Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.3 - Analyzing Graphs of Functions - 2.3 Exercises - Page 196: 69


a) $s=-16t^2+120t$ b) See graph c) $-8$ d) Negative slope e) $y=-8t+240$ f) See graph

Work Step by Step

a) The function representing the position is: $s=-16t^2+v_0t+s_0$ where $v_0$ is the velocity and $s_0$ is the height from which the object is thrown. We are given: $s_0=0$ $v_0=120$ The equation of the position is: $s=-16t^2+120t$ b) Graph the function. c) Determine the average rate of change from $t_1=3$ to $t_2=5$: $\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(5)-s(3)}{5-3}$ $=\dfrac{[-16(5^2)+120(5)]-[-16(3^2)+120(3)]}{2}$ $=\dfrac{-16}{2}$ $=-8$ d) Notice that the slope of the secant line is negative. e) Determine the equation of the secant using the slope $m=-8$ and the point $(3,s(3))=(3,-16(3^2)+120(3))=(3,216) $: $y-216=-8(t-3)$ $y=-8t+24+216$ $y=-8t+240$ f) Graph the secant line:
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