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1959 AHSME Problems/Problem 8

Revision as of 06:05, 24 September 2018 by Dajeff (talk | contribs) (Created page with "The <math>x</math> value at which the minimum value of this quadratic occurs is <math>-\frac{-6}{2\cdot1}=3</math>. The minimum value of the quadratic is therefore at <cmath>3...")
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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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