Ask Question
18 January, 11:55

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?

+1
Answers (1)
  1. 18 January, 13:20
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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? ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers