Find an equation in standard form of the parabola passing through these points: (0,-,-,0)

Find an equation in standard form of the parabola passing through these points: (0,-,-,0)

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  1. The standard form of a parabola is y=ax²+bx+c
    use the three given points to find the three unknown constants a, b, and c:
    -2=a+b+c1
    -2=4a+2b+c 2
    -4=9a+3b+c3
    equation 2 minus equation 1: 3a+b=04
    equation 3 minus equation 2: 5a+b=-25
    equation 5 minus equation 4: 2a=-2, so a=-1
    plug a=-1 in equation 4: -3+b=0, so b=3
    Plug a=-1, b=3 in equation 1: -2=-1+3+c, so c=-4
    the parabola is y=-x²+3x-4

    double check: when x=1, y=-1+3-4=-2
    when x=2, y=-4+6-4=-2
    when x=3, y=-9+9-4=-4
    Yes.

  2. One form of the equation of a parabola is
    y = ax² + bx + c

    The curve passes through (0,-6), (-1,-12) and (3,0). Therefore
    c = - 6                       (1)
    a - b + c = -12            (2)
    9a + 3b + c = 0         (3)

    Substitute (1) into (2) and into (3).
    a - b -6 = -12
    a - b = -6                  (4)
    9a + 3b - 6 = 0
    9a + 3b = 6              (5)

    Substitute a = b - 6 from (4) into (5).
    9(b - 6) + 3b = 6
    12b - 54 = 6
    12b = 60
    b = 5
    a = b - 6 = -1

    The equation is
    y = -x² + 5x - 6 

    Let us use completing the square to write the equation in standard form for a parabola.
    y = -[x² - 5x] - 6
       = -[ (x - 2.5)² - 2.5²] - 6
       = -(x - 2.5)² + 6.25 - 6

    y = -(x - 2.5)² + 0.25
    This is the standdard form of the equation for the parabola.
    The vertex us at (2.5, 0.25).
    The axis of symmetry is x = 2.5
    Because the leading coefficient is -1 (negative), the curve opens downward.
    The graph is shown below.

    y = -(x - 2.5)² + 0.25
    [tex]Find an equation in standard form of the parabola passing through these points: (0,-,-,0)[/tex]

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