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Math Help - Parabola?

  1. #1
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    Parabola?

    Write the equation of the parabola that meets each set of conditions.

    The parabola has a horizontal axis and passes through the origin and points at (-1,2) and (3,-2).

    Now I am not sure how to do this would I use the midpoint formula to find the vertex but If I do I end up with (1,0)
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  2. #2
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    The general formula for a parabola is y=ax^2 + bx + c

    Since it goes through the origin, you can sub in (0,0) to the equation

    So it becomes:  0=c So we know that so far the equation is  y= ax^2 + bx .

    Now you have 2 variables "a" and "b" and you have 2 points. Sub those 2 points in to find a and b.
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  3. #3
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    Quote Originally Posted by homeylova223 View Post
    Write the equation of the parabola that meets each set of conditions.

    The parabola has a horizontal axis and passes through the origin and points at (-1,2) and (3,-2).

    Now I am not sure how to do this would I use the midpoint formula to find the vertex but If I do I end up with (1,0)
    1. Let V(h, k) denote the vertex of the parabola and p the distance between the vertex and the focus then the genaral equation of the parabola is:

    (y-k)^2=4p(x-h)

    2. You know 3 points of the parabola: (0, 0), (-1, 2), (3, -2) and you have to determine 3 unknowns: h, k, p

    3. Plug in the coordinates of the points into the general equation and solve for (h,k,p).
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  4. #4
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    Hello, homeylova223!

    Write the equation of the parabola.

    The parabola has a horizontal axis and passes through the origin
    and the points at (-1,2) and (3,-2).

    Now I am not sure how to do this.
    Would I use the midpoint formula to find the vertex? . What?

    A horizontal parabola has the form: . x \;=\;ay^2 + by + c

    We know three points on the parabola.
    Substitute their values and solve the system of equations.


    (0,0)\!:\;\;a(0^2) + b(0) + c \:=\:0 \quad\Rightarrow\quad \boxed{c \:=\:0}

    (\text{-}1,2)\!:\;\;a(2^2) + b(2) + 0 \:=\:\text{-}1 \quad\Rightarrow\quad 4a + 2b \:=\:\text{-}1 .[1]

    (3,\text{-}2)\!:\;\;a(\text{-}2)^2 + b(\text{-}2) + 0 \:=\:3 \quad\Rightarrow\quad 4a - 2b \:=\:3 .[2]

    Add [1] and [2]: . 8a \,=\,2 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{4}}

    Substitute into [1]: . 4(\tfrac{1}{4}) + 2b \:=\:-1 \quad\Rightarrow\quad \boxed{b \:=\:\text{-}1}


    The equation is: . x \;=\;\tfrac{1}{4}y^2 - y

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