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

To find the object's height after 10 seconds as well as the rate of change for the interval from 5 seconds to 8 seconds, we will use the given height function:

\[
y = -5(x - 4)^2 + 180
\]

### Part A: Height after 10 seconds

We need to substitute \( x = 10 \) into the height function.

\[
y(10) = -5(10 - 4)^2 + 180
\]

Calculate \( 10 - 4 \):

\[
10 - 4 = 6
\]

Now substitute back into the equation:

\[
y(10) = -5(6)^2 + 180
\]

Calculate \( 6^2 \):

\[
6^2 = 36
\]

Now multiply:

\[
-5 \times 36 = -180
\]

So now we have:

\[
y(10) = -180 + 180 = 0
\]

The object's height after 10 seconds is **0 meters**.

### Part B: Rate of change from 5 seconds to 8 seconds

To find the rate of change of height over the interval from \( x = 5 \) to \( x = 8 \), we first need to calculate the height at both \( x = 5 \) and \( x = 8 \).

1. **Calculate height at \( x = 5 \)**:

\[
y(5) = -5(5 - 4)^2 + 180
\]
\[
= -5(1)^2 + 180
\]
\[
= -5 \times 1 + 180
\]
\[
= -5 + 180 = 175
\]

So, \( y(5) = 175 \) meters.

2. **Calculate height at \( x = 8 \)**:

\[
y(8) = -5(8 - 4)^2 + 180
\]
\[
= -5(4)^2 + 180
\]
\[
= -5 \times 16 + 180
\]
\[
= -80 + 180 = 100
\]

So, \( y(8) = 100 \) meters.

3. **Calculate the rate of change from 5 to 8 seconds**:

The rate of change is given by the formula:

\[
\text{Rate of change} = \frac{y(8) - y(5)}{8 - 5}
\]

Substituting the values we calculated:

\[
\text{Rate of change} = \frac{100 - 175}{8 - 5}
\]
\[
= \frac{-75}{3} = -25
\]

The rate of change of height from 5 seconds to 8 seconds is **-25 meters per second**.

### Final Answers:

- Height after 10 seconds: **0 meters**
- Rate of change from 5 seconds to 8 seconds: **-25 meters per second**.