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Math Help - A Question About a Parabola

  1. #1
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    A Question About a Parabola

    First of all, this is my first post here, and I hope to fit in with y'all here with Math.
    I'm in grade 10 Math in Canada, and I just recently wrote a test on the Quadratic Function.
    I didn't do as great as I wanted to, as I an usually a 90+ student. One question on the test really puzzled me, and I was shoping someone could help me with it.

    Determine the equation of the parabola with x-intercepts -6 and 2, and a y intercept of 4. Express your answer in standard form.

    Thanks!
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  2. #2
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    Quote Originally Posted by ty2391 View Post
    First of all, this is my first post here, and I hope to fit in with y'all here with Math.
    I'm in grade 10 Math in Canada, and I just recently wrote a test on the Quadratic Function.
    I didn't do as great as I wanted to, as I an usually a 90+ student. One question on the test really puzzled me, and I was shoping someone could help me with it.

    Determine the equation of the parabola with x-intercepts -6 and 2, and a y intercept of 4. Express your answer in standard form.

    Thanks!
    Hello, Ty2391,

    the equation of your parabola is:
    y=a\cdot (x+6)(x-2). Expand the RHS:

    y=ax^2+4ax-12a. Now you know that the y-intercept is: (0, -12a)
    -12a = 4\ \Longleftrightarrow\ a=-\frac{1}{3}

    Thus your equation is:

    y=-\frac{1}{3}x^2-\frac{4}{3}x+4

    EB
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  3. #3
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    Hello, ty2391!

    Another approach . . .


    Determine the equation of the parabola with x-intercepts -6 and 2, and a y-intercept of 4.

    The general form of the parabola is; . y \:=\:ax^2 + bx + c

    Since the y-intercept is 4, the point (0,4) satisfies the equation.
    . . We have: . 4 \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad c = 4

    The equation (so far) is: . y \:=\:ax^2 + bx + 4


    Here is a sketch of the parabola . . .
    Code:
                          |
                      *   |
                  *   :   *
                *     :   | *
               *      :   |  *
                      :   |
          ----*-------+---+---*----
             -6      -2   |   2
                          |

    We know that the axis of the parabola is halfway between the x-intecepts.
    . . That is, the vertex lies on the vertical line x = -2

    We also know that the vertex is at x \:=\:\frac{\text{-}b}{2a}
    . . Then: . \frac{\text{-}b}{2a}\:=\:-2\quad\Rightarrow\quad b \,= \,4a

    So the equation is: . y \:=\:ax^2 + 4ax + 4


    Since (2,0) satisfies the equation, we have:
    . . 0 \:=\:a\cdot2^2 + 4a\cdot2 + 4\quad\Rightarrow\quad 12a \,=\,\text{-}4\quad\Rightarrow\quad a = \text{-}\frac{1}{3}

    Therefore, the equation is: . \boxed{y \;= \;-\frac{1}{3}x^2 - \frac{4}{3}x + 4}

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  4. #4
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    Thanks a lot both of you!
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