Simple Formula for Curve of DJ Crossfader volume : Dipped Below are the common set of curves that are used in a DJ mixer on the crossfader. I have a software equivalent and am using the "Transition" type curve but would like to include some different curve types. 
On both the x and y axis' my range is from 0-1. What I usually do is have a function that gives me one side of the curve. (i.e. just the red from the diagrams below) Then as I need two levels (the red and the blue) for a given x value I invert the x value and feed it to the same formula (something like invertedX = x * -1 + 1)
I need the formula for the Dipped curve in the diagram below.
Extra credit goes to those who can give me the formulas to the other curves 


*

*Intermediate

*Constant Power

*I'm pretty sure Slow Fade, Slow Cut and Fast Cut are all the same formula with just a parameter difference or two.


I have Transition (the easiest of them all)

 A: I was wrestling with some of these same questions myself earlier and never really found completely solid answers for what makes a good curve.  I can, however, share my functions which I wound up using.  They're all constant-power ones, but with different levels of fade/cut.
To understand what makes a curve constant-power, you have to understand that the signal is a sound-pressure level signal and that power goes as sound pressure squared.  So if we have input signal $w_1$ and $w_2$ and we're attenuating the signal by multiplying $w_1 \cdot f(x)$ and $w_2 \cdot f(1-x)$ then $f$ is constant-power if $f^2(x)+f^2(1-x) = 1$.  So, in this case, the easiest way would be to make $f(x) = \cos(\frac{\pi}{2}x)$.
In fact, we can generalize this and say that any function $g$ with range [0,1] for domain [0,1] can be used to produce a constant-power crossfade function $f(x)=\cos(\frac{\pi}{2}g(x))$.  However, it'll make the most sense if $g(0) = 0$, $g(0.5)=0.5$, $g(1)=1$, and $g$ is monotonic.  With this in mind, I more looked at functions $h$ where $h(-1) = -1$, $h(0)=0$, and $h(1)=1$ and then just did a simple linear transform between [-1,1] and [0,1].  So, the first thing I tried was $h(x)=x^{2n+1}$ for non-negative integers n.  This turned out to work quite well.  $n=0$ gives the constant power curve you show above and then as I go to $n=1,3,10$ I get curves a lot like slow fade, slow cut, and fast cut (although not identical since these are all constant power).  There's obviously a lot of room to adjust the sharpness by using other values for n.
So my final function is $f(x)=\cos(\frac{\pi}{4}((2x-1)^{2n+1}+1))$.  As you can see from the graphs below, they're quite similar to slow fade, slow cut, and fast cut, except that the plateau in the middle is at about 0.7 (actually $\frac{1}{\sqrt{2}}$) rather than at 1 and the track which doesn't fall off rises to 1 at the outside edge.  You might think that going down to 0.7 in the middle would have a big sound impact on the other track, but it really doesn't.  Obviously, you could convert them into the exact functions by doing $\sqrt{2} \min(f(x),1/\sqrt{2})$.  But I've tried them out in a software fader in Pure Data without doing that adjustment, and they sound pretty good to my ears.  I was also happy because it meant that I could switch between the four different curves with only a single parameter so that keeps the logic simple.
Below: Curves for Constant Power (n=0), Constant Power Slow Fade (n=1), Constant Power Slow Cut (n=3), and Constant Power Fast Cut (n=10).




A: Intermediate is clearly  $y = 1 - x$ and $y = x$; should be similarly related to transition.
I want to say that Dipped is a parabola (not sure it is; hard to tell); in which case it would be
$ y = (x-1)^2 $ and $ y = x^2 $
But there are many parabolas that fit to the points (0,0) and (1,1), or (0,1) and (1,0).
The name "Power" seems to imply Power in Sound, which would mean logarithms.  I could imagine fitting logarithms into that, but perhaps it would be easier to just use 1-dipped.  That is, $ y = 1 - (x-1)^2 $ and $ y = 1 - x^2 $
The fade/cut/cut could be cubic formulas, shifted upwards +0.5.  So they'd be variations of $x^3 + 0.5$ stretched vertically/horizontally.
I'll get back to this when I'm less sleepy, heh.
A: For the slow fade may be you can take the Gaussian curve $y = e^{-x^{2}}$ and here is the diagram below.

And for the transition you could try $y=-|x|$.
