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Stuck help find points

catwalk

New member
Sep 29, 2020
4
curve.PNG
I know this is the equation I have to use a(x - h) 2 + k, but dont know what points to use and how to covert to vertex form
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
Do you know what "best fit" means? There are several different types of "best fit". Which have you learned?
 

catwalk

New member
Sep 29, 2020
4
Do you know what "best fit" means? There are several different types of "best fit". Which have you learned?
No, im not even sure what best fit means. I guess whatever is the simplest
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
Where did you get this problem? Did some evil teacher assign a problem about "best fit" without giving any instruction related to that?

(Don't you just hate when they do that?)

Have you tried looking up "best fit" in your textbook or online?
 

catwalk

New member
Sep 29, 2020
4
Where did you get this problem? Did some evil teacher assign a problem about "best fit" without giving any instruction related to that?

(Don't you just hate when they do that?)

Have you tried looking up "best fit" in your textbook or online?
Haha yeah thats what happened.

I've looked online but still don't know what to do
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
716
The x-values of each data point seem to be easy ... 0, 2, 4, 5, 7, 9, 10, 12, and 14

You’ll just have to make your best eyeball estimate for the y-values.

Once you get a reasonable list of coordinates, do a google search for an online quadratic regression calculator and see what that can do for you.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
Since you say you want a quadratic of the form a(x-h)^2+ k (notice the "^" for the exponent. That's clearer than just "x2".) You need to determine a, h, and k, three numbers. For that, you need three equations. The simplest thing to do is to choose three "data points" to get those equations. I recommend three widely spaced points so as not to exagerate any "anomaly" (like those two successive points with almost the same y which cannot happen in parabola). To the far left we have (0, 0), in the center, (7, 350), and on the right, (14, 80). In order that the parabola pass through those points we must have a(0- h)^2+ k= ah^2+k= 150, a(7- h)^2+ k= 350, and a(14-h)^2+ k= 80. Solve those equations for a, h, and k. Subtracting the first and third equations from the second eliminates "k" giving a(7- h)^2- ah^2= 200 and a(7-h)^2- a(14- h)^2= 270. Calculating the squares we have 49a- 14ha+ ah^2- ah^2= (49- 14h)a= 200 and 49a- 14ah+ ah^2- 196a+ 28ah- ah^2= a(14h- 147)= 270. Divide the second equation by the first to eliminate a, (14h- 147)/(49- 14h)= 270/200 and solve that equation for h.

A more "sophisticated" method would the "least squares" method which is harder but uses all of the data. With y= a(x-h)^2+ k, for each data point (xi, yi) the "error" is yi- a(xi- h)^2- k, the difference of the data value and the computed value. For example, for the first ^data point, (0, 0) the "error" is 0- a(0- h)^2- k= -ah^2- k. For (7, 350) the "error" is 350- a(7- h)^2- k. If we were to simply add those we might have a negative error cancelling a positive error that we do not want to happen. So square each and then sum. To find the smallest possible error, take the derivative of that sum of squares with respect to a, h, and k and set the derivatives equal to 0. Again that gives three equations to solve for a, h, and k.