For the following table of data

        x                 y
       -5                -40
       -4                  0
       -3                 18
       -2                 20
       -1                 12
        0                  0
        1                -10
        2                -12
        3                  0
        4                 32
        5                 90
a)Find the model y=f(x).

b)Find the derivative model.

c)Find the local linearization,y=mx+b, of the model at the point(4,32).

Hello Fredewq,

The results below are based on:
 - entering the data you provided into Microsoft Excel
 - generating a scatter chart of the X/Y data provided
 - adding a trendline to the chart (third order polynomial)
 - displaying the formula and Rsquared value
I can explain how to do this yourself if you request.

To answer your question

#1 - Y(X) = x^3 + x^2 - 12x
or        = x * (x * (x + 1) - 12)
Actually Excel indicated there should be a constant 2E-13 value in the
formula, but I believe that to be round off error in the calculations.
Rsquared was 1 (as expected).

A quick check
  x=0, y=0 check
  x=1, y= 1+1-12 = -10 OK
  x=3, y= 27+9-36 = 0 OK
#2 - the derivative of Y(X) is
  3*x^2 + 2*x - 12
Basically - the derivative of a polynomial term is...
 constant factor times
 original exponent times
 x^(original exponent - 1)
or for 1*x^3, 1*3*x^2.

#3 - the slope at X=4, Y=32 is the same as the derivative at that
point or
  3*4*4 + 2*4 - 12 = 44
as a quick check, let's compute the values at X=3.9 and X=4.1 to find
 Y(3.9) = 27.7 (close to 27.6 = 32-4.4) OK
 Y(4.1) = 36.5 (close to 36.4 = 32+4.4) OK

For some on line resources, you can try search phrases such as
  polynomial derivative
to get explanations of this branch of mathematics.

  --Maniac

Thank you very much. Using a TI-83 helped in the round off.