Difference between revisions of "1950 AHSME Problems/Problem 41"
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==Solution== | ==Solution== | ||
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| + | The vertex of a parabola is at <math>x=-\frac{b}{2a}</math> for <math>ax^2+bx+c</math>. Because <math>a>0</math>, the vertex is a minimum. Therefore <math>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)}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 16:07, 1 May 2012
Problem
The least value of the function $ax^2\plus{}bx\plus{}c$ (Error compiling LaTeX. Unknown error_msg) with 
is:
Solution
The vertex of a parabola is at 
for 
. Because , the vertex is a minimum. Therefore 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 40 |
Followed by Problem 42 | |
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| All AHSME Problems and Solutions | ||
