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 Cubic Quandary (Posted on 2014-09-13)
Prove that graphs of all cubic functions are rotationally symmetrical about their point of inflection, or provide a counter-example.

 No Solution Yet Submitted by Dustin No Rating

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 Solution without calculus. Comment 2 of 2 |
An odd function is rotationally symmetric about the origin.  If we can show that there is a translation of a general cubic that creates an odd function we are done.

Given f(x) = ax³+bx²+cx+d
Find the point (h,k) such that y=f(x-h)+k is an odd function.

f(x-h)+k = a(x-h)³+b(x-h)²+c(x-h)+d+k
y = ax³+(b-3ah)x²+(3ah²-2bh+c)x+(-ah³+bh²-ch+d+k)

To be odd the coefficient of x² and the constant term must both be zero.
b-3ah=0

h=b/(3a)
k=b³/(27a²)-b³/(9a²)+bc/(3a)-d

So (h,k) is the point of rotational symmetry.

I suppose one could argue I did not show this is the point of inflection.  The term is from Calculus, but even Geometrically we can see this is the point we are talking about.

 Posted by Jer on 2014-09-16 10:01:59
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