Symmetrical curve equation in range of [0,1] I would like to define a function $f(x,a)$ that has the following properties:
$f(0,a)=0$, excluding $a=\infty$
$f(1,a)=1$, excluding $a=-\infty$
$f(x,0)=x$
$f(0,a\to\infty)\to1$
$f(1,a\to-\infty)\to0$
The midpoint along the length of $f(x,a)$ within the range of $0\le x\le 1$ will be along the line of $g(x)=1-x$.
$g(x)$ is also the line of symmetry for $f(x,a)$ within the range of $0\le x\le 1$.
A line tangential to the midpoint along the length of $f(x,a)$ within the range of $0\le x\le 1$ will always have a slope of 1.
$f(1-x,a)=1-f(x,a)$ within the range of $0\le x\le 1$. I think this should define the symmetry I'm trying to imply.  
I've tried to think how best to describe the problem mathematically without drawing a picture, but I feel it necessary to give some visual cue all the same. Consider the following curves: $y=x^{1/3}$, $y=x^{1/2}$, $y=x$, $y=x^2$, $y=x^3$ within the range of $0\le x\le 1$. For all of those curves within those bounds they provide a mapping of $x$ to $y$ that doesn't exit the $(0,0),(1,1)$ square. Same is true for any $y=x^n$ curve. However, none of the curves are symmetrical around $g(x)$ excluding $y=x$.
$y=\sqrt{1-(x-1)^2}$ gives a curve that is symmetrical in the way I'm looking for. However, playing with constants hasn't given me any results that would lead me to having some way to change its shape easily with only a second variable ($a$). I also don't expect, or need, lines tangential at $f(0,a)$ and $f(1,a)$ to be $\infty$ and $0$, respectively.  
$a$ is meant to be a coefficient to control whether the curve dips under or over $f(x,0)$ and by how much; negative values of $a$ imply a curve under $f(x,0)$. The curve I expect for $f(x,\infty)$ is similar to the curve $y=x^{1/\infty}$.  
My questions are thus:
1) Is there a name for this type of curve and as such already defined somewhere?
2) Where should I go from here in determining the curve? What's my next step?
 A: Your function is impossible because $f(0,a) = 0$ for any $a \ne \infty$ implies $\lim_{a\to\infty} f(0,a) = 0 \ne 1$, however I think you want a function $f$ on $[0,1] \times (-\infty,\infty)$ such that:
$f(0,a) = 0$ for any $a$
$f(1,a) = 1$ for any $a$
$f(x,0) = x$ for any $x \in [0,1]$
$f(x,a) \to 1$ as $a \to \infty$ for any $x \in (0,1)$
$f(x,a) \to 0$ as $a \to -\infty$ for any $x \in (0,1)$
$x \mapsto f(x,a)$ is differentiable on $[0,1]$ for any $a$
$f(1-x,a) = 1-f(x,a)$
(The last two conditions will imply both the midpoint and the tangent conditions.)
Now here is a possible family (there are infinitely many possible families):
Let $f = (x,a) \mapsto (1-(1-x)^{2^a})^{2^{-a}}$
You can check that all the properties hold, and $f(x,1) = \sqrt{1-(1-x)^2}$, and $x \mapsto f(x,a)$ has gradient $\infty$ and $0$ at $0$ and $1$ respectively when $a>0$.
By the way, this function is modeled on what is called the $p$-norm. Here $p = 2^a$.
A: This is too long for a comment so...
Did you take a look at http://en.wikipedia.org/wiki/P-norm? It is symmetric and hence would produce the kind of curve you want except reflected about $x=\frac{1}{2}$. By the way it's unrelated to linear algebra. There cannot be a method to deduce it because any family of curves that starts at the origin, stays within the lower triangle and is perpendicular to the diagonal can be reflected across to get a valid solution.
Another family is $f = (x,a) \mapsto \frac{1}{2^a-1}\left(\frac{2^a}{(2^a-1)*(1-x)+1}-1\right)$, which has gradients at $0$ and $1$ changing gradually. This was obtained by using different parts of the curve $x \mapsto 1/x$, translated and reflected to pass through the endpoints, and then modified so that the range for $a$ is $(-\infty,\infty)$. Note that for $a=0$ you need to either use limits or define $f$ separately. This is corresponds to the degeneracy of the hyperbola into two straight lines.
By the way, you may want to read about splines, which is a generic way to construct curves that satisfy constraints such as passing through a sequence of points and having certain derivatives there. The earlier family I gave is not a spline, but this one above is of a special type of spline called a Bezier curve.
