I am looking for a continuous function $y=f(x,\alpha)$ for the interval $0\le x \le 1$ such that $0\le y \le 1$ and $y(0,\alpha)=0$ and $y(1,\alpha) = 1$ and $y(\alpha,\alpha) = 1-\alpha$ and $dy/dx|_{x=\alpha}=1$. It is effectively symmetrical about the line $y=1-x$. Typically $0.2\le \alpha \le 0.8$.
It looks like a parabola but rotated so that the new y axis is along $y=1-x$, but since I need to be able to compute $y$ given $x$, I have not yet been able to work it out.
The exact form is not too important, but since this is to go in a OpenGL vertex shader I am trying to come up with something that is as simple as possible in terms of math operations.

The solution by @bubba is quite nice, but if by "It looks like a parabola but rotated", you don't mean that it necessarily is exactly a parabola, then you can get a slightly simpler expression by using a rotated hyperbola instead. The general form for a hyperbola with axis-aligned asymptotes is $$y=p\frac{x+q}{x+r}.$$ Plugging in your conditions gives $$p=\frac{(1-\alpha)^2}{1-2\alpha}, \quad q=0, \quad r=\frac{\alpha^2}{1-2\alpha},\\ y=\frac{(1-\alpha)^2x}{(1-2\alpha)x+\alpha^2}.$$ Here are the graphs of the function for $\alpha=0.1,0.2,\ldots,0.9$: