Hello, Frontier!

A parabola has x-intercepts at (-2, 0) and (6, 0).

Its maximum y-value is 12.

Write the standard form of the equation of this parabola. Did you make a sketch? Code:

| (2,12)
| *
*| : *
* | : *
* | : *
| :
- - * - - + - + - - - - * - -
-2 | 2 6

From the symmetry, we see that the vertex is at (2, 12).

The general equation of a parabola is: .$\displaystyle f(x) \;=\;ax^2 + bc + c$

We know three points on the parabola: .$\displaystyle (-2,0),\;(2,12),\;(6,0)$

. . Use these point to create a system of equations.

$\displaystyle \begin{array}{ccccccccc}f(\text{-}2) = 0\!: & a(\text{-}2)^2 + b(\text{-}2) + c &=& 0 & \Rightarrow & 4a - 2b + c &=& 0 & [1] \\

f(2) = 12\!: & a(2^2) + b(2) + c &=& 12 & \Rightarrow & 4a + 2b + c &=& 12 & [2] \\

f(6) = 0\!: & a(6^2) + b(6) + c &=&0 & \Rightarrow & 36a + 6b + c &=& 0 & [3] \end{array}$

Subtract: [2] - [1]: .$\displaystyle 4b \:=\: 12 \quad\Rightarrow\quad \boxed{b \:=\:3}\;\;[4]$

Subtract: [3] - [2]: .$\displaystyle 32a + 4b \:=\: -1 $

Substitute [4]: .$\displaystyle 32a + 4(3) \:=\:-12 \quad\Rightarrow\quad 32a \:=\:-24 \quad\Rightarrow\quad\boxed{ a \:=\:\text{-}\tfrac{3}{4}}\;\;[5]$

Substitute [4] and [5] into [1]: .$\displaystyle 4\left(\text{-}\tfrac{3}{4}\right) - 2(3) + c \:=\:0 \quad\Rightarrow\quad\boxed{ c \:=\:9}$

Therefore: .$\displaystyle f(x) \;=\;\text{-}\tfrac{3}{4}x^2 + 3x + 9$

Edit: Dinkydoe has a much better solution!'.