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.

Testing Collinear points

This is what I got when I tried using systems of equations to find possible POS (points of symmetry) for three collinear evenly spaced points. the points are (-2,-2) (0,0) and (2,2).

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?

Testing non-collinear points

I plotted points (-2,-2) (2,2) and (5,1) then plugged in x values from -7 to 18 to find the y values of points of symmetry at those x values. I then graphed the (x,y) coordinates of the points.
 

The red markers show where the original three points are, and each is a possible POS.

I'm assuming that when the points can be connected by a curved line, the possible POS form a shape opening in the same direction as the curve. for example, the points in the graph above could be connected by a downward facing shape, and the trend of the POS is also a downward opening curve.

Abstract

My project investigates what the constraints are for the location of the point of symmetry of a cubic function give 3 points. I looked into this by putting 3 points on graphs in different places and patterns, and solving for given x values. I did the first few systems of equations by hand then used Wolfram Alpha. I found that in some situations there was a y value for every x input, for others there were limits.

New Stuff

I did a ton of new calculations using the points (-2,-2) (2,2) and (5,1).
There was so much writing that I didn't have time to type it up here, so I will show you in class instead.

Much Better...? or not.

Whoops! I found my mistake, it was a simply careless error. (-3)^3 does not equal -9!

I plotted points at (-2,-2) (0,0) and (2,2). Then I tried plugging in an x value for the point of symmetry, and solved from there. The base equation I'm using is: f(x) = a(x-h)^3 +b(x-h) +k
I used the information from the coordinates of each point and made and equation with each:
f(2):      2 = a(2-1)^3 +b(2-1) +k    simplify to: 2 = a+b+k
f(-2):   -2 = a(-2-1)^3 +b(-2-1) +k   simplify to: 2 = -27a-3b+k+4
f(0):      0 = a(0-1)^3 +b(0-1) +k   simplify to: 2 = -a-b+k+2

Then I used system of equations to eliminate variables by combining the equations.
a+b+k = -27a-3b+k+4
a+b = -27a-3b+4
28a+4b =4
> 14a+2b = 2

a+b+k = -a-b+k+2
> 2a+2b = 2

14a+2b = 2a+2b
12a = 0
a = 0
Dammit, not again! I still can't figure out what I did wrong. I'll be after school tomorrow I guess. Sorry I haven't gotten as far on this project as I would have liked, and as I should have.

Still having issues

I plotted points at (-2,-2) (0,0) and (2,2). Then I tried plugging in an x value for the point of symmetry, and solved from there. The base equation I'm using is: f(x) = a(x-h)^3 +b(x-h) +k
I used the information from the coordinates of each point and made and equation with each:
f(2):      2 = a(2-1)^3 +b(2-1) +k    simplify to: 2 = a+b+k
f(-2):   -2 = a(-2-1)^3 +b(-2-1) +k   simplify to: 2 = -9a-3b+k+4
f(0):      0 = a(0-1)^3 +b(0-1) +k   simplify to: 2 = -a-b+k+2

Then I used system of equations to eliminate variables by combining the equations.
a+b+k = -9a-3b+k+4
a+b = -9a-3b+4
10a+4b = 4
> 5a+2b = 2

a+b+k = -a-b+k+2
> 2a+2b = 2

5a+2b = 2a+2b
3a = 0
>> a = 0  Wait, what? this is wrong; if a = 0 then the function will be just linear.. where did I mess up?

10a+4b = 4
10(0)+4b = 4
>> b = 1

2 = a+b+k
2 = 1+k
>> k = 1

coordinates for point of symmetry: (1,1)
equation for this function: f(x) = (x-1)+1
                                      >> f(x) = x   
How do I keep the function cubic when I solve for k?

Step one: a little trouble

My first step was going to be to put three colinear points on a graph, and find a series of points of symmetry. I then discovered that I'm not sure just how to find the point of symmetry, when given an x value and 3 other points.

Just the Beginning

My project:
"Can you take three given points on the coordinate plane and describe the set of x, y pairs that could potentially represent the point of symmetry of a cubic that contains those three points?"

 My first step was to draw a few cubic functions with the online function grapher. Here's what I made:

As you can see, I marked three points that both cubic functions go through then I marked the point of symmetry of both functions. Since any given three points cannot constrain a cubic function, there will always be multiple (and in fact infinite) possibilities for the point of symmetry.