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1 January, 21:25

Consider the function f (x) = x2+bx-16, where b is a constant.

Part A

If the function has an axis of symmetry at x=5, what is the value of b?

Part B

If b=-6, what are the zero (s) of the function?

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Answers (1)
  1. 1 January, 21:48
    0
    F (x) = x^2+bx-16

    x=5

    F (5) = 5^2+b (5) - 16

    =25+5b-16=0

    =5b+9=0

    =5b=-9

    =b = - 9/5

    By remainder theorem
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