1950 AHSME Problems/Problem 41

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

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

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

Page

Toolbox

Search

1950 AHSME Problems/Problem 41

Contents

Problem

Solution

Solution 2

See Also