# Creating a Cubic Equation with Coordinates

• Apr 19th 2010, 04:12 PM
Nerdilicious
Creating a Cubic Equation with Coordinates
Solved. Thanks!
• Apr 19th 2010, 04:18 PM
pickslides
To make a cubic model $y = ax^3+bx^2+cx+d$ you will need at least $4$ points. With these points you can find $a,b,c,d$ using basic algebra or with matrices.

What other information do you have?
• Apr 19th 2010, 04:21 PM
Nerdilicious
Solved.
• Apr 19th 2010, 04:26 PM
pickslides
Before you attempt substitution do you have a start year (or initial point) for the data?

Otherwise $x=0$ will mean the year $0$ and maybe not representative of what you are trying to model.
• Apr 19th 2010, 04:29 PM
Nerdilicious
No, the data starts at 1950. Could I possibly translate the data so that 1950 is my start date?
• Apr 19th 2010, 04:40 PM
pickslides
Quote:

Originally Posted by Nerdilicious
No, the data starts at 1950. Could I possibly translate the data so that 1950 is my start date?

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 $(0, 554.8)$

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

Using $(10, 657.5)$

$y = ax^3+bx^2+cx+d$ becomes $657.5 = a(10)^3+b(10)^2+c(10)+d= \dots$
• Apr 19th 2010, 05:49 PM
Nerdilicious
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, 05:53 PM
pickslides
Quote:

Originally Posted by Nerdilicious
So that would make d=554.8, right?
Then would I substitute that in the other equations, or should I use matrices?

Correct. You need to substitute either way. After this you can decide if you want to use simple algebra or find the inverse of a $4\times 4$ matrix.
• Apr 19th 2010, 06:05 PM
Nerdilicious
Thank you so much =D
You were a huge help!