Thread: Need help coming up with an equation for three points

Title pretty much says it all. I have three points: (1, 4.077), (2, 11.667), and (3, 945). I need to come up with an equation that explains a relationship between these three points and should predict what the y value with be at x=4. Also, the equation should always be increasing. Much thanks to anyone who can help me out with this.

Applying a quadratic regression, I got:



R^2 is 1, so it's about as accurate as it gets.

With this x=4 results in 2804.076

A what?

What do your mean, "A what?". What's a quadratic regression?.

Using data to derive a quadratic equation. Do you have a calculator?.

I used QuadReg on mine.

I have a calculator, but mine doesn't have a button on it that says Quadreg. Is some sort of hidden function? Or is it only on really high end calculators?

TI-83's do it. What is yours?.

Casio. fx-9750G Plus.

Originally Posted by G-Rex

Title pretty much says it all. I have three points: (1, 4.077), (2, 11.667), and (3, 945). I need to come up with an equation that explains a relationship between these three points and should predict what the y value with be at x=4. Also, the equation should always be increasing. Much thanks to anyone who can help me out with this.

if you're not afraid of a little tedious calculations, you can do it by hand.

assume that a quadratic passes through all three points. let this quadratic be of the form

since we pass though the point (1,4.077) we know that f(1) = 4.077

similarly, we know that f(2) = 11.667 and f(3) = 945

thus we can construct a system of three equations:

.....................(1)
................(2)
.....................(3)

now solve for a,b, and c and you will have your curve

Originally Posted by Jhevon

if you're not afraid of a little tedious calculations, you can do it by hand.

assume that a quadratic passes through all three points. let this quadratic be of the form

since we pass though the point (1,4.077) we know that f(1) = 4.077

similarly, we know that f(2) = 11.667 and f(3) = 945

thus we can construct a system of three equations:

.....................(1)
................(2)
.....................(3)

now solve for a,b, and c and you will have your curve

Let me teach you how to do it with the program "Graph 4.3".
1)Click FUNCTION.
2)INSERT POINT SERIES.
3)Label the points and click OK.
4)Right-click on the left hand side which is the point series.
5)Select INSERT TREDLINE.
6)Select the least-squares approximation you want.
7)You can also gaze upon the "r^2" value which tells you how good approximation is, if r^2=1, then it is perfect.

Originally Posted by ThePerfectHacker

Let me teach you how to do it with the program "Graph 4.3".
1)Click FUNCTION.
2)INSERT POINT SERIES.
3)Label the points and click OK.
4)Right-click on the left hand side which is the point series.
5)Select INSERT TREDLINE.
6)Select the least-squares approximation you want.
7)You can also gaze upon the "r^2" value which tells you how good approximation is, if r^2=1, then it is perfect.

Nice! Thanks for that. it gives the same answer galactus got from his TI-83

Actually, I used a TI-92.