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?
I think I see several mistakes. Did you say that three cubed is nine? It looks like you did.
ReplyDeleteI agree that you don't want solutions where a=0, because the function will no longer be cubic. That should happen only in situations where a cubic cannot contain the given points. (I think you can imagine such a situation.)
In your solving, you arbitrarily chose to put the point of symmetry at x=1. You should make that clear in the work. So you are saying, given these three points, can the point of symmetry go at 1 and if it does, what is the corresponding H value? You should always put some words with the equations to show what you are trying to solve for.
As I imagine you noticed, (1,1) is collinear with the other three points. So that would suggest that the point of symmetry cannot go at H=1. First correct the mistake with three cubed, although I am not sure if that will fix it all. But start there.