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2 April, 19:36

Consider the function f (x) = x^2+bx-49 where b is a constant. if the function has an axis of symmetry at x = 8, what is the value of b?

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  1. 2 April, 19:54
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    For this case we have the following equation:

    f (x) = x ^ 2 + bx-49

    Deriving we have:

    f ' (x) = 2x + b

    We match zero:

    0 = 2x + b

    We clear x:

    x = - b / 2

    The axis of symmetry is at x = 8, therefore:

    x = - b / 2 = 8

    Clearing b:

    b = - 2 * (8)

    b = - 16

    Answer:

    the value of b is:

    b = - 16
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