Thank you for these solutions to Pierce Geoghegan, Tarbert Comprehensive School; Hyeyoun Chung, St Paul's Girls' School, London; Dorothy Winn, Madras College, St Andrew's, Scotland and Yatir Halevi. The graph of the cubic y = x3−6x2+9x+1 has rotational symmetry about the point(2,3) and using the translation u=x−2, v=y−3 the function in the new coordinates becomes v = u3− 3u which is symmetric about the origin.

The graph of the cubic y=2x3+3x2+5x+4 has rotational symmetry about the point of inflection (-1/2, 2) and using the translation u=x+1/2, v=y−2 the function in the new coordinates becomes v = 2u3+7u/2 which is symmetric about the origin.

A NON CALCULUS METHOD FROM PIERCE

In this proof I will show that all cubic polynomials are rotationally symmetric. Firstly, I'll discuss some transformations in the plane.

Suppose we are given the graph of the cubic polynomial f(x) = ax3 + bx2 +cx1 + dx0. What kind of transformations can we perform on it that do not alter the shape of the graph?

Firstly, consider the transformation T1 which maps (x,f(x)) onto (x,f(x)+m), for some constant real number m. This function has the effect of moving the entire graph either up or down in the plane, depending on whether m is positive or negative, without changing the shape of the graph. Secondly, consider the transformation T2 which maps (x,f(x)) onto (x,f(x+h)) for some constant real number h. This has the effect of moving the graph of the cubic either left or right in the co-ordinate plane, depending on whether h is positive or negative, again without changing the shape of the graph. What use are these transformations? The answer lies in the fact that if a function satisfies the equation f(−x)=−f(x) then it is rotationally symmetric about the origin (with an angle of rotation equal to pi radians.) In the above general equation of the cubic
−f(x)
= −ax3 − bx2 −cx1 − dx0
f(−x)
= −ax3 + bx2 −cx1 + dx0.
We can see that if the x2 and x0 terms weren't there, then the cubic would satisfy the equation f(−x)=−f(x) and so the graph would be rotationally symmetric. Is there any way to get rid of these terms by transformations which do not alter the shape of the graph? The answer is that there is using the two transformations T1 and T2 which I outlined at the start. Firstly, we can get rid of the x2 term as follows. Consider the transformation T2
f(x+h)
= a(x+h)3 + b(x+h)2 + c(x+h)1 +d(x+h)0
= ax3 +(3ah+b)x2+ (3ah2 + 2bh+c)x1 +(ah3 + bh2 + d)x0
.
We can choose h so that the co-efficient of x2 is zero i.e (3ah+b) = 0 which implies that h=(−b/3a). So by applying the transformation T2 with h=(−b/3a) i.e by shifting the graph in the right/left direction in the plane by h = (−b/3a) units we have removed the x2 term. Now by applying the transformation T1 to this new curve with m=−(ah3 +bh2+d) we can remove the x0 term. The overall translation may be defined as follows: Map (x,f(x)) onto (x, f(x+h)−(ah3+bh2+d)) where h=−b/3a. After this translation what is the equation of the graph? The equation is
g(x)
= f(x+h)− (ah3+bh2+d)
= ax3 + (3ah2 + 2bh+ c)x1.
where h=−b/3a. This curve, g(x), is exactly the same shape as the original graph (since it was derived from transformations which did not alter the shape of the graph).
g(−x)
= −ax3 −(3ah2 + 2bh+c)x1
−g(x)
= −ax3 −(3ah2 + 2bh+ c)x1
therefore g(−x)=−g(x) and the graph is rotationally symmetric so the original graph was also rotationally symmetric. It is interesting to note that since we have effectively moved the graph back (−b/3a) units and down (ah3 +bh2+d) units, and the origin is now our centre of rotation, the original centre of rotation must have been at ((−b/3a),(ah3 +bh2 +d) ).

A SOLUTION USING CALCULUS FROM YATIR:

If the graph of a cubic function has a rotational symmetry, then after the rotation the minimum becomes that maximum and vice versa. In that case the point-of-symmetry must be the midpoint between the minimum and the maximum. I will prove that this point is also the point of inflection, and if the function doesn't have a maximum and minimum then this point will serve as the point-of-symmetry. First, to find the point:
y
=ax3+bx2+cx+d
y '
=3ax2+2bx+c.
When y ' =0 we have 3ax2+2bx+c=0 so the turning points are given by:
x1
=(−b+
√
 
(b2−3ac)/(3a)
 
x2
=(−b−
√
 
(b2−3ac)/(3a)
 
.
The midpoint between the turning points is given by :
xmid
=(x1+x2)/2 = −b/(3a)
ymid
=(2b3−9abc+27a2d)/(27a2).
The point of inflection of the function is a point at when the second derivative y ' ' =6ax+2b is zero which happens when x=−b/(3a) which is exactly the same point. Once that we have found our potential point, lets prove that it acts as point of symmetry of rotation. In order to do that, we'll have to translate the graph of the function until its point-of-symmetry is the origin (or move the axes until the origin becomes (xmid,ymid)). Then if f(−x)=−f(x) it is the point of symmetry because a function has rotational symmetry around the origin if f(−x)=−f(x). Lets call the new axes u and v (instead of of x and y. Our original function was
y=ax3+bx2+cx+d     (1)
The transformation formulas are:
v
=y−ymid = y−(2b3−9abc+27a2d)/(27a2)
u
=x−xmid=x+b/(3a).
Let's plug them in (1) to get:
y = a
 u−  b
3a

 3

 
 +b
 u−  b
3a

 2

 
 +c
 u−  b
3a

 +d = au3 +
 3ac−b2
3a

 u +
 2b3−9abc+27a2d
27a2

 .
After simplifications we get:
v=g(u) = au3+
  3ac−b2
3a

 u.
The graph of this function has rotational symmetry about the origin because g(−u)=−g(u) and hence the general cubic polynomial has rotational symmetry. (Notice that the constant term turns out to be zero because our new function passes through the origin.)