## The “Center” of a Polynomial Graph

The article Polynomial Graphs and Symmetry by Geoff Goehle and Mitsuo Kobayashi in the latest The College Mathematics Journal caught my interest. The authors “give conditions under which” a polynomial function “has rotational or horizontal symmetry.”  It occurred to me that if a polynomial function had a point of rotational symmetry then that point would be the “best” point from which to view its graph.  The question naturally arose of the location of the best point from which to view a polynomial function with horizontal symmetry  or the location of the best point from which to view any polynomial.

So, my problem became: How should I center the graph  of a polynomial function in a rectangular viewing area if I am not interested in its absolute location, that is not interested in its roots or local maxima or minima and the like?  Whatever method I come up with should result, in the case of rotational symmetry, the point of symmetry being in the center of the viewing rectangle and in the case of horizontal symmetry the axis of symmetry splitting the rectangle vertically in two.

Let a general polynomial function be given by $f(x)=a_{n}x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$.  As Goehle and Kobayashi conclude, if a polynomial has a center of symmetry then the x-coordinate of that point called $s$ is $\frac{-a_{n-1}}{na_n}$.  This value $s$ is called the center of gravity of the function (As is well known, $s=\frac{-a_{n-1}}{a_n}$ is the sum of the roots of the function.).  The authors refer to this paper by F.Irwin and H.Wright for an explanation.  For any polynomial, symmetric or not, I will use the value of $s$ to center the graph in the horizontal direction.

With this idea in mind I thought to “center” the graph in the vertical direction by finding the center of gravity of the y-values of the local minimas and maximas of the polynomial. This is relatively easy to do.  Find $f'(x)$.  Solve $f'(x)=0$ for all $n-1$ solutions including the complex ones.  Call them $r_1,r_2,...,r_{n-1}$.  Then calculate the center of gravity of the $f(r_i)$s by adding up the $f(r_i)$s and dividing by $n-1$.  Adjust the graph vertically by this amount.  Note that complex roots are ok since they appear as conjugates and the imaginary parts add to zero.  This also works with complex roots substituted into the original polynomial since $f(\bar{z})=\bar{f(z)}$.

Here are some sample graphs.  The method certainly works for polynomials with rotational symmetry but I don’t know if polynomials with horizontal symmetry centered by this method are aesthetically pleasing.  What do you think?

All cubic polynomials have rotational symmetry.  Here is one.  The one tenth factor allows the graph to have x-axis scales and y-axis scales of about the same size.  The blue graph is “centered” and the green graph is the original.

Cubic Polynomial Graph with Axes Showing

And without the axes showing.

Cubic Polynomial without Axes

Here is a symmetric quartic polynomial.

Symmetric Quartic Polynomial

And finally here is an asymmetric quintic polynomial.

Asymmetric Quintic Polynomial