TITLE: Does there exist a formula to find the coefficients of this parabola?
QUESTION [0 upvotes]: For equation $y=ax^2+bx+c$, assume I know the value of the coefficient $a$, I know the value of $y$ at the parabola's vertex (though I do not yet know the $x$ at that point), and I am given a point $(x_z, y_z)$ somewhere on the parabola, though I do not have further information about this point, such as slope at the point, nor do I know if this point's location relative to the vertex is consistent from evaluation to evaluation. 
How can I find the coefficients $b$ and $c$, given only this information? If not possible, is there information I might be able to gather about my parabola, in order to be able to find the coefficients $b$ and $c$?
This happens to be work I am doing for a client, but you are welcome to answer it as homework and teach me instead of giving me the answer, if it is trivial and I am overlooking something simple.

REPLY [1 votes]: Knowing that, at the vertex, $x_{vertex}=-b/{2a}$, I can substitute that value in my quadratic at the vertex, for the first equation in my system, resulting in $y_{vertex}=-b^2/{4a}+c$. 
The next equation for my system is $y_z=ax_z^2+bx_z+c$. 
Solving for c in the first equation and substituting that in the second, I am left with $y_z=ax_z^2+bx_z+y_{vertex}+b^2/{4a}$, and rearranging (for use in the quadratic equation) gives me $0={1\over4a}b^2+x_zb+(y_{vertex}+ax_z^2-y_z)$, or $b = {-x_z \pm \sqrt{x_z^2-4*{1\over4a}*(y_{vertex}+ax_z^2-y_z)} \over 2*{1\over4a}}.$
This is simplified to $b = \pm{\sqrt{4a(y_z-y_{vertex})}}-2ax_z$. 
This means that $c$ equals ${{y_z+ax_z^2-x_z{\sqrt{4a(y_z-y_{vertex})}}}}-y_{vertex}\pm{y_{vertex}}$
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