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: 41


$ y=\frac{137}{16}$ is the minimum, when $ x=\frac{8}{3}$

Work Step by Step

Given $$ y= -4 x^{2}+3 x+8 $$ So, we have \begin{aligned} y&-4 x^{2}+3 x+8 \\ &=-4\left(x^{2}-\frac{3}{4} x\right)+8\\ & =-4\left(x^{2}-\frac{3}{4} x+\frac{9}{64}-\frac{9}{64}\right)+8\\ & =-4\left(x^{2}-\frac{3}{4} x+\frac{9}{64}\right)+8+\frac{9}{16}\\ & =-4\left(x-\frac{3}{8}\right)^{2}+\frac{137}{16}\\ \end{aligned} The lowest value for the squared term is $0$. So, we see that $ y=\frac{137}{16}$ is the minimum, when $ x=\frac{8}{3}$.
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