Finding a simple spline-like interpolating function

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 $\alpha=0.1,0.2,\ldots,0.9$:

I constructed a quadratic Bezier curve (somewhat like your b-spline idea), and then converted it to $y = f(\alpha, x)$ form. The result is:

$$y = \frac{1 + 4 \alpha (x-1)-2 x+\sqrt{(1-4 \alpha)^2+8 (1-2 \alpha) x}}{2-4 \alpha}$$

My guess is that this isn't very good for your purposes because of the square root.

Also, this formula only works for values of $\alpha$ with $0.25 \le \alpha \le 0.75$.

Here are the graphs of these functions for $\alpha = 0.25, 0.35, 0.45, 0.55, 0.65, 0.75$

They are all (rotated) parabolas.

The $0.25 \le \alpha \le 0.75$ restriction could be lifted by using a rational quadratic curve, instead. But the square root won't go away, so I want to know if you can live with that before I spend time working out the details of the rational case.