To find the object's height after 10 seconds as well as the rate of change for the interval from 5 seconds to 8 seconds, we will use the given height function:
\[ y = -5(x - 4)^2 + 180 \]
Part A: Height after 10 seconds
We need to substitute \( x = 10 \) into the height function.
\[ y(10) = -5(10 - 4)^2 + 180 \]
Calculate \( 10 - 4 \):
\[ 10 - 4 = 6 \]
Now substitute back into the equation:
\[ y(10) = -5(6)^2 + 180 \]
Calculate \( 6^2 \):
\[ 6^2 = 36 \]
Now multiply:
\[ -5 \times 36 = -180 \]
So now we have:
\[ y(10) = -180 + 180 = 0 \]
The object's height after 10 seconds is 0 meters.
Part B: Rate of change from 5 seconds to 8 seconds
To find the rate of change of height over the interval from \( x = 5 \) to \( x = 8 \), we first need to calculate the height at both \( x = 5 \) and \( x = 8 \).
- Calculate height at \( x = 5 \):
\[ y(5) = -5(5 - 4)^2 + 180 \] \[ = -5(1)^2 + 180 \] \[ = -5 \times 1 + 180 \] \[ = -5 + 180 = 175 \]
So, \( y(5) = 175 \) meters.
- Calculate height at \( x = 8 \):
\[ y(8) = -5(8 - 4)^2 + 180 \] \[ = -5(4)^2 + 180 \] \[ = -5 \times 16 + 180 \] \[ = -80 + 180 = 100 \]
So, \( y(8) = 100 \) meters.
- Calculate the rate of change from 5 to 8 seconds:
The rate of change is given by the formula:
\[ \text{Rate of change} = \frac{y(8) - y(5)}{8 - 5} \]
Substituting the values we calculated:
\[ \text{Rate of change} = \frac{100 - 175}{8 - 5} \] \[ = \frac{-75}{3} = -25 \]
The rate of change of height from 5 seconds to 8 seconds is -25 meters per second.
Final Answers:
- Height after 10 seconds: 0 meters
- Rate of change from 5 seconds to 8 seconds: -25 meters per second.