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?
I agree with Toby's approach of differentiating a cubic function and plugging in the mid-point of roots as suggested in the comment section. That certainly serves as a proof.
I have come up with a way to take this observation intuitively (or just from another perspective).
Quadratic Function
First, consider a quadratic function $y=x^2-1$, which has roots at $x=\pm1$, and vertex at $(0,-1)$.
Next, consider the second derivative, $y''=2$, which is the rate of change of the slope with respect to $x$. Note that it is a constant. That means the slope of the function always grows at the same rate.
Therefore, if we start from the vertex $(0,-1)$, the slope increases its magnitude at 2 per unit of $x$ to both left and right hand sides. The graph is symmetric and the vertex is right at the middle between the two roots.
Cubic Function
Now, we shall apply the same logic to cubic functions. Let's take $y=x^3-x$ as an example. It has roots at $x=0,\pm1$. For illustration purpose, we may focus on $0\leq x\leq1$.
Let $(k, k^3-k)$ be the turning point, for some $0<k<1$.
The second derivative is $y''=6x$. This time, the rate of change of slope depends on $x$. In fact, the slope changes faster as $x$ grows, viz. as we go further to the right hand side. It means that the slope changes more mildly to the left of $x=k$.
Thus, it takes longer $x$-distance for the function to move from $(k, k^3-k)$ to the root on the left than the one on the right. Therefore, you should expect the turning point to be closer to the root at $x=1$.
Summary
In short, for a cubic function, the turning point does not lie halfway between two roots because the slope changes at different rates on the two sides.
Furthermore, you may be curious about higher-degree polynomial functions. It depends on the second derivative.
For quartic functions, it may or may not possess this property.
https://amsi.org.au/ESA_Senior_Years/SeniorTopic2/2e/2e_2content_2.html
https://olvereducation.weebly.com/7f---families-of-quartic-functions.html
$f(x)$ quadratic has just a turning point at some $x_0$ (whatever its roots were real or not) and the line $x=x_0$ is an axis of symmetry which explains what you say. In the case of $f(x)$ cubic having two real roots it has necessarily the third root also real and have two turning points $x_1$ and $x_2$ such that, in general, $$r_1\lt x_1\lt r_2\lt x_2\lt r_3$$ where the $r_i$ are the roots. It is easy to verify that the lines $x=x_1$ and $x=x_2$ does not defines axis of symmetry of the corresponding arcs.
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.
