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 using the given function:

### **Step 1: Calculate height at \(x = 10\)**

The height function is given by:

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

Now, substituting \(x = 10\):

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

Calculating inside the parentheses first:

\[
10 - 4 = 6
\]

Substituting that into the equation:

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

Calculating \(6^2\):

\[
6^2 = 36
\]

Now substitute back:

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

Calculating \(-5(36)\):

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

Substituting this value into the equation:

\[
y = -180 + 180
\]

So the height at \(x = 10\) is:

\[
y = 0
\]

### **Answer for Height:**
The object’s height after 10 seconds is **0 meters**.

---

### **Step 2: Calculate the rate of change from \(x = 5\) seconds to \(x = 8\) seconds**

To find the rate of change, we can use the formula:

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

First, we need to calculate \(y(8)\) and \(y(5)\).

#### **Calculate \(y(8)\)**

Substituting \(x = 8\) back in:

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

Calculating inside the parentheses first:

\[
8 - 4 = 4
\]

Now substituting back:

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

Calculating \(4^2\):

\[
4^2 = 16
\]

Now substituting:

\[
y(8) = -5(16) + 180
\]

Calculating \(-5(16)\):

\[
-5 \times 16 = -80
\]

So:

\[
y(8) = -80 + 180 = 100
\]

#### **Calculate \(y(5)\)**

Now for \(x = 5\):

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

Calculating inside the parentheses:

\[
5 - 4 = 1
\]

Substituting back:

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

Calculating \(1^2\):

\[
1^2 = 1
\]

Now substituting:

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

Calculating:

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

#### **Calculate the Rate of Change**

Now we can find the rate of change:

\[
\text{Rate of Change} = \frac{y(8) - y(5)}{8 - 5} = \frac{100 - 175}{8 - 5}
\]

Calculating the difference:

\[
= \frac{-75}{3} = -25
\]

### **Answer for Rate of Change:**
The rate of change from 5 seconds to 8 seconds is **-25 meters per second**.