If I have an x intercept and a y intercept, how might I find the vertex of a parabola? I have looked all over, and I have found different things here and there for figuring pretty much everything but that. 
What I have is an x chord and a y chord, but I need to find the vertex. 
This isn't for school or anything, it is for an algorithm for a program I am writing.
I never got over college algebra even when I was in college and I'm certain I have learned this before, but it has been literally years since then.
So, how do I find the vertex if I have an x chord and a y chord and that is all? And what mucks it up even more is I need a reliable means of retrieving the x and y chords back again.
Realistically I can either use one of my points as x intercept and one y intercept, or I can use one as x intercept and the other the vertex. But in either case I need a reliable means of moving back and forth from the points given to a said point and back. 
I have two points and I need to save one, and be able to reliably get the two original points back at will from that saved single point. 
As you can see this is not an easy task and hence my brain pain. lol
Any assistance would be greatly appreciated.

Please allow me to broaden it a bit further.
I have a list of sorts which resembles a parabola. 
What I really have is two points within a "shaded" region which is a parabola or bell curve.
I need to some how store these two points via some equation with a single point, and with that single point be able to retrieve the original 2 points.
I know there is a method of doing this, but the math to do it is over my head. I'm plenty smart, I just haven't learned it yet.
If it helps at all here is a python code snipet demonstrating the values which would equate to whole number chords on a graph. My program will take two of those values, equate them to chords on a graph, figure a singular value to uniquely represent those two chords, save that value, and then later on retrieve those original chords(and values) for later calculation.
def main():
    clist = []

    for x in range(0, 37):
        clist.append("")

    for a in range(0, 10):
        for s in range(0, 10):
            for d in range(0, 10):
                for f in range(0, 10):
                    if clist[a+s+d+f] == "": 
                        clist[a+s+d+f] = str(a) + str(s) + str(d) + str(f) 
                    else: 
                        clist[a+s+d+f] = clist[a+s+d+f] + "," + str(a) + str(s) + str(d) + str(f)

    count = 0
    for x in range(0, 37):
        datalst = clist[x].split(',')
        datacnt = len(datalst)
        count += datacnt
        print(str(x) + ", " + str(datacnt) + ", " + clist[x])

The results of which are:
http://pastebin.com/SjtBMUqa
(formatting bad, but you get the picture)
 A: Usually two points do not determine a parabola because the space of parabolas has a typical point $f(x)=Ax^2+Bx+C$ and as you can count there are three parameters to fix: $A,B,C$. However, if we are given the very special point $(h,k)$ called the vertex and we are given another point $(a,b)$ such that $k \neq b$ and $h \neq a$ then these two points fix a parabola as follows: first the fact $(h,k)$ is the vertex indicates that
$$ f(x) = A(x-h)^2+k $$
where $A$ is a constant. Continuing, $f(a)=b$ shows $A(a-h)^2+k=b$ so $A=\frac{b-k}{(a-h)^2}$. In summary, the pair of points $(h,k)$ and $(a,b)$ fix the parabola:
$$ y=f(x) = \frac{b-k}{(a-h)^2}(x-h)^2+k. $$
All of this said, you should realize I'm cheating. These are not just any old pair of points on the parabola. No, $(h,k)$ is the vertex which means that it is at a horizontal tangent and so implicit within the statement $(h,k)$ is the vertex is both the condition the point be on the graph $k=Ah^2+Bh+C$ and the condition that the vertex have a horizontal tangent $0=2Ah+B$.
