I'm looking for some kind of ease-in function. Within the range of $0 < x < 1$, $0 < y < 1$ and depending on a parameter $0 < g < 1$ it should have the following properties: $$ f(0) = 0, f(1) = 1, f'(0) = 0, f'(1) = \infty $$ $$ f(x, g = 0) = x, f(x, g = 1) = 0$$ $$ f\left(\frac{1}{2} + g\right) = \frac12 - g, f'\left(\frac{1}{2} + g\right) = 1$$ So basically it is symmetric about y = 1 - x and crosses that line perpendicularly. I tried to stitch it together from two 3rd degree polynomials, which almos does the job, but not exactly.
Biggest issue is that it becomes negative as the parameter gets larger. I'm not quite sure what to look for, so any help would be appreciated!
Consider the equation for a quarter-arc of a square superellipse: $$x^n+y^n=1$$ The arc is symmetric about $x=y$ and passes it perpendicularly. At $n=1$ it generates a straight line ($g=0$) and as $n\to\infty$ it generates a constant one function ($g=1$). If we now take $n=\frac1{1-g}$ and flip the curve vertically we get an easing curve that satisfies most of the desired properties: $$y=1-(1-x^{1/(1-g)})^{1-g}$$ It does not satisfy $f'(1)=1$, but if the arc is symmetric then given $f'(0)=0$, $f'(1)$ would have to be $\infty$.