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
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)
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)
Answers
GPT-4o mini
To analyze the given function \( y = -5(x - 4)^2 + 180 \) for the height of the object, we will perform the following steps:
### Calculate the object's height after 10 seconds
1. Substitute \( x = 10 \) into the function:
\[
y = -5(10 - 4)^2 + 180
\]
2. Simplify the expression:
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]
Thus, the object's height after **10 seconds** is **0 meters**.
### Calculate the rate of change from 5 seconds to 8 seconds
To find the rate of change, we will calculate the height of the object at \( x = 5 \) and \( x = 8 \), and then use those values to find the average rate of change.
1. **Calculate height at \( x = 5 \)**:
\[
y = -5(5 - 4)^2 + 180
\]
\[
y = -5(1)^2 + 180
\]
\[
y = -5(1) + 180
\]
\[
y = -5 + 180
\]
\[
y = 175
\]
2. **Calculate height at \( x = 8 \)**:
\[
y = -5(8 - 4)^2 + 180
\]
\[
y = -5(4)^2 + 180
\]
\[
y = -5(16) + 180
\]
\[
y = -80 + 180
\]
\[
y = 100
\]
3. **Calculate the average rate of change from \( x = 5 \) to \( x = 8 \)**:
The average rate of change is given by the formula:
\[
\text{Average Rate of Change} = \frac{y(8) - y(5)}{8 - 5}
\]
Substituting the values calculated:
\[
\text{Average Rate of Change} = \frac{100 - 175}{8 - 5}
\]
\[
\text{Average Rate of Change} = \frac{-75}{3}
\]
\[
\text{Average Rate of Change} = -25
\]
Thus, the rate of change of the object's height from **5 seconds to 8 seconds** is **-25 meters per second**. This indicates that the object's height is decreasing at that interval.
### Calculate the object's height after 10 seconds
1. Substitute \( x = 10 \) into the function:
\[
y = -5(10 - 4)^2 + 180
\]
2. Simplify the expression:
\[
y = -5(6)^2 + 180
\]
\[
y = -5(36) + 180
\]
\[
y = -180 + 180
\]
\[
y = 0
\]
Thus, the object's height after **10 seconds** is **0 meters**.
### Calculate the rate of change from 5 seconds to 8 seconds
To find the rate of change, we will calculate the height of the object at \( x = 5 \) and \( x = 8 \), and then use those values to find the average rate of change.
1. **Calculate height at \( x = 5 \)**:
\[
y = -5(5 - 4)^2 + 180
\]
\[
y = -5(1)^2 + 180
\]
\[
y = -5(1) + 180
\]
\[
y = -5 + 180
\]
\[
y = 175
\]
2. **Calculate height at \( x = 8 \)**:
\[
y = -5(8 - 4)^2 + 180
\]
\[
y = -5(4)^2 + 180
\]
\[
y = -5(16) + 180
\]
\[
y = -80 + 180
\]
\[
y = 100
\]
3. **Calculate the average rate of change from \( x = 5 \) to \( x = 8 \)**:
The average rate of change is given by the formula:
\[
\text{Average Rate of Change} = \frac{y(8) - y(5)}{8 - 5}
\]
Substituting the values calculated:
\[
\text{Average Rate of Change} = \frac{100 - 175}{8 - 5}
\]
\[
\text{Average Rate of Change} = \frac{-75}{3}
\]
\[
\text{Average Rate of Change} = -25
\]
Thus, the rate of change of the object's height from **5 seconds to 8 seconds** is **-25 meters per second**. This indicates that the object's height is decreasing at that interval.