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Math Help - Equation of a parabola when two x-intercepts are given?

  1. #1
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    Equation of a parabola when two x-intercepts are given?

    I am having trouble solving this problem. Any help would be appreciated!

    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.

    If its max y-value is 12, the y-value of the vertex is 12, right? I know I should be able to use one of the given x values to find what I need to write the equation, but I just don't know where to start with this...
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  2. #2
    Senior Member Dinkydoe's Avatar
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    The standard form of the parabola is: f(x) = ax^2+bx+c

    Since f(-2)= f(6) = 0 we have f(x) = c(x+2)(x-6) = c(x^2-4x-12) for some constant c.

    We know f(x) reaches maxima/minima when f'(x) = c(2x-4) = 0.

    Can you figure out c with this information and write f(x) in standard form?
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  3. #3
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    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: . f(x) \;=\;ax^2 + bc + c


    We know three points on the parabola: . (-2,0),\;(2,12),\;(6,0)
    . . Use these point to create a system of equations.


    \begin{array}{ccccccccc}f(\text{-}2) = 0\!: & a(\text{-}2)^2 + b(\text{-}2) + c &=& 0 & \Rightarrow & 4a - 2b + c &=& 0 & [1] \\<br />
f(2) = 12\!: & a(2^2) + b(2) + c &=& 12 & \Rightarrow & 4a + 2b + c &=& 12 & [2] \\<br />
f(6) = 0\!: & a(6^2) + b(6) + c &=&0 & \Rightarrow & 36a + 6b + c &=& 0 & [3] \end{array}


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

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

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


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



    Edit: Dinkydoe has a much better solution!'.
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