What is the minimum value of x^2 - 8x + 32
1 Answer | Add Yours
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
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes