Some facts about cubic functions and quartic functions motivate this question:
Every cubic function $f$ has exactly one inflection point $P$, and the graph $y=f(x)$ is symmetric about $P$. In particular, if $f$ has two local extrema $A$ and $B$, then $P$ is the midpoint of $AB$. Phrased in another way, the ratio $\frac{PA}{PB}$ is an invariant.
A quartic function has either $0$ or $2$ inflection points. Suppose it has two inflection points $P$ and $Q$. The secant line passing through these two points will intersect the curve at two other points $A$ and $B$. Assume these four points are in the order $A, P, Q, B$ from left to right, then $AP:PQ:QB = 1:\frac{1+\sqrt{5}}{2}:1$. Phrased in another way, the ratios $\frac{AP}{PQ}$ and $\frac{AP}{QB}$ are invariants. There is also a fact about the areas bounded between the secant line and the curve.
Now consider a quartic function with three local extrema (and hence two inflection points). I wonder if there exists any invariant concerning the distances between these $5$ points, or their coordinates, or perhaps some areas bounded between lines and the curve. I'd be glad to see any such result.
I've done some computations and found some relations among the slopes, so I decided to post this as an answer, which can hopefully attract others to explore the relations among the lengths and areas as well.
Suppose $f(x) = ax^4 + bx^3 + cx^2 + dx + e$ with $a \neq 0$ has three local extrema $A, O, B$ and two inflection points $P, Q$. Label these five points $A, P, O, Q, B$ from left to right.
First, let me point out the following trivial facts by virtue of Vieta's formulas: the sum of all the (real and imaginary) roots of $f$ is $-\frac{b}{a}$, the sum of the $x$-coordinates of $A, O$ and $B$ is $-\frac{3b}{4a}$, and the sum of the $x$-coordinates of $P$ and $Q$ is $-\frac{b}{2a}$, so in particular \begin{equation} \tag{1} \frac{x_A+x_O+x_B}{x_P+x_Q} = \frac{3}{2}. \end{equation} In other words, the average of the $x$-coordinates of the critical points equals the average of the $x$-coordinates of the inflection points.
Since here we are concerned only with slopes (and similarly for lengths and areas), WLOG we can assume $O$ is the origin by performing a translation. A necessary (but not sufficient) condition for this is $f(0) = 0$ and $f'(0) = 0$, so that $d = e = 0$.
Then $$f(x) = ax^4 + bx^3 + cx^2 = x^2(ax^2+bx+c),$$ $$f'(x) = 4ax^3 + 3bx^2 + 2cx = x(4ax^2+3bx+2c),$$ $$f''(x) = 12ax^2 + 6bx + 2c = 2(6ax^2 + 3bx + c).$$
The critical points are $$O(0, 0),$$ $$A\Bigg(\frac{-3b-\sqrt{9b^2-32ac}}{8a}, \frac{\Big(3b+\sqrt{9b^2-32ac}\Big)^2 \Bigl(-3b^2+16ac-b\sqrt{9b^2-32ac}\Bigl)}{2048a^3}\Bigg),$$ $$B\Bigg(\frac{-3b+\sqrt{9b^2-32ac}}{8a}, \frac{\Big(3b-\sqrt{9b^2-32ac}\Big)^2 \Bigl(-3b^2+16ac+b\sqrt{9b^2-32ac}\Bigl)}{2048a^3}\Bigg),$$
and the inflection points are $$P\Bigg(\frac{-3b-\sqrt{9b^2-24ac}}{12a}, \frac{\Big(3b+\sqrt{9b^2-24ac}\Big)^2 \Bigl(-3b^2+20ac-b\sqrt{9b^2-24ac}\Bigl)}{3456a^3} \Bigg),$$ $$Q\Bigg(\frac{-3b+\sqrt{9b^2-24ac}}{12a}, \frac{\Big(3b-\sqrt{9b^2-24ac}\Big)^2 \Bigl(-3b^2+20ac+b\sqrt{9b^2-24ac}\Bigl)}{3456a^3} \Bigg).$$
Denote by $m_{XY}$ the slope of the line passing through $X$ and $Y$, and by $m_X$ the slope of the tangent line at $X$. Then
$$m_{AB} = \frac{9b^3-32abc}{64a^2},$$ $$m_{PQ} = \frac{b^3-4abc}{8a^2},$$
which implies that $m_{AB} - m_{PQ} = \frac{b^3}{64a^2}$.
$$m_{OP} + m_{OQ} = \frac{3b^3-14abc}{24a^2},$$ $$m_{OA} + m_{OB} = \frac{9b^3-40abc}{64a^2},$$ $$m_P + m_Q = \frac{b^3-4abc}{4a^2}.$$
Each of the above $5$ quantities is a linear combination of $\frac{b^3}{a^2}$ and $\frac{bc}{a}$, so it is easy to find many ratios that are independent of the coefficients of $f$. The difficulty lies in finding simple and elegant ones such as the following two:
\begin{equation} \tag{2} m_{PQ} = \frac{1}{2}(m_P + m_Q). \end{equation} \begin{equation} \tag{3} \frac{m_{AB}-m_{OA}-m_{OB}}{m_{PQ}-m_{OP}-m_{OQ}} = \frac{3}{2}. \end{equation}
I'm looking forward to more insights from this community regarding other invariant quantities.