Question

An object is launched from a platform. Its height, y (in meters), x seconds after the launch is modeled by the function y=−5(x−4)2+180
.

What is the object’s height after 10 seconds?

What is the rate of change for the interval 5 seconds to 8 seconds?

Show your work.

(2 points)

Answer 1

y(10) = −5(10−4)^2 + 180 = −5(6)^2 + 180 = −5(36) + 180 = −180 + 180 = 0 m.

Average rate of change on [5,8]:
y(5) = −5(5−4)^2 + 180 = −5(1) + 180 = 175 m
y(8) = −5(8−4)^2 + 180 = −5(16) + 180 = 100 m
Rate = (y(8) − y(5)) / (8 − 5) = (100 − 175) / 3 = −75 / 3 = −25 m/s.

So after 10 s the height is 0 m, and the average rate of change from 5 s to 8 s is −25 m/s (height decreasing).