I am trying to build an equation where I could start at (x,y) which are known and create a parabola from that starting point. I have no idea where it intercepts the X or Y.
I know where I want the line on the other side to go down at (the other root) well I know I want it to be (??) units between the two lines. I know roughly how high it should go.
Any idea of how I could do that.
EX: http://crappygraphs.com/user_graphs/?id=7101 I know the (x,y) of 1 and I know the (x,y) of (x) and I know the (x,y) of (y). I just need to make something follow this path.
Here's your problem worked with the vertex of the parabola (with vertical line of symmetry) and another point on the parabola specified:
Suppose your vertex is at $(2,3)$ and that $(3,1)$ is another point on the parabola.
The equation of the parabola has the form $$ y=a(x-2)^2+3. $$
Since $(3,1)$ is on the parabola $$ 1=a(3-2)^2+3\quad\iff\quad 1=a+3\quad\iff a=-2. $$
So your equation is $y=-2(x-2)^2+3$.