polynomials - Quadratics: Intuitive relation between discriminant and derivative at roots

While working with quadratics that have real roots, I realized an interesting fact:

The slope of a quadratic at its roots is equal to $\pm \sqrt{D}$ where $D=b^2-4ac$

Proof:

$$f(x) = ax^2 + bx +c$$

$$f'(x) = 2ax+b$$

Roots:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So, if we try to find the slope at any root ($r$):

$$f’(r) = \pm \sqrt D$$

where the sign ($\pm$) can be determined by whether the root is on the right of the vertex or the left.

If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.

What geometric/intuitive approach can be applied to explain this interesting phenomenon?