## Algebra and Trigonometry 10th Edition

a) $s=-16t^2+80$ b) See graph c) $-48$ d) Negative slope e) $y=-48t+112$ f) See graph
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: $s_0=80$ $v_0=0$ The equation of the position is: $s=-16t^2+80t$ b) Graph the function. c) Determine the average rate of change from $t_1=1$ to $t_2=2$: $\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(2)-s(1)}{2-1}$ $=\dfrac{[-16(2^2)+80]-[-16(1^2)+80]}{1}$ $=\dfrac{-48}{1}$ $=-48$ d) Notice that the slope of the secant line is negative. e) Determine the equation of the secant using the slope $m=-48$ and the point $(1,s(1))=(1,-16(1^2)+80)=(1,64)$: $y-64=-48(t-1)$ $y=-48t+48+64$ $y=-48t+112$ f) Graph the secant line: