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?

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