Find the minimum or maximum y-value for f(x)=3x^2 + 12x + 8.

Question

Find the minimum or maximum y-value for f(x)=3x^2 + 12x + 8.

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Mackenzie 1 week 2021-09-15T17:52:02+00:00 1 Answer 0

Answers ( )

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    2021-09-15T17:53:18+00:00

    Finding x:

    x = – b / 2a, where b is the coefficient of the x term and a is the coefficient of the x2 term.

    x = – 12 / (2 * 3)

    x = -2

    Finding y:

    To find y, substitute for x in the given function

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

    y = 12 – 24 + 4

    y = -16

    Vertex:

    The vertex is (-2, -16)

    Since the coefficient of the x2 term is positive, we have a minimum.

    The minimum is at -16.

    Your answer:

    MIN,-16

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