<s>[INST] Will the perpendicular bisector between the line connecting two cubic roots of the same arc never intersect its turning point? The quadratic graph: $$ f(x) = (x+2)(x+1)$$
would have a midpoint between its roots at $x = -1.5$. This line would intersect its turning point.


However the cubic graph: $$ f(x) = (x+1)(x-2)(x+3)$$
would have a midpoint between the roots -1 and -3 at $x = -2$ , which does not intersect the turning point as shown below.

Is there any particular proof or intuition for this?
 [/INST] We can take as a requirement that the perpendicular bisector of two adjacent zeroes of a polynomial pass through the "turning point" [relative extremum] between them and see what polynomials may satisfy it.  For a quadratic polynomial $ \ x^2 + a_1 x  + a_0 \ $ (it suffices to use a monic polynomial, since a "leading coefficient" $ \ a_2 \ $ will only affect the "vertical scale" of the function curve), we will place  the zeroes at $ \ x \ = \ c \ \ , \ $ $ d \ = \ c + 2 \Delta \   \ , \ $  with the turning point at $ \ x \ = \ c + \Delta \ = \ d - \Delta \ \ . \ $  The value of the quadratic function at the turning point is then
$$   (c + \Delta)^2 \  + \ a_1·(c + \Delta) \ + \ a_0 \ \ = \ \ (d - \Delta)^2 \  + \ a_1·(d - \Delta) \ + \ a_0 $$
$$ \Rightarrow \ \ c^2 \ + \ 2c \Delta \ + \  \Delta^2 \ + \ a_1  c \ + \ a_1 \Delta \ + \ a_0 \ \ = \ \ d^2 \ - \ 2d \Delta \ + \  \Delta^2 \ + \ a_1  d \ - \ a_1 \Delta \ + \ a_0  $$
$$ \Rightarrow \ \ c^2 \ + \ 2c \Delta \  + \ a_1 c \ + \ a_1\Delta \ + \ a_0  \ \ = \ \ d^2 \ - \ 2d \Delta  \ + \ a_1d \ - \ a_1\Delta \ + \ a_0 \ \ ,  $$
or, since $ \ c^2 + a_1c + a_0 \ = \ d^2 + a_1d + a_0 \ = \ 0 \ \ , \ $
$$  2c \Delta  \ + \ a_1\Delta   \ \ = \ \  -  2d \Delta   \ - \ a_1\Delta \ \ .  $$
As we are taking $ \ \Delta \ \neq \ 0 \ \ , \ $ we obtain $ \ a_1 \ = \ -(c + d) \ \ , \ $ with no dependence on the separation between the adjacent zeroes.  The location of the turning point is given by setting the first derivative equal to zero, thus, $ \ 2  x + a_1 \ = \ 0 \ \Rightarrow \ x \ = \ -\frac12·a_1 \ = \ \frac{c \ + \ d}{2} \ \ , \  $  which is just where we set it to be by the bisector requirement.  (Unsurprisingly, a little manipulation would show this to be the $ \ x-$coordinate of the "vertex" of the quadratic parabola.)
If we make a similar calculation for a general monic cubic polynomial $ \ x^3 + a_2 x^2 + a_1 x  + a_0 \ \ , \ $ the bisector requirement produces
$$   (c + \Delta)^3 \  + \ a_2·(c + \Delta)^2 \   + \ a_1·(c + \Delta) \ + \ a_0 $$ $$ = \ \ (d - \Delta)^3 \  + \ a_2·(d - \Delta)^2 \  + \ a_1·(d - \Delta) \ + \ a_0 $$
$$ \Rightarrow \ \ 2 a_2 (d + c)   \ + \ 2 a_1   \ + \ 3 (c^2 + d^2) \ - \ 3 (d - c) \Delta  \ + \ 2 \Delta^2 \ \ = \ \ 0 \ \ , \ \ \Delta \ \neq \ 0 \ \ , $$
indicating that there is not only a dependence upon the coefficients of the polynomial, but also upon the separation of adjacent zeroes.  The turning point(s) of the curve for the cubic polynomial (when they exist) are found from
$$ 3 x^2 \ + \ 2 a_2 x \ + \ a_1 \ \ = \ \ 0 \ \ \Rightarrow \ \  x \ \ = \ \ \frac13 · \left(-a_2 \ \pm \ \sqrt{a_2^2 \ - \ 3 a_1} \right) \ \ , \ $$
which do not have this relation to the separation of its zeroes.  So it would be only for very contrived conditions at best that the perpendicular bisector between adjacent zeroes could pass through a turning point of the cubic polynomial curve.
This dependence on separation of zeroes will persist for all higher-degree polynomials as well, except in particular circumstances.  For instance, the quartic polynomial $ \ x^4 \ - \ (c^2 + d^2) x^2 \ + \ c^2 d^2 \ \ \ $ $ = \ (x^2 - c^2)·(x^2 - d^2) \ \ , \ \ c^2 \ < \ d^2 \ \   $  has its zeroes at $ \ x \ = \ \pm c \ \ , \ \ \pm d \ \ $ and its turning points at
$$ 4x^3 \ - \ 2·(c^2 + d^2) x \ \ = \ \ 2x · [ \ 2x^2 \ - \ (c^2 + d^2) \ ] \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ 0 \ \ , \ \ \pm \sqrt{\frac{c^2 \ + \ d^2}{2}} \ \ . \ $$
The turning point at $ \ x \ = \ 0 \ $ is certainly half-way between $ \ x \ = \ -c \ $ and $ \ x \ = \ +c \ \ , \ \  $ but the bisectors between adjacent zeroes, $ \ \pm \frac{c \ + \ d}{2} \  , \ $ are closer to the $ \ y-$axis than the turning points $ \ \pm \sqrt{\frac{c^2 \ + \ d^2}{2}} \ \ $ by the "root-mean-square-arithmetic-mean inequality" (for which equality only holds when $ \ c \ = \ d \ \ ) \ . \ $  So it is not even sufficient for the polynomial to be an even function, much less of even degree, to have bisectors between successive zeroes pass through its turning points; only quadratic polynomials have a simple enough symmetry.
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