Question

answer attempt 2 out of 2
f(-2)= -34, meaning that -2 seconds after the object was launched, the object was -34 feet above the ground. this interpretation does not make sense in the context of the problem.
f(0.5)= 51, meaning that 0.5 seconds after the object was launched, the object was 51 feet above the ground. this interpretation makes sense in the context of the problem.
f(6)= -466, meaning that 6 seconds after the object was launched, the object was -466 feet above the ground. this interpretation does not make sense in the context of the problem.
based on the observations above, it is clear that an appropriate domain for the function is real numbers in a≤x≤b for ≤x≤

Explanation:

Step1: Analyze the context

The height of an object above the ground can't be negative in a real - world sense and time before launch ($t=-2$) is not relevant. Time after launch when height is positive makes sense.

Step2: Determine the domain

The domain should start from the time of launch ($t = 0$) and end before the height becomes negative again. Since $f(0.5)=51$ (positive height) and $f(6)= - 466$ (negative height), the domain should be from $0$ to some value less than $6$. A reasonable upper - bound could be the time when the object hits the ground (height = 0). But based on the given values, a simple estimate for the domain of non - negative height values starts at $0$ and we can assume an upper bound of $6$ for now as we don't have more information to find the exact time it hits the ground.

Answer:

$0\leq x\leq6$