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

(Created page with "==Problem== The least value of the function <math> ax^2\plus{}bx\plus{}c</math> with <math>a>0</math> is: <math>\textbf{(A)}\ -\dfrac{b}{a} \qquad \textbf{(B)}\ -\dfrac{b}{2a}...")
 
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==Problem==
 
==Problem==
  
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The least value of the function <math> ax^2\plus{}bx\plus{}c</math> with <math>a>0</math> is:
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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
 
<math>\textbf{(A)}\ -\dfrac{b}{a} \qquad
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\textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad
 
\textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad
 
\textbf{(E)}\ \text{None of these}</math>
 
\textbf{(E)}\ \text{None of these}</math>
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==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>.
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==Solution 2 ==
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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>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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==See Also==
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{{AHSME 50p box|year=1950|num-b=40|num-a=42}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 12:43, 4 April 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 $\boxed{\textbf{(D)}\ \dfrac{4ac-b^2}{4a}}$.

~ cxsmi

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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