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.
it's possible that given those three points, the point of symmetry cannot be at H=1. Maybe you should try a different H value.
ReplyDeleteYou can also try a different set of points. Move just one of them and see what happens when you try to solve.
ReplyDeleteYou may have created a special case with your particular set of points. But that's okay because it's good to be able to talk about the special cases too.