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pre-cal

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what is the vertex and directrix of the parabola y=-1/2x^2 + 10 - 46

  • pre-cal -

    I'm assuming your equation is this:

    y = -1/2x^2 + 10x - 46

    The above equation would then be in the format:
    y = ax^2 + bx + c
    ...where a, b, and c are the coefficients of each term in this format.

    Therefore, in your equation, a = -1/2, b = 10, and c = -46

    To find the vertex, use this:
    -b/2a for x
    (4ac - b^2)/4a for y

    You can also find y by substituting the value for x into the equation and solving for y.

    Vertex will be in the format (x,y).

    I'll let you take it from here.

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