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Math Help - Strange Parabola

  1. #1
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    Strange Parabola

    Find the equation of the parabola
    Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
    could u please help me solve using this equation substituting the x and y values
    x^2 + Dx + Ey + F

    Because i always end up with this answer x^2 - 4x + 3 = 0 hence not a parabola
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the equation of the parabola
    Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
    could u please help me solve using this equation substituting the x and y values
    x^2 + Dx + Ey + F

    Because i always end up with this answer x^2 - 4x + 3 = 0 hence not a parabola
    Hello,

    (1. the x-value of the vertex must be the mean of -1 and 5, that means 2, because the first two points have the same y-value, they are situated symmetrical to the axis)

    plug in the x- and y-values into the given equation:

    (-1,0):\ 0=1-D+F
    (5,0):\ 0=25+5D+F
    (1,8):\ 0=1+D+8E+F

    This is a system of simultaneous equations. I don't know which method you use. But the solution is: d = -4, E = 1, F = -5

    Thus the equation of the parabola is:

    0=x^2-4x+y-5

    EB
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the equation of the parabola
    Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
    could u please help me solve using this equation substituting the x and y values
    x^2 + Dx + Ey + F

    Because i always end up with this answer x^2 - 4x + 3 = 0 hence not a parabola
    x^2 +Dx +Ey +F is not a parabola.
    x^2 +Dx +Ey +F = 0 is. -----------(i)

    At (-1,0),
    (-1)^2 +D(-1) +E(0) +F = 0
    1 -D +F = 0 ------------------(1)

    At (5,0),
    5^2 +D(5) +E(0) +F
    25 +5D +F = 0 -------------(2)

    (2) minus (1),
    24 +6D = 0
    So, D = -24/6 = -4 ----***
    Substitute that into (1),
    1 -(-4) +F = 0
    So, F = -5 ------------***

    At (1,8),
    1^2 +D(1) +E(8) +F = 0
    1 +D +8E +F = 0 ----------------(3)
    Plug D and F in,
    1 +(-4) +8E +(-5) = 0
    8E -8 = 0
    So, E = 8/8 = 1 -----***

    Therefore, the parabola is x^2 -4x +y -5 = 0 --------answer.
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!

    Let's start from the beginning . . .


    Find the equation of the parabola with a vertical axis
    and passes through (-1,0),\:(5,0),\:(1,8)

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

    \begin{array}{ccc}(-1,0): \\ (5,0): \\ (1,8):\end{array}\begin{array}{ccc}0 \:=\:a(\text{-1})^2 + b(\text{-}1) + c \\ 0 \:=\:a(5^2) + b(5) + c \\ 8 \:=\:a(1^2) + b(1) + c\end{array}\begin{array}{ccc}\Rightarrow \\ \Rightarrow \\ \Rightarrow\end{array}\begin{array}{ccc}a - b + c \:=\:0 \\ 25a + 5b + c \:=\:0 \\ a + b + c \:=\:8\end{array}\begin{array}{ccc}(1)\\(2)\\(3)\e  nd{array}


    Subtract (1) from (2): . 24a + 6b\:=\;0\;\;\;(4)
    Subtract (3) from (2): . 24a + 4b \:=\:\text{-}8\;\;(5)

    Subtract (5) from (4): . 2b = 8\quad\Rightarrow\quad\boxed{b = 4}

    Substitute into (4): . 24a + 6(4)\:=\:0\quad\Rightarrow\quad\boxed{a = -1}

    Substitute into (3): . -1 + 4 + c \:=\:8\quad\Rightarrow\quad\boxed{c = 5}


    Therefore, the equation of the parabola is: . \boxed{y \:=\:-x^2 + 4x + 5}


    Note: . -x^2 + 4x + 5 \:=\:0 is not a parabola.
    . . . . . It is an equation . . . with two solutions: x \:=\:-1,\,5
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