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: 52

Answer

a) $8.125$ft b) $x \approx 16.33 ft$

Work Step by Step

a) Given: $f(x)=-\dfrac{32}{(20)^2}x^2+x+5$ or, $f(x)=-\dfrac{-2}{25}x^2+x+5$ But $a=-\dfrac{-2}{25} \lt 0$ shows that the function has a maximum value. The maximum value occurs at $x=-\dfrac{b}{2a}=-\dfrac{1}{2(\dfrac{-2}{25})}=6.25$ Thus, the maximum value of the function $f(x)$ at $x=6.25$ is: $f(6.25)=-\dfrac{-2}{25}(6.25)^2+6.25+5$ This gives: $f(6.25)=8.125$ ft b) Consider $f(x)=0$ $-\dfrac{2}{25}x^2+x+5=0$ $x=\dfrac{-1\pm \sqrt {1+\dfrac{8}{5}}}{2(\dfrac{-2}{25})}$ This gives: $x \approx 16.33 ft$ or, $x \approx -3.83 ft$ Since, $x \gt 0$ so we will only consider the positive solution.
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