Given

\(y = x^2 - 32x + 256\)

Carefully analyzing the equation it's clear that the value of x^2 will finally be added to 256. In order to have less least value for y , x has to be minimum.

Option E fits.

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 0

We can clearly see the the coefficient of the linear term (-32) and the constant term (256) can be obtained by adding -16 to itself and multiplying -16 by itself respectively.

Therefore there exists one real root, which is 16.

And the equation will have a value of zero for x=16

Since all values in the choices are positive and zero, zero will be the minimum value for \(y\).

The answer is Choice E.

OR

The discriminant of this equation is \(b^2-4ac\)

\(b^2 = (-32)^2 = (2 \times -16)^2 = (2)^2 \times (-16)^2 = 4 \times 256\)

\(4ac= 4 \times 1 \times 256\)\( = 4 \times 256\)

We note that \(b^2 = 4ac\) so the discriminant is zero. So there should be one real root, the parabola is touching the x-axis at just one place. So the equation will be equal to zero at that point.

So Choice E is correct.

y' = 2x-32
x=16
=> y min at x=16
=> y=0