Solved. Thanks!

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- Apr 19th 2010, 03:12 PMNerdiliciousCreating a Cubic Equation with Coordinates
Solved. Thanks!

- Apr 19th 2010, 03:18 PMpickslides
To make a cubic model $\displaystyle y = ax^3+bx^2+cx+d$ you will need at least $\displaystyle 4$ points. With these points you can find $\displaystyle a,b,c,d$ using basic algebra or with matrices.

What other information do you have? - Apr 19th 2010, 03:21 PMNerdilicious
Solved.

- Apr 19th 2010, 03:26 PMpickslides
Before you attempt substitution do you have a start year (or initial point) for the data?

Otherwise $\displaystyle x=0$ will mean the year $\displaystyle 0$ and maybe not representative of what you are trying to model. - Apr 19th 2010, 03:29 PMNerdilicious
No, the data starts at 1950. Could I possibly translate the data so that 1950 is my start date?

- Apr 19th 2010, 03:40 PMpickslides
I would do that making

(1950, 554.8)

(1960, 657.5)

(1985, 1070.0)

(1995, 1220.5)

into

(0, 554.8)

(10, 657.5)

(35, 1070.0)

(45, 1220.5)

Now one by one substitute these values into the equation in post #2.

Using $\displaystyle (0, 554.8)$

$\displaystyle y = ax^3+bx^2+cx+d$ becomes $\displaystyle 554.8 = a(0)^3+b(0)^2+c(0)+d= \dots$

Using $\displaystyle (10, 657.5)$

$\displaystyle y = ax^3+bx^2+cx+d$ becomes $\displaystyle 657.5 = a(10)^3+b(10)^2+c(10)+d= \dots$ - Apr 19th 2010, 04:49 PMNerdilicious
So that would make d=554.8, right?

Then would I substitute that in the other equations, or should I use matrices? - Apr 19th 2010, 04:53 PMpickslides
- Apr 19th 2010, 05:05 PMNerdilicious
Thank you so much =D

You were a huge help!