find the VERTEX of the parabola f(x) = x^2 – 16x + 63

Question

find the VERTEX of the parabola f(x) = x^2 – 16x + 63

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Rylee 3 weeks 2021-09-09T07:43:58+00:00 1 Answer 0

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    0
    2021-09-09T07:45:29+00:00

    Answer:

    vertex = (8, – 1 )

    Step-by-step explanation:

    Given a parabola in standard form, f(x) = ax² + bx + c : a ≠ 0

    Then the x- coordinate of the vertex is

    x_{vertex} = – \frac{b}{2a}

    f(x) = x² – 16x + 63 ← is in standard form

    with a = 1 and b = – 16, thus

    x_{vertex} = – \frac{-16}{2} = 8

    Substitute x = 8 into f(x)

    f(8) = 8² – 16(8) + 63 = 64 – 128 + 63 = – 1

    vertex = (8, – 1 )

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