Question

Massive stands in the balcony and throws a ball to her dog who is that ground level the Paul heightened meters act seconds after massive threw it his model by the equation y equals negative x squared plus 6X + 7 what is the ball's high after 5 seconds what is the rate of change for the interval 2 to 5 seconds

Answer 1

To find the height of the ball after 5 seconds, we will use the given equation:

\[
y = -x^2 + 6x + 7
\]

where \( y \) represents the height of the ball in meters, and \( x \) represents the time in seconds.

### Step 1: Calculate the height after 5 seconds

Substituting \( x = 5 \) into the equation:

\[
y(5) = - (5)^2 + 6(5) + 7
\]

Calculating step-by-step:

\[
y(5) = -25 + 30 + 7
\]
\[
y(5) = 12
\]

So, the height of the ball after 5 seconds is **12 meters**.

### Step 2: Calculate the rate of change from 2 to 5 seconds

The rate of change of the height of the ball over the interval from 2 to 5 seconds is given by the average rate of change formula:

\[
\text{Average Rate of Change} = \frac{y(b) - y(a)}{b - a}
\]

Where \( a = 2 \) and \( b = 5 \).

First, we calculate \( y(2) \):

\[
y(2) = - (2)^2 + 6(2) + 7
\]
\[
y(2) = -4 + 12 + 7
\]
\[
y(2) = 15
\]

Now we can calculate the rate of change:

\[
\text{Average Rate of Change} = \frac{y(5) - y(2)}{5 - 2} = \frac{12 - 15}{5 - 2} = \frac{-3}{3} = -1
\]

So, the average rate of change of the ball's height from 2 to 5 seconds is **-1 meter per second**.

### Summary:
- The ball's height after 5 seconds is **12 meters**.
- The rate of change of the ball's height from 2 to 5 seconds is **-1 meter per second**.