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18 April, 03:02

For the equation y=3x^2+12x-4, what is the minimum value?

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  1. 18 April, 03:29
    0
    (-2, - 16)

    Step-by-step explanation:

    the minimum is at the vertex which can be found using - b/2a for the x-value of the vertex.

    y = 3x^2 + 12x - 4

    = ax^2 + bx - c

    b = 12, a = 3

    -12 / (2 (3))

    -12/6

    x = - 2

    y = 3 (-2) ^2 + 12 (-2) - 4

    y = - 16

    (-2, - 16)
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