Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write

Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write the equation of the parabola in vertex form.

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  1. As it has a maximum value the coefficient of x^2 will be negative

    The vertex will be at (-8,2)  so in vertex form it is

    y = a(x + 8)^2 + 2
    and as it passes through (-7,-1) we have:

    -1  = a(-7+8)^2 + 2

    -1 = a + 2  so a = -3

    answer is  y = -3(x + 8)^2 + 2

    in standard form this is y = -3x^2 -48x  - 190

  2. Its standard form of equation: (x-h)^2=-4p (y-k), (h,k)=(x,y) coordinates of the vertex
    For given parabola: Vertex = (-8, 2)
    (x-(-8))² = -4p(y - 2)

    solve for 4p using coordinates of given point ( -7, -1)
    -7+8  = 4p (-1 - 2)
    1 = 4p(-3)
    4p = -1/3

    So, Equation of given parabola:
    (x+8)² = -(y-2) / 3

    Hope this helps!

  3. we know that

    If an axis of symmetry is [tex]x=-8[/tex] and has a maximum height of [tex]2[/tex]

    then

    Is a vertical parabola open down with vertex at [tex](-8,2)[/tex]

    the equation in vertex form is equal to

    [tex]y=a(x-h)^{2}+k[/tex]

    where

    (h,k) is the vertex

    substitute

    [tex]y=a(x+8)^{2}+2[/tex]

    Find the value of a

    the parabola passes through the point [tex](-7,-1)[/tex]

    substitute in the formula

    [tex]-1=a(-7+8)^{2}+2[/tex]

    [tex]-1=a(1)^{2}+2[/tex]

    [tex]-1=a+2[/tex]

    [tex]a=-3[/tex]

    therefore

    the answer is

    the equation in vertex form is  [tex]y=-3(x+8)^{2}+2[/tex]

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