7/21/2019 Symmetries of Polynomial Graphs in Two Dimension

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Symmetries of Polynomial Graphs in Two Dimension 

Two kinds of symmetries are relevant in this discussion:

  Rotational symmetry, which requires a centre of rotation (e.g., the origin with associated

coordinates (0,0)), an angle of rotation (e.g., 180o,

), and an orientation of rotation (e.g.,

clockwise or anti-clockwise).

  Reflective symmetry, which requires a line of reflection (e.g., y = a, y = mx + c).

We restrict ourselves to polynomials in one variable, x , where the degree of the polynomial is the

largest exponent of x . Non-integer and negative powers are not explored here, likewise for

multivariable and infinite polynomials.

Example 1

3x2 + 2x + 1 is a polynomial of degree 2

A polynomial may be drawn in the Cartesian coordinates using the regular notation y = f(x)

Example 2 

If y = f(x) = 3x2 + 2x + 1 then the following graph represents this polynomial

In this short piece we hope to demonstrate the following:

 

Quadratic polynomials have reflective symmetry about the line normal to their extrema.

  Cubic polynomials have rotational symmetry of  about their points of inflection.

  Linear polynomials have rotational symmetry of  about any point on their curve. They also

have reflective symmetry through the line normal to their curve.

  Constants have rotational symmetry of  about any point on their curve. They also have

reflective symmetry through the line normal to their curve.

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Quadratic Polynomials

Given an arbitrary quadratic polynomial  =   + +  one can determine the extrema

by equating the first derivative to zero

 ’ = 2 +  

2 + = 0 

=  2 

The associated y coordinate is given by     =  

+   +  

= 4  

2 +  

= 4 +  

Let us call the coordinates of the extrema (a,b).

If we define a new function = +  then it is clear that we have translated our original

quadratic such that the extrema now lie at the origin (0,0) rather than (a,b).

 = (   2) + (  

2) + 4 +

 

 =       +  

4 +  2 + 

4 

 =   + 

4 +  

2 + 

4 

 =  

Recall, a function F(x) has reflective symmetry through the y-axis if F(-x) = F(x)

 =  

=  =   ∎ 

Alternatively

If we show  + =  (where a is the x-coordinate of the extrema and e is an arbitraryvalue) then it is also clear that the function has reflective symmetry through the normal to the

extrema.

  ( 2 + ) = ( 

2 + ) + (  2 + ) + =  

4  22 + +  

2 +  

  ( 2  ) = ( 

2  ) + (  2  ) + =  

4 + 22 +  

2 +  

  ( 2 + ) = ( 

2  ) ∎ 

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Derivation of  + =  

Suppose that  has reflective symmetry through the line =  

Let = + .

 has reflective symmetry about the y-axis, hence =  

So  + = +  So  = +  

To show the converse simply reverse the steps.

Cubic Polynomials 

Given an arbitrary cubic polynomial   =  +   + +  one can determine the points

of inflections by equating the second derivative to zero

 ’’ = 6 + 2 

6 + 2 = 0  =  3 

The associated y coordinate is given by     =  

+   +  

+  

=   27 +  

9  3 +  

Let us call the coordinates of the point of inflection (a,b).

If we define a new function = +  then it is clear that we have translated our original

cubic such that the point of inflection now lies at the origin (0,0) rather than (a,b).

 = (   3) + (  

3) + (   3) +  

27 +   9  

3 +  

 =       +

3    27 +   2

3   +   9 +

3 + +   27   

9 + 3 

 

 =    + 3    

27 +   23   +  

9 + 3 +  

27    9 +

3  

 =

 

2

3   +

3   +  

 =  +  3  

Recall, a function F(x) has rotational symmetry of  about the origin if F(-x) = - F(x)

 =  +  3

 

=    3

=

 +  

3  

=   ∎ 

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Alternatively

If a function has rotational symmetry of  about a point (a,b) then 2 + = 2Derivation

Suppose  has rotational symmetry of  about (a,b) then g(x) = f(x+a)-b has rotational symmetry

of  about (0,0)

As such  =  

Hence

  + =   +      + = + +    + + +  = 2

   + +  = 2 

→    + 2  = 2 

To show the converse simply reverse the steps. 

Utilisation

  =  +   + +  

=  3 

=   27 +  

9  3 +

 

=   227 

3 +  

  + 2  

=   +   + + + (2 ( 3) ) + ( 2 (  

3) ) + (2 ( 3) )+  

=  +  + +   827  4

3  2   + 49 + 4

3  +   23   +  

=  + 2   827  4

3  2 + 49 + 4

3  +   23  

= 2   8

27 + 4

9  2

3 

= 2   427 + 29  3 

= 2 +   227 

3 

= 2 ∎ 

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Linear Polynomials and Constants

For the constant  =  it is trivial to show reflective symmetry;  = = .

Rotational symmetry is likewise trivial:

Let  = + –   with (a,c) being an arbitrary point on the curve. 

 = 0 

=  

For  = +  

Let = + –  with (a, f(a)) being an arbitrary point on the curve.

 = + +   =

 

 =  =

 

=  

Reflective symmetry may be demonstrated as follows.

For  = +  

Let  = +  with (a, f(a)) being an arbitrary point on the curve.

 = + +   =  

Now rotate  clockwise by an angle of θ = tan−  ∗ (or the equivalent appropriate angle when

A<0) to give a constant line  = 0. Such a line has reflective symmetry from above.

*If θ = tan−  → = tan → =    

cos =   1√   + 1 ;sin =  

√   + 1 

The rotation matrix for a clockwise rotation about the origin is given by

( cos sinsin   cos) 

Hence we have

( cos sinsin   cos)    

′ =   √   + 1 +  

√   + 1 

=   √   + 1 +  

√   + 1 = 0 

Points to be explored 

 

Under what conditions do such symmetries exist for polynomials of degree greater than 3,multivariable polynomials, polynomials in 3 dimensions, etc.