Difference between revisions of "1950 AHSME Problems/Problem 41"

Revision as of 13:16, 19 April 2015

Problem

The least value of the function $ax^2 + bx + c$ with $a>0$ is:

$\textbf{(A)}\ -\dfrac{b}{a} \qquad \textbf{(B)}\ -\dfrac{b}{2a} \qquad \textbf{(C)}\ b^2-4ac \qquad \textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad \textbf{(E)}\ \text{None of these}$

Solution

The vertex of a parabola is at $x=-\frac{b}{2a}$ for $ax^2+bx+c$ . Because $a>0$ , the vertex is a minimum. Therefore $a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c=a(\frac{b^2}{4a^2})-\frac{b^2}{2a}+c=\frac{b^2}{4a}-\frac{2b^2}{4a}+c=-\frac{b^2}{4a}+c=\frac{4ac}{4a}-\frac {b^2}{4a}=\frac{4ac-b^2}{4a} \Rightarrow \mathrm{(D)}$ .

1950 AHSC (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-The least value of the function <math> ax^2\plus{}bx\plus{}c</math>  with <math>a>0</math> is:
+The least value of the function <math>ax^2 + bx + c</math> with <math>a>0</math> is:
 <math>\textbf{(A)}\ -\dfrac{b}{a} \qquad

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Difference between revisions of "1950 AHSME Problems/Problem 41"

Revision as of 13:16, 19 April 2015

Problem

Solution

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