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