# Does there exist a formula to find the coefficients of this parabola?

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.

• though I do not yet know the x at that point You do know that $x=-b/2a$, though, so that's one equation. You get one more from the next condition Then solve the system for $b,c$. – dxiv Aug 7 '18 at 3:01
• I think the easiest way to work with this type of problems is to complete a square and look for a parabola in the from $y=a(x+b/2a)^2+c$. This way you immediately know $c$ because that's $y$ of the vertex and you only need to find $b$ using the second point. – Vasya Aug 7 '18 at 3:12
• @CWilson my 2 unknowns are x and b No, $\,y_{vertex}=a(-b/2a)^2+b(-b/2a)+c\,$ has the unknowns $\,b, c\,$. – dxiv Aug 7 '18 at 3:22
• @dxiv So, I think what you are saying is, I can use the identity of a parabola, to show that the y value at the vertex is equal to $-b^2/{4a}+c$. If so, then yes, my problem becomes easy to solve. – CWilson Aug 7 '18 at 3:33
• @dxiv Thank you for your help, I needed to get this solved for work. I posted my solution as a thank you. If this is trivial or duplicate, please close, with my thanks. – CWilson Aug 7 '18 at 5:34

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 ${{y_z+ax_z^2-x_z{\sqrt{4a(y_z-y_{vertex})}}}}-y_{vertex}\pm{y_{vertex}}$
• See, if you write equation as $y=a(x+b/2a)^2+d$, $d=y_{vertex}$ because the squared part will be zero. – Vasya Aug 8 '18 at 2:53