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Math Help - I need your help bruddas! ASAP! STAT! NOW!

  1. #1
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    Angry I need your help bruddas! ASAP! STAT! NOW!

    For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3

    Find the Y coordinate of the turning point.
    The equation of the parabola
    The coordinates of the x-intercepts.

    -----------------------------------------------------------------

    The parabola has a turning point at (Z, -8); it intersects the y-axis at y=10 and one of the x-intercepts is x=5.

    Find the value of Z.
    Find the equation of the parabola.

    -----------------------------------------------------------------

    The graph of y = a(h - x)^3 + k cuts the x-axis at x = - 4 and the y axis and y= 28. It has a dilation of 1.

    Find the position of the SPOI.
    -----------------------------------------------------------------

    Thanks
    Help would be appreciate.

    Cheers
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  2. #2
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    Quote Originally Posted by mibamars View Post
    For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3
    What have you done?

    The Range restriction is in 'y', so this must open up or down. None of that left or right stuff.

    You then have y - k = a(x-h)^2

    You are given k = 3 and a < 0 and h = -4

    y - 3 = a(x+4)^2

    Substitute x = 0 and find the value for 'a' that gives y = -7/3.

    Let's see what you get.
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by mibamars View Post
    For the parabola whose range is y <_ 3 , whose x-coordinate of the turning point -4 and whose y-intercept is y= - 2 1/3

    Find the Y coordinate of the turning point.
    The equation of the parabola
    The coordinates of the x-intercepts.
    Try to post one question at a time. It leads us to believe you just want questions answered for homework when you post so many at once within one thread.

    I assume turning point is where the graph turns, hence where a max or min is at. This is also known as a critical point. The graph is a parabola, and it has a maximum value at y=3 thanks to the given range. Therefore, that maximum must be the critical point at x=-4. The coordinates of the critical point are then (-4, 3).

    The parabola takes on this equation: y=a(x+c)^2+b
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  4. #4
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    Hello, mibamars!

    Here's the first one . . .


    For the parabola whose range is y \leq 3,
    whose x-coordinate of the turning point is -4,
    and whose y-intercept is -\frac{7}{3}

    Find:
    (1) the y-coordinate of the turning point.
    (2) the equation of the parabola
    (3) the coordinates of the x-intercepts.

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

    Make a sketch . . .
    Code:
                          |
               (-4,3)     |
                  *       |
               *     *    |
             *         *  |
            *           * |
          - - - - - - - - + - - - - -
           *             *|
                          |
                          |
          *               *
                          |
    Since \left(0,\,-\frac{7}{3}\right) is on the parabola: . -\frac{7}{3} \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad \boxed{c \:=\:-\frac{7}{3}}

    The equation (so far) is: . y \;=\;ax^2 + bx - \frac{7}{3}


    Since the range (values of y) is < 3,
    . . and the x-coordinate of the turning point is -4,
    . . clearly the vertex is (-4,3).

    {\color{red}\text{(1) The }y\text{-coordinate of the turning point is }3.}



    The x-coordinate of the vertex is given by: . \frac{-b}{2a}
    . . So we have: . \frac{-b}{2a} \:=\:-4\quad\Rightarrow\quad 8a - b \:=\:0 . [1]

    Since (-4,3) is on the parabola: . 3 \:=\:a(\text{-}4)^2 + b(\text{-}4) - \frac{7}{3}
    . . which simplifies to: . 16a - 4b \:=\:\frac{16}{3}\quad\Rightarrow\quad 4a - b \:=\:\frac{4}{3} . [2]

    Subtract [2] from [1]: . 4a \:=\:-\frac{4}{3}\quad\Rightarrow\quad\boxed{ a \:=\:-\frac{1}{3}}

    Substitute into [1]: . 8\left(\text{-}\frac{1}{3}\right) - b \:=\:0\quad\Rightarrow\quad\boxed{ b \:=\:-\frac{8}{3}}

    {\color{red}\text{(2) The equation of the parabola is: }\:y \;=\;-\frac{1}{3}x^2 - \frac{8}{3}x - \frac{7}{3}}



    For the x-intercepts: . -\frac{1}{3}x^2 - \frac{8}{3}x - \frac{7}{3} \;=\;0

    Multiply by -3: . x^2 + 8x + 7 \:=\:0

    . . Factor:. (x+7)(x+1) \:=\:0\quad\Rightarrow\quad x \:=\:-7,-1


    {\color{red}\text{(3) The }x\text{-intercepts are: }\:(-7,0), (-1,0)}

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