Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 252: 31

Answer

a. $f(x) = 3(x-2)^2 +1$ b. See graph attached below c. Minimum value = 1

Work Step by Step

$f(x) = 3x^2 -12x + 13$ a. $f(x) = 3( x^2 -4x + 4) + 13 - 3(4)$ $f(x) = 3(x-2)^2 +1$ b. See graph attached below c. The minimum value is at the vertex, which is at (2, 1) (from standard form) Minimum value = 1
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