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Difference between revisions of "1950 AHSME Problems/Problem 41"
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==Solution 2 ==
 
==Solution 2 ==
  
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Using calculus, the derivative of the function is <math>2ax + b</math>, which has a critical point at <math>\frac{-b}{2a}</math>. The second derivative of the function is <math>2a</math>; since <math>a > 0</math>, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain <math>\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}</math>.
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Using calculus, we find that the derivative of the function is <math>2ax + b</math>, which has a critical point at <math>\frac{-b}{2a}</math>. The second derivative of the function is <math>2a</math>; since <math>a > 0</math>, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain <math>\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:17, 19 January 2024

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 $a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c=a(\frac{b^2}{4a^2})-\frac{b^2}{2a}+c=\frac{b^2}{4a}-\frac{2b^2}{4a}+c=-\frac{b^2}{4a}+c=\frac{4ac}{4a}-\frac {b^2}{4a}=\frac{4ac-b^2}{4a} \Rightarrow \mathrm{(D)}$.

Solution 2

Using calculus, we find that the derivative of the function is $2ax + b$, which has a critical point at $\frac{-b}{2a}$. The second derivative of the function is $2a$; since $a > 0$, this critical point is a minimum. As in Solution 1, plug this value into the function to obtain $\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}$.

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

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