# Chapter 2 - 2.3 - Analyzing Graphs of Functions - 2.3 Exercises - Page 196: 67

a) $s=-16t^2+64t+6$ b) See graph c) $16$ d) Positive slope e) $y=16t+6$ 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 where the object is thrown. We are given: $s_0=6$ $v_0=64$ The equation of the position is: $s=-16t^2+64t+6$ b) Graph the function. c) Determine the average rate of change from $t_1=0$ to $t_3$: $\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(3)-s(0)}{3-0}$ $=\dfrac{[-16(3^2)+64(3)+6]-[-16(0^2)+64(0)+6]}{3}$ $=\dfrac{48}{3}$ $=16$ d) Notice that the slope of the secant line is positive. e) Determine the equation of the secant using the slope $m=16$ and the point $(0,s(0))=(0,-16(0^2)+64(0)+6)=(0,6)$: $y-6=16(t-0)$ $y=16t+6$ f) Graph the secant line:

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