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