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Thread: Writing equations from pieces of info

  1. #1
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    Writing equations from pieces of info

    Given:
    Direction opens up with minimum at -2
    Passes through (0,0) and (2,0)

    Write the equation of the parabola.

    Let y = a(x - p) + q

    The q value of vertex is -2, so

    y = a(x - p) - 2

    What do I do next? Pass through the points? How to solve this?

    Thanks.
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  2. #2
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    Quote Originally Posted by shenton View Post
    Given:
    Direction opens up with minimum at -2
    Passes through (0,0) and (2,0)

    Write the equation of the parabola.

    Let y = a(x - p) + q

    The q value of vertex is -2, so

    y = a(x - p) - 2

    What do I do next? Pass through the points? How to solve this?

    Thanks.
    Yes.

    For (0, 0):
    $\displaystyle 0 = a(-p)^2 - 2$

    For (2, 0):
    $\displaystyle 0 = a(2 - p)^2 - 2$

    The first says $\displaystyle ap^2 = 2$ and the second says $\displaystyle ap^2 - 4ap + 4a - 2 = 0$.

    Putting $\displaystyle ap^2 = 2$ into the second equation gives:
    $\displaystyle -4ap + 4a = 4a(-p + 1) = 0$

    so either a = 0 (a contradiction, if this were true then we won't have a parabola!) or $\displaystyle -p + 1 = 0$

    Thus p = 1.

    Thus $\displaystyle ap^2 = a = 2$ gives a = 2.

    So your parabola is:
    $\displaystyle y = 2( x - 1)^2 - 2$

    -Dan
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  3. #3
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    Hello, shenton!

    Given: opens up with minimum at -2
    Passes through (0,0) and (2,0)

    Write the equation of the parabola.

    The general equation of a "vertical" parabola is: .$\displaystyle y \:=\:ax^2 + bcx + c$

    We have three points: .$\displaystyle (0,0),\;(2,0)$ . . . and the vertex: $\displaystyle (1,-2)$
    . . If you plot the two points, you see they are x-intercepts.
    . . The axis of symmetry lies halfway between them at $\displaystyle x = 1$
    . . And that is how I located the vertex.

    Substitute the points into the general equation.

    $\displaystyle \begin{array}{ccc}(0,0): \\ (2,0): \\ (1,\text{-}2):\end{array}
    \begin{array}{ccc}0 \;= & a(0^2) + b(0) + c \\ 0 \:= & a(2^2) + b(2) + c \\ \text{-}2 \:= & a(1^2) + b(1) + c \end{array}\begin{array}{ccc}\Rightarrow\\ \Rightarrow\\ \Rightarrow\end{array}\begin{array}{ccc} \boxed{c \:=\:0} \\ 4a + 2b \:=\:0 \\ a + b\:=\:\text{-}2\end{array}\begin{array}{ccc}(1)\\(2)\\(3)\end{a rray}$

    Divide (2) by 2: .$\displaystyle 2a + b \:=\:0$
    . . Subtract (3): .$\displaystyle \text{-}a - b \:=\:2$
    . .And we have: .$\displaystyle \boxed{ a = 2}$

    Substitute into (3): .$\displaystyle 2 + b\:=\:\text{-}2\quad\Rightarrow\quad\boxed{b = \text{-}4}$


    Therefore, the equation of the parabola is: .$\displaystyle y \:=\:2x^2 - 4x$

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