Determine where peak or valley of polynomial graph without calculus I notice that if I graph $y=(x+2)^2(x-4)^2$ that the midpoint of the roots occurs at $x=1$.  I note that this is also where the local maximum occurs.
If I graph $y=(x-5)^2(x^2)$ the midpoint of the roots is 2.5 where the local maximum occurs.
I also know how to find the exact values of the local max / min by taking the derivative and setting to zero.
My question is, for those in pre-calculus who have not yet learned about derivatives, is there a relationship between where the roots occur and the local max / min values?  Certainly for the 2 examples I chose, there seems to be a relationship, but I can't see the mathematical reasoning for this. Calculus explains this, but could I predict the local max/min by simply finding the mid-point of the roots?
 A: There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$

A: You can start with a parabola.  If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$.  The factor $a$ does not change the location of the roots or the vertex.  By using the quadratic formula or calculus you can find that the vertex is at $x=\frac 12(b+c)$, the midpoint between the roots.  Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.  
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct.  If you look at more general quartics you can have different behavior.  If we consider $xy=x^2(x-1)^2+\frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$.  A graph is below.

