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
𝑥
seconds after Mesa threw it is modeled by the equation y=−x2+6x+7
𝑦
=
−
𝑥
2
+
6
𝑥
+
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 determine the ball’s height after 5 seconds using the equation \( y = -x^2 + 6x + 7 \), we can substitute \( x = 5 \) into the equation:

\[
y = -(5)^2 + 6(5) + 7
\]
\[
y = -25 + 30 + 7
\]
\[
y = 12
\]

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

Next, we’ll find the rate of change of the ball’s height from 2 seconds to 5 seconds. The rate of change is calculated using the formula:

\[
\text{Rate of Change} = \frac{y_2 - y_1}{x_2 - x_1}
\]

First, we need to find the height of the ball at \( x = 2 \):

\[
y_1 = -(2)^2 + 6(2) + 7
\]
\[
y_1 = -4 + 12 + 7
\]
\[
y_1 = 15
\]

Now we have:
- For \( x_1 = 2 \), \( y_1 = 15 \)
- For \( x_2 = 5 \), \( y_2 = 12 \)

Now plug these values into the rate of change formula:

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

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