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

Expert Answers
justaguide eNotes educator| Certified Educator

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