Tuesday, November 23, 2010

Conclusion

For this project, I worked to find the x and y coordinates of the point of symmetry of a cubic function that goes through 3 given points. To investigate I used three different placements of points- collinear and evenly spaced, collinear and not evenly spaced, and neither collinear nor evenly spaced.
I found that there are infinite points of symmetry for cubic functions going through non-collinear and not evenly spaced points. When graphed, the possible points of symmetry (POS) make a curve very similar to a parabola, only deviating from this shape at and between the given points. (see “Testing non-collinear points) This is very reminiscent of when we graphed the possible vertices of a quadratic function, when given 2 points- then, the POS formed a linear-like shape, with a similar phenomenon of deviation at and between the points given.
For the evenly spaced collinear points, there was only one possible point of symmetry; this was at the middle point. When I found and graphed the POS according to wolfram’s k values, the function seemed linear, except at x = 1, when for some reason the k value of the POS was dependant on the A value of the function. I didn’t have time to investigate this fully- if I did I would try to draw this, to figure out if there was some physical reason it wouldn’t work.
My last step was to try to create a universal formula to determine the constraints of the possible points of symmetry of any cubic function going through three given points. To do this I assigned each of the coordinates of the points with a variable, then I tried to simplify the formula to make it understandable. I gave each point a variable like this for the three points: (x1, y1) (x2, y2) and (x3, y3) I then plugged these values into the normal equations like normal, trying to find h and k in terms of x1 x2 x3 and y1 y2 and y3. The formulas looked like this:
y1=a(x1-h)3+b(x1-h)+k
y2=a(x2-h)3+b(x2-h)+k
y3=a(x3-h)3+b(x3-h)+k
Unfortunately I didn’t get very far combining these by hand, and Wolfram apparently doesn’t do symbolic algebra, so I didn’t end up with a final answer. I did fill a few pages with writing though.

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