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. in 4 sentences.
Answers
GPT-4o mini
To find the object's height after 10 seconds, we substitute \( x = 10 \) into the height function:
\[
y = -5(10 - 4)^2 + 180 = -5(6)^2 + 180 = -5(36) + 180 = -180 + 180 = 0.
\]
Thus, the object’s height after 10 seconds is 0 meters.
For the rate of change from 5 to 8 seconds, we calculate the height at these times:
At \( x = 5 \):
\[
y = -5(5 - 4)^2 + 180 = -5(1)^2 + 180 = -5 + 180 = 175.
\]
At \( x = 8 \):
\[
y = -5(8 - 4)^2 + 180 = -5(4)^2 + 180 = -5(16) + 180 = -80 + 180 = 100.
\]
The rate of change from 5 to 8 seconds is then calculated as:
\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{3} = \frac{-75}{3} = -25 \, \text{meters per second}.
\]
\[
y = -5(10 - 4)^2 + 180 = -5(6)^2 + 180 = -5(36) + 180 = -180 + 180 = 0.
\]
Thus, the object’s height after 10 seconds is 0 meters.
For the rate of change from 5 to 8 seconds, we calculate the height at these times:
At \( x = 5 \):
\[
y = -5(5 - 4)^2 + 180 = -5(1)^2 + 180 = -5 + 180 = 175.
\]
At \( x = 8 \):
\[
y = -5(8 - 4)^2 + 180 = -5(4)^2 + 180 = -5(16) + 180 = -80 + 180 = 100.
\]
The rate of change from 5 to 8 seconds is then calculated as:
\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{3} = \frac{-75}{3} = -25 \, \text{meters per second}.
\]