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 300: 55


See below.

Work Step by Step

Let's compare $f(x)=2x^2+12x$ to $f(x)=ax^2+bx+c$. We can see that a=2, b=12, c=0. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot 2}=-3.$ Hence the minimum value is $f(-3)=2(-3)^2+12(-3)=-18.$
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