How to derive equation for second order interpolation of soft knee cutoff in a compressor?
If you look at the formula and plot it, you get what you see in the other link on dsp.ee. But if you add the inflexion points and plot the consituent functions, it should become more clear:

The function would have been a linear one, with two slopes, but between the points defined by $T-W/2$ and $T+W/2$, there is a curve uniting the two segments. This means that the conditions become:

a simple, quadratic formula (to avoid too much computation)
the function's value at the first point should be equal to the point, itself, and
the derivatives at those points should be the actual slopes

With these, there are many ways to interpolate, so I can't say exactly how they did, but these must have been their starting points.
For example, considering two slopes of $3/4$ and $1/4$ at $[-1,1]$, the quadratic function $f(x)=ax^2+bx+c$ that would unite those would have to have:
$$\left\{
\begin{aligned}
f'(-1)&amp;=2a(-1)+b&amp;=\dfrac34 \\
f'(1)&amp;=2a(1)+b&amp;=\dfrac14
\end{aligned}\tag{1}
\right.$$
The derivative makes $c$ disappear, but its value at $-1$ is $3/4$, so the solving the equations above results in the function:
$$f(x)=\dfrac34+\dfrac{x}{2}-\dfrac{x^2}{8}\tag{2}$$
and this is how it looks like:

The slopes at $[-1,1]$ are a match, which means the two slopes are now connected by a quadratic formula. The whole graph would be, from left to right: the red slope until $-1$, then the quadratic blue until $1$ and then the green slope; just like in the 1st picture.

I'll add that you can go the other way, just as well: from the derivative, up.
You know that the signal has a slope in the beginning, and one at the end which means that their derivatives give you fixed values (in the example above they were $3/4$ and $1/4$). Uniting the points will be a slope of the form $ax+b$, as seen in (1), starting from $-1$ with the value $3/4$, and ending up at $1$ with $1/4$. Since it's a line, it will pass through the middle point, $m$, and it will have a slope $s$, both given by:
$$\begin{align}
m=\dfrac{\dfrac14+\dfrac34}{2}&amp;=\dfrac12 \\
s=\dfrac{\dfrac14-\dfrac34}{2}&amp;=-\dfrac14 \\
\Rightarrow\quad g(x)=m+sx&amp;=\dfrac12-\dfrac14x\tag{3} \\
\end{align}$$

Integrate this to get the quadratic, to which you add the constant term $3/4$ -- the value of the function at $-1$:
$$\int_x{g(x)\mathrm{d}x}=\color{blue}{\dfrac34}+\dfrac{x}{2}-\dfrac{x^2}{8}\tag{4}$$
(2) and (4) are the same. q.e.d.