1954 AHSME Problem 42

Consider the graphs of (1)\qquad y=x^2-\frac{1}{2}x+2 and (2)\qquad y=x^2+\frac{1}{2}x+2 on the same set of axis. These parabolas are exactly the same shape. Then:

\begin{array}{l} \textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.