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Finding the equation to a non-linear set of points

by Joshk80k
Tags: equation, nonlinear, points
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Joshk80k
#1
May7-10, 09:31 PM
P: 17
1. The problem statement, all variables and given/known data

Determine the equation of the following points:

(12. 270)
(14, 300)
(16, 330)
(18. 380)

2. Relevant equations

3. The attempt at a solution

I realized right away that this is not a linear set of points, so it wouldn't work out to just find the slope between two points and to find the intercept. I know the answer won't be in the format y = mx+b, but I'm not really sure how to solve this.

I guess what's bothering me is the fact that the equation jumps 20 after a seemingly normal pattern. Any tips as to what direction I should take this problem next?
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rock.freak667
#2
May8-10, 12:20 AM
HW Helper
P: 6,207
Do you know how to do linear regression analysis? Because that sounds like what you want to do.
Joshk80k
#3
May8-10, 12:55 AM
P: 17
No, I don't know how to do that - I'll look it up though, thanks =)

zachzach
#4
May8-10, 06:00 AM
P: 258
Finding the equation to a non-linear set of points

This is either a linear fit or a piecewise function of two lines.
HallsofIvy
#5
May9-10, 06:37 AM
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P: 39,339
A "linear regression" line or "least squares line" is a line that does NOT go through all the points but comes "closest" in the sense of the average distance from each point to the line being smallest.

You could, as zachach suggests, use a piecewise linear function- although you might need 3, not two pieces- a line from (12. 270) to (14, 300), a line from (14, 300) to (16, 330), and a line from (16, 330) to (18. 380).

It is also true that there exists a unique n-1 or less degree polynomial through any given n points. Here you have four points so there exist a unique cubic (or less) polynomial whose graph passes through those four points.

Write the polynomial as [math]y= ax^3+ bx+ c+ d[/math] and put in the given x and y values to get four linear equations to solve for a, b, c, and d.


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