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What is the minimum value of x^2 - 8x + 32
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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
Posted by justaguide on September 4, 2013 at 5:27 PM (Answer #1)
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