Will the perpendicular bisector between the line connecting two cubic roots of the same arc never intersect its turning point?

Question

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.

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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?

Answer 1

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

Answer 2

$f(x)$ 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.