College Algebra (10th Edition)

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

Chapter 4 - Review Exercises - Page 317: 14


See below.

Work Step by Step

Let's compare $f(x)=3x^2-6x+4$ to $f(x)=ax^2+bx+c$. We can see that a=3, b=-6, c=4. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-6}{2\cdot 3}=1.$ Hence the minimum value is $f(1)=3(1)^2-6(1)+4=1.$
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