Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.3 The Complex Plane; De Moivre's Theorem - 9.3 Assess Your Understanding - Page 595: 74


Minimum value: $-3.2.$

Work Step by Step

Let's compare $f(x)=5x^2-12x+4$ to $f(x)=ax^2+bx+c$. We can see that a=5, b=-12, c=4. a>0, hence the graph opens up, hence it's vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-12}{2\cdot 5}=1.2.$ Hence the minimum value is $f(1.2)=5(1.2)^2-12(1.2)+4=7.2-14.4+4=-3.2.$
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