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Thread: Finding the curve of best fit

  1. #1
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    Question Finding the curve of best fit

    Hi,

    I hope someone can help me with the following question:

    Finding the curve of best fit-screen-shot-2017-06-30-11.07.08-am.png

    And this is the data:

    Finding the curve of best fit-screen-shot-2017-06-30-11.07.35-am.png

    I used the online graphing technology program called, MyCurveFit, and I got the following data for the following functions:
    Quadratic: y = 395.3571 - 11.28571*x + 0.1285714*x^2
    Exponential(basic): y = 57.99277 + 339.1588*e^(-0.03634446*x)
    Linear: y = 3263.235*x - 6380067

    These equations that I listed above are pretty far off from what it seems to the textbook's, as follows:
    Finding the curve of best fit-screen-shot-2017-06-30-11.13.14-am.png

    Can anyone tell why my equations are so far off from the textbook's?
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  2. #2
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    Re: Finding the curve of best fit

    What data did you put into this program? When x= 0, the quadratic "approximation" you list gives y= 395.397, the exponential gives y= 397.1515, and the linear gives y= - 6380067. None of those are really close to 400 and the last is simply ludicrous. Have you tried [b]doing the problem yourself rather than just handing it over to a program?
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  3. #3
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    Re: Finding the curve of best fit

    Yeah, I just tried solving it for myself and I got the correct formulas for exponential and linear. I get a different formula for quadratic, however. I know that the quadratic equation form is y=ax^x + bx + c. I ended up getting the quadratic formula of y = -1.56x^2 - 5.2 + 400... which is pretty different from the textbook's solution even after I placed mine into vertex form.

    The reason why I'm trying to learn how to use a graphing program is because when data sets become more complex, it will be a lot easier to use a program. I hope that you can help teach me.
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