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What is the minimum value of x^2 - 8x + 32

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What is the minimum value of x^2 - 8x + 32

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Posted (Answer #1)

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The minimum value of x^2 - 8x + 32 has to be determined.

x^2 - 8x + 32

= x^2 - 8x + 16 + 16

= (x - 4)^2 + 16

The least value that (x - 4)^2 can take on is 0.

This gives the least value of x^2 - 8x + 32 as 16

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