0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $b,c$. $\endgroup$ – dxiv Aug 7 '18 at 3:01
  • $\begingroup$ 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. $\endgroup$ – Vasya Aug 7 '18 at 3:12
  • 1
    $\begingroup$ @CWilson my 2 unknowns are x and b No, $\,y_{vertex}=a(-b/2a)^2+b(-b/2a)+c\,$ has the unknowns $\,b, c\,$. $\endgroup$ – dxiv Aug 7 '18 at 3:22
  • 1
    $\begingroup$ @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. $\endgroup$ – CWilson Aug 7 '18 at 3:33
  • 1
    $\begingroup$ @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. $\endgroup$ – CWilson Aug 7 '18 at 5:34
1
$\begingroup$

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 ${{y_z+ax_z^2-x_z{\sqrt{4a(y_z-y_{vertex})}}}}-y_{vertex}\pm{y_{vertex}}$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.