    (8.73,54)
    I already tried multiple times with a lot of different forms of fuctions but I didn't succeed into getting the right numbers out, even if I did get close with a square root function.
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  2. #2
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    e^(i*pi)'s Avatar
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    Looks like a standard linear function to me (and excel)

    The graph on the left is with the first two values removed as they look like outliers
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Looks like a standard linear function to me (and excel)

    The graph on the left is with the first two values removed as they look like outliers
    O.K, but I tried the formula that was written in your left graphic with x=4 and it actually did quite a difference (1,something I don't remember)... Wouldn't there be something else?
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  4. #4
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    Quote Originally Posted by Liberius View Post
    Hey, could anyone tell me what is the rule of that function?
    Here are the numbers found in the table (numbers with periods are approximated by the original problem):
    (x.y)
    (0.100)
    (1.100)
    (2.90)
    (3.85,86)
    (4.82,68)
    (6.77,63)
    (8.73,54)
    I already tried multiple times with a lot of different forms of fuctions but I didn't succeed into getting the right numbers out, even if I did get close with a square root function.
    The first three values are not clear.
    Are these the numbers?:
    Code:
     
    0.00, 100
    1.00, 100
    2.00,  90
    3.85,  86
    4.82,  68
    6.77,  63
    8.73,  54
    using this values
    Code:
     
    3.85,  86
    4.82,  68
    6.77,  63
    8.73,  54
    this results:
     <br />
-1.23x^3 + 24.44x^2 -160.95x + 413.43<br />

    .
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