If $y=x^2+px+q$, then if the least possible value of $y$ is zero $q$ is equal to:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{p^2}{4}\qquad\textbf{(C)}\ \frac{p}{2}\qquad\textbf{(D)}\ -\frac{p}{2}\qquad\textbf{(E)}\ \frac{p^2}{4}-q$
The least possible value of $y$ is given at the $y$ coordinate of the vertex. The $x$- coordinate is given by 
\[\frac{-p}{(2)(1)} = \frac{-p}{2}\] Plugging this into the quadratic, we get
\[y = \frac{p^2}{4} - \frac{p^2}{2} + q\]
\[0 = \frac{p^2}{4} - \frac{2p^2}{4} + q\]
\[0 = \frac{-p^2}{4} + q\]
\[q = \frac{p^2}{4} = \boxed{B}\]
~JustinLee2017