What equation produces this curve? I'm working on an engineering project, and I'd like to be able to input an equation into my CAD software, rather than drawing a spline.
The spline is pretty simple - a gentle curve which begins and ends horizontal.
Is there a simple equation for this curve?
Or perhaps two equations, one for each half?
I can also work with parametric equations, if necessary.


Note: Now that I've finished this derivation, I see it's the same as Jaume Oliver Lafont's answer. Perhaps some extra justifications will be useful.
In looking over @Giffyguy's comments on other answers, it seems the real goal is to have a curve with minimum curvature or maximum smoothness. I'll take this to mean the maximum absolute value of the second derivative should be minimized. The way to minimize the second derivative everywhere is to make it everywhere constant, which means we're dealing with a parabola
$$y(x) = ax^2 + bx + c$$
with constraints
$$y(0) = 0$$
$$y(60) = -20$$
for the right half, and
$$y'(60) = 0.$$
The first constraint means $c = 0$. The second and third constraints result in, respectively,
$$a(60)^2 + b(60) = -20.$$
$$2a(60) + b = 0$$
Solving results in $a = \frac{1}{180}$ and $b = \frac{2}{3}$. So,
$$y = \frac{1}{180}x^2 - \frac{2}{3}x = \frac{x(x-120)}{180}$$
The second derivative has a constant value of $1/90$, which is smaller than the maximum value of the second derivative of the cosine functions ($1/60$) and the cubic polynomial ($\approx 1/72$). The function spanning the full width would be
$$y(x) = \left\{
\begin{array}[ll]
\\
-\frac{x(x+120)}{180} & x \lt 0 \\
\frac{x(x-120)}{180} & x \geq 0 \\
\end{array}
\right.$$
