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! (Itwasntme)

Printable View

- Jul 30th 2012, 08:09 AMwoahitzmeFinding Equation of a Parabola?
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! (Itwasntme) - Jul 30th 2012, 08:30 AMrichard1234Re: Finding Equation of a Parabola?
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 $\displaystyle c$. - Jul 30th 2012, 08:32 AMWilmerRe: Finding Equation of a Parabola?
- Jul 30th 2012, 08:34 AMrichard1234Re: Finding Equation of a Parabola?
- Jul 30th 2012, 08:50 AMProve ItRe: Finding Equation of a Parabola?
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. - Jul 30th 2012, 08:55 AMWilmerRe: Finding Equation of a Parabola?
- Jul 30th 2012, 09:02 AMwoahitzmeRe: Finding Equation of a Parabola?
- Jul 30th 2012, 09:11 AMwoahitzmeRe: Finding Equation of a Parabola?
Nevermind, I got the answer, thank you guys!