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Lesechka [4]
4 years ago
7

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?
Mathematics
1 answer:
Jlenok [28]4 years ago
3 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
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