What is the minimum value of x^2 - 8x + 32, Please use Calculus.

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The minimum value of an expression y = f(x) is at the point where y' = 0 and if the solution of f'(x) = 0 is x = a, f''(a) is positive.

For y = x^2 - 8x + 32

y' = 2x - 8

2x - 8 = 0

=> x = 4

y'' = 2 which is always positive.

The minimum value of x^2 - 8x + 32 is equal to (4)^2 - 8*4 + 32 = 16 - 32 + 32 = 16

**The minimum value of x^2 - 8x + 32 is 16**

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