Answer
a) $s=-16t^2+72t+6.5$
b) See graph
c) $8$
d) Positive slope
e) $y=8t+6.5$
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:
$s_0=6.5$
$v_0=72$
The equation of the position is:
$s=-16t^2+72t+6.5$
b) Graph the function.
c) Determine the average rate of change from $t_1=0$ to $t_2=4$:
$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(4)-s(0)}{4-0}$
$=\dfrac{[-16(4^2)+72(4)+6.5]-[-16(0^2)+72(0)+6.5]}{3}$
$=\dfrac{32}{4}$
$=8$
d) Notice that the slope of the secant line is positive.
e) Determine the equation of the secant using the slope $m=4$ and the point $(0,s(0))=(0,-16(0^2)+72(0)+6.6)=(0,6.5) $:
$y-6.5=8(t-0)$
$y=8t+6.5$
f) Graph the secant line:
