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