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Math Help - develope a linear, extonential and quadratic model using the following table

  1. #1
    Junior Member surffan's Avatar
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    develope a linear, extonential and quadratic model using the following table

    So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
    Age Value
    1 14000
    2 9100
    3 6200
    4 4000
    3 3000

    I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
    Thank you
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  2. #2
    MHF Contributor
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    Quote Originally Posted by surffan View Post
    So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
    Age Value
    1 14000
    2 9100
    3 6200
    4 4000
    3 3000

    I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
    Thank you
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  3. #3
    Junior Member surffan's Avatar
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    i'm giving it a try but the problem is that i cant get it to match the graph, to get the exponential function to start at around 17500 and end at 0 for y.
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  4. #4
    Junior Member surffan's Avatar
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    I developed the exponential, its just the quadratic I have to do now. I'm pretty lost I must admit
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  5. #5
    MHF Contributor
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    Quote Originally Posted by surffan View Post
    I developed the exponential, its just the quadratic I have to do now. I'm pretty lost I must admit
    did you perform a quadratic regression using the data in your table?
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  6. #6
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by surffan View Post
    So I was given the age of a car and the value at each year, with this information I'm suppose to develope a linear, exponential and quadratic equation.
    Age Value
    1 14000
    2 9100
    3 6200
    4 4000
    3 3000

    I developed the linear equation y= -2750x+17500 by calculating the slope and drawing a line through the points on the graph to get the y intercept. I'm having trouble with the other two forms, how can I draw a linear equation that will match my graph? And for quadratic I know that the vertex is going to be the same as the y intercept. If someone can please help me understand how to form these equations I would really appreciate it.
    Thank you
    Hi Surffan,

    If you're trying to do this manually, good luck with that.

    I used a TI-84+ and got a linear regression equation of y = -2710x + 15390, which is a little better than your calculation. The correlation coefficient is .9659370337.

    Using the same calculator data, the quadratic regression equation was 607.1428571x^2-6352.857143x+19640 which is pretty darn good. The correlation coefficient is .9985988704.

    My exponential regression equation was equally good at y=20130.14771(.676862578)^x. The correlation coefficient is .9979550889
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  7. #7
    Junior Member surffan's Avatar
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    i used that link given in the first response to create a table and use my data. Wow much easier than trying to figure it all out manually
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