Determine the equation of the parabola:
Given two points at (-4,9) and (12,9). I figured out the axis of symmetry is x=4. Also given the y-intercept, which is (0,-3). Figure out the equation?
Thanks for the help!
Write the parabola in the form $\displaystyle y = a(x-b)^2 + c$ where $\displaystyle a,b,c$ are constants and $\displaystyle a \neq 0$.
It is symmetric about $\displaystyle x = 4$, so letting $\displaystyle b = 4$ yields our desired parabola.
Now our parabola is in the form $\displaystyle y = a(x-4)^2 - c$. Subsitutute the given values of (x,y) to obtain a system of equations in terms of $\displaystyle a$ and $\displaystyle c$.
The quadratic has the equation $\displaystyle \displaystyle \begin{align*} y = a\,x^2 + b\,x + c \end{align*}$. Substituting the y-intercept $\displaystyle \displaystyle \begin{align*} (0, -3) \end{align*}$ yields $\displaystyle \displaystyle \begin{align*} c = -3 \end{align*}$, giving $\displaystyle \displaystyle \begin{align*} y = a\,x^2 + b\,x - 3 \end{align*}$.
Substitute each of the other points and solve the two equations simultaneously for a and b.