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30 June, 11:30

F (x) = x^2-6x+11 how can you find the vertex of this quadratic function?

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Answers (2)
  1. 30 June, 14:01
    0
    (3,2)

    Step-by-step explanation:

    a = 1

    b = 6

    c = 11

    Vertex of x = 6 / (2*1)

    6 / (2*1) = 3

    Back to f (x) = x^2-6x+11

    Substitute x with 3

    3^2-6 (3) + 11

    9-18+11 = 2

    y = 2
  2. 30 June, 14:48
    0
    (3,2)

    Step-by-step explanation:

    Vertex of x-coordinate = - b/2a

    b = - 6 and a = 1

    Vertex of x-coordinate = - (-6) / (2*1)

    = 6/2

    = 3

    Vertex of y-coordinate = f (3)

    = 3^2 - 6 (3) + 11

    = 9 - 18 + 11

    = 2
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