Question

And object is long from a platform it's height why and meters X seconds after the launch is modeled by the function y = -5 (x - 4)^2 + 180
What is an object height after 10 seconds
What is the rate of change for the interval 5 seconds to 8 seconds

Answer 1

To analyze the height of the object described by the function \( y = -5 (x - 4)^2 + 180 \), we will first determine the object's height after 10 seconds and then the rate of change between 5 seconds and 8 seconds.

**1. Finding the height after 10 seconds:**

We substitute \( x = 10 \) into the equation:

\[
y = -5(10 - 4)^2 + 180
\]
Calculating the value inside the parentheses first:

\[
10 - 4 = 6
\]

Now square it:

\[
6^2 = 36
\]

Now substitute this back into the equation:

\[
y = -5(36) + 180
\]
Calculating:

\[
y = -180 + 180 = 0
\]

The height of the object after 10 seconds is \( 0 \) meters.

**2. Finding the rate of change from 5 seconds to 8 seconds:**

To determine the rate of change, we first find the height at both seconds.

- **At \( x = 5 \):**

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

\[
5 - 4 = 1 \quad \Rightarrow \quad 1^2 = 1
\]
Substituting:

\[
y = -5(1) + 180 = -5 + 180 = 175
\]

So, the height at \( x = 5 \) seconds is \( 175 \) meters.

- **At \( x = 8 \):**

\[
y = -5(8 - 4)^2 + 180
\]
Calculating:

\[
8 - 4 = 4 \quad \Rightarrow \quad 4^2 = 16
\]
Substituting:

\[
y = -5(16) + 180 = -80 + 180 = 100
\]

So, the height at \( x = 8 \) seconds is \( 100 \) meters.

Now, we find the rate of change from \( x = 5 \) to \( x = 8 \):

Rate of change = \(\frac{\text{Change in height}}{\text{Change in time}} = \frac{y(8) - y(5)}{8 - 5}\)

Substituting the values we found:

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

Thus, the rate of change of height from 5 seconds to 8 seconds is \( -25 \) meters per second.

**Summary:**
- Height after 10 seconds: \( 0 \) meters
- Rate of change from 5 to 8 seconds: \( -25 \) meters per second