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=-x^2+6x+7. (1 point)

What is the ball's height after 5 seconds?

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

Answer 1

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

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

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

Next, we need to find the rate of change for the interval from 2 to 5 seconds. The rate of change can be found by calculating the average rate of change between the two points, which is given by the formula:

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

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

First, we calculate the height of the ball at \( t = 2 \):

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

Now, we have the values at \( t = 2 \) and \( t = 5 \):
- \( f(2) = 15 \)
- \( f(5) = 12 \)

Now, apply these values to the rate of change formula:

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

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