1952 AHSME Problems/Problem 13

Problem

The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:

$\textbf{(A) \ }x=-p \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$

$\textbf{(E) \ }x=\frac{-p}{2}$

Solution

The minimum value of this parabola is found at its turning point, on the line $\boxed{\textbf{(E)}\ x=\frac{-p}{2}}$ . Indeed, the turning point of any function of the form $ax^2+bx+c$ has an x-coordinate of $\frac{-b}{2a}$ . This can be seen at the average of the quadratic's two roots (whose sum is $\frac{-b}{a}$ ) or (using calculus) as the value of its derivative set equal to $0$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1952 AHSME Problems/Problem 13

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Solution

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