Question

Mesa stands on a balcony and throws a ball to her dog, who is at ground level. The ball’s height (in meters) x
seconds after Mesa threw it is modeled by the equation y=−x2+6x+7
.(1 point)
What is the ball’s height after 5 seconds?
meters

What is the rate of change for the interval 2 to 5 seconds?

meters per second

Answer 1

To find the ball's height after 5 seconds using the given equation \(y = -x^2 + 6x + 7\), we need to substitute \(x = 5\) into the equation.

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

\[
y = -5^2 + 6(5) + 7
\]
\[
y = -25 + 30 + 7
\]
\[
y = 12 \, \text{meters}
\]

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

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

The rate of change over an interval is calculated by finding the difference in height at the two points divided by the difference in time.

First, we find the height at \(x = 2\):

\[
y = -2^2 + 6(2) + 7
\]
\[
y = -4 + 12 + 7
\]
\[
y = 15 \, \text{meters}
\]

Now we have:
- Height at \(x = 2\) seconds: 15 meters
- Height at \(x = 5\) seconds: 12 meters

Now we calculate the rate of change:

\[
\text{Rate of change} = \frac{y(5) - y(2)}{5 - 2}
\]
\[
= \frac{12 - 15}{5 - 2}
\]
\[
= \frac{-3}{3}
\]
\[
= -1 \, \text{meter per second}
\]

So, the rate of change for the interval from 2 to 5 seconds is **-1 meter per second**.