hi,
I am taking with my solution on this question, because other posts have provided different solutions within quant concept skill set of GRE
Withous calculus use and derivative of \(y = x^2 - 32x + 256\) set equal to zero, it's practical to factorize the equation firstly
\(y = x^2 - 32x + 256 = (x-16)^2\), when \(y=0\), x-intercept will be 16 and this is the only intercept here
hence, x-intercept accomodates also the minimum value of \(y\) and it's
0sandy wrote:
If \(y = x^2 - 32x + 256\), then what is the least possible value of y ?
A. 256
B. 32
C. 16
D. 8
E. 0
Drill 2
Question: 12
Page: 512