1959 AHSME Problems/Problem 8

Problem

The value of $x^2-6x+13$ can never be less than:

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 13$

Solution

The $x$ value at which the minimum value of this quadratic occurs is $-\frac{-6}{2\cdot1}=3$ . The minimum value of the quadratic is therefore at $3^2-6\cdot3+13$ $=9-18+13$ $=4.$ So, the answer is $\boxed{\textbf{(A)} \ 4}$ .

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1959 AHSME Problems/Problem 8

Problem

Solution