Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set - Page 459: 81

Answer

a) Please see the table. b) 5 seconds c) approximately 304 feet

Work Step by Step

a) $x=0$ $y=-16x^2+20x+300$ $y=-16*0^2+20*0+300$ $y=0+0+300 = 300$ $x=1$ $y=-16x^2+20x+300$ $y=-16*1^2+20*1+300$ $y=-16+20+300$ $y=304$ $x=2$ $y=-16x^2+20x+300$ $y=-16*2^2+20*2+300$ $y=-16*4+40+300$ $y=-64+340$ $y=276$ $x=3$ $y=-16x^2+20x+300$ $y=-16*3^2+20*3+300$ $y=-16*9+60+300$ $y=-144+360$ $y=216$ $x=4$ $y=-16x^2+20x+300$ $y=-16*4^2+20*4+300$ $y=-16*16+380$ $y=-196+380$ $y=184$ $x=5$ $y=-16x^2+20x+300$ $y=-16*5^2+20*5+300$ $y=-16*25+400$ $y=-400+400$ $y=0$ $x=6$ $y=-16x^2+20x+300$ $y=-16*6^2+20*6+300$ $y=-16*36+120+300$ $y=-576+420$ $y=-156$ b) The ground has a height of 0, and we are at that height at $x=5$ c) The maximum value for the height was 304 feet (at 2 seconds)
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