The red markers show where the original three points are, and each is a possible POS. This graph seems to indicate that the function isn't cubic, and is linear. this means that instead of infinite POS, this graph only has one, at (0,0).
If the points are collinear but not evenly spaced:
I used Wolfram Alpha with the points (-2,-2) (2,2) and (3,3)
to solve for k based on inputs of h.
h: k:
-4 -4
-3 -3
-2 -2
-1 -1
0 0
1 6A+1 Why is this here? I can't see any physical reasons this should be.
2 2
3 3
4 4
when h = 1, k = 6A+1 Why is this here? I can't see any physical reasons this should be. is it impossible to have a POS here?
I don't understand this one. If you have three evenly spaced collinear points, then the point of symmetry can ONLY be at the middle point, right? isn't that what you found before when you were solving and A kept coming out to be zero? The graph for this should be a single point. I guess I am not sure what you graphed.
ReplyDeletealso, did you do three collinear points that are NOT evenly spaced?
ReplyDeleteI found that if A came out to be zero, the k value was still left as a linear function of h. this means that when I solved the systems of equations for each x input there was still a k output, but when I solved for the equation of that POS it ended up as a linear function, since the cubed part was eliminated by the A=0.
ReplyDelete