College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 4 - Section 4.3 - Quadratic Functions and Their Properties - 4.3 Assess Your Understanding - Page 301: 85


See below.

Work Step by Step

Let's compare $f(x)=-4x^2+4000x$ to $f(x)=ax^2+bx+c$. We can see that a=-4, b=4000, c=0. $a\lt0$, hence the graph opens down, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{4000}{2\cdot(-4)}=500.$ Hence the maximum value is $f(500)=-4(500)^2+4000(500)=1000000.$
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