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