*Use cubic regression to fit a curve through the points (-3,-9),(-1,21),(1,7), and (3,-15).*

(1) If you are allowed to use technology (a graphing calculator, Excel spreadsheet, or web sites such as Wolfram Alpha) you find the answer to be `ax^3+bx^2+cx+d=y` where `a=3/4,b=(-13)/4,c=(-31)/4, d=(69)/4` with `r^2=1` .

(2) To compute by...

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*Use cubic regression to fit a curve through the points (-3,-9),(-1,21),(1,7), and (3,-15).*

(1) If you are allowed to use technology (a graphing calculator, Excel spreadsheet, or web sites such as Wolfram Alpha) you find the answer to be `ax^3+bx^2+cx+d=y` where `a=3/4,b=(-13)/4,c=(-31)/4, d=(69)/4` with `r^2=1` .

(2) To compute by hand, set up the following system of equations:

`a(-3)^3+b(-3)^2+c(-3)+d=-9`

`a(-1)^3+b(-1)^2+c(-1)+d=21`

`a(1)^3+b(1)^2+c(1)+d=7`

`a(3)^3+b(3)^2+c(3)+d=-15`

or

`-27a+9b-3c+d=-9`

`-a+b-c+d=21`

`a+b+c+d=7`

`27a+9b+3c+d=-15`

Solve this system -- you could use substitution, linear combinations, Gaussian elimination, etc...

Using linear combinations, subtract the second, third, and fourth equations from the first equation to get:

`-26a+8b-2c=-30`

`-28a+8b-4c=-16`

`-54a-6c=6`

We use combinations on this system:

`2a+2c=-14` or `a+c=-7`

`-54a-6c=6` or `-9a-c=1`

Again using combinations we get `-8a=-6 => a=3/4` .

Using back substitution we get `c=-7-3/4=-31/4`

etc...

Then the answer is :`ax^3+bx^2+cx+d=y` where `ax^3+bx^2+cx+d=y` where `a=3/4,b=(-13)/4,c=(-31)/4,d=(69)/4```