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?

Answer 1

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