Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 36


$ y=4$

Work Step by Step

Given $$ y=x^{2}+2 x+5$$ So, we have \begin{aligned} y&=x^{2}+2 x+5\\ &=x^{2}+2 x+1 -1+5\\ &=(x^{2}+2 x+1) +4\\ &=(x+1)^{2}+4\\ \end{aligned} So, we see that $ y=4$ is the minimum, when $ x=-1$
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