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'Cubic Spin' printed from https://nrich.maths.org/
Thank you for these solutions to Pierce from Tarbert Comprehensive
School; Hyeyoun from St Paul's Girls' School, London; Dorothy from
Madras College, St Andrew's, Scotland and Yatir from Israel.
The graph of the cubic $y =x^3-6x^2+9x+1$ 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 $v = u^3 - 3u$ which is
symmetric about the origin.
The graph of the cubic $y=2x^3+3x^2+5x+4$ 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 $v =
2u^3+7u/2$ 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 $f(x)= ax^3
+ bx^2 +cx^1 + dx^0$. What kind of transformations can we perform
on it that do not alter the shape of the graph?
Firstly, consider the transformation $T_1$ 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 $T_2$ 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
$$\eqalign{ -f(x)&= -ax^3 - bx^2 -cx^1 - dx^0\cr f(-x)&=
-ax^3 + bx^2 -cx^1 + dx^0.}$$
We can see that if the $x^2$ and $x^0$ 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
$T_1$ and $T_2$ which I outlined at the start. Firstly, we can get
rid of the $x^2$ term as follows. Consider the transformation $T_2$
$$\eqalign{ f(x+h)&= a(x+h)^3 + b(x+h)^2 + c(x+h)^1 + d(x+h)^0
\cr &= ax^3 +(3ah+b)x^2+ (3ah^2 + 2bh+ c)x^1 +(ah^3 + bh^2 +
d)x^0}.$$
We can choose $h$ so that the co-efficient of $x^2$ is zero i.e
$(3ah+b) = 0$ which implies that $h=(-b/3a)$. So by applying the
transformation $T_2$ 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 $x^2$ term. Now by applying the transformation $T_1$ to
this new curve with $m=-(ah^3 +bh^2+d)$ we can remove the $x^0$
term. The overall translation may be defined as follows: Map
$(x,f(x))$ onto $(x, f(x+h)-(ah^3 +bh^2+d))$ where $h=-b/3a$. After
this translation what is the equation of the graph? The equation is
$$\eqalign{ g(x)&= f(x+h)- (ah^3+bh^2+d)\cr &= ax^3 +
(3ah^2 + 2bh+ c)x^1.}$$
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).
$$\eqalign {g(-x)&= -ax^3 -(3ah^2 + 2bh+ c)x^1 \cr -g(x)&=
-ax^3 -(3ah^2 + 2bh+ c)x^1}$$
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 $(ah^3 +bh^2+d)$ units, and the
origin is now our centre of rotation, the original centre of
rotation must have been at $((-b/3a), (ah^3 +bh^2 +d) )$.
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:
$$\eqalign {y&=ax^3+bx^2+cx+d \cr y'&=3ax^2+2bx+c.}$$
When $y'=0$ we have $3ax^2+2bx+c=0$ so the turning points are given
by:
$$\eqalign{ x_1&=(-b+\sqrt {(b^2-3ac)/(3a)}\cr
x_2&=(-b-\sqrt {(b^2-3ac)/(3a)}.}$$
The midpoint between the turning points is given by :
$$\eqalign{ x_{mid}&=(x_1+x_2)/2 =-b/(3a) \cr y_{mid}
&=(2b^3-9abc+27a^2d)/(27a^2).}$$
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 $(x_{mid},y_{mid})$).
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
$$y=ax^3+bx^2+cx+d \quad (1)$$
The transformation formulas are:
$$\eqalign{v&=y-y_{mid}= y-(2b^3-9abc+27a^2d)/(27a^2)\cr
u&=x-x_{mid}=x+b/(3a).}$$
Let's plug them in (1) to get:
$$y= a\left(u- {b\over 3a}\right)^3+b\left(u-{b\over
3a}\right)^2+c\left(u-{b\over 3a}\right)+d =au^3 +
\left({3ac-b^2\over 3a}\right)u + \left( {2b^3-9abc+27a^2d\over
27a^2}\right).$$
After simplifications we get:
$$v=g(u)= au^3+\left({3ac-b^2\over 3a}\right)u.$$
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.)