Sure, let's determine the domain of the function [tex]\( f(x) = -16x^2 + 14400 \)[/tex] for all viable [tex]\( x \)[/tex] values based on the context of the problem.
The function [tex]\( f(x) = -16x^2 + 14400 \)[/tex] models the height of a food packet dropped from a helicopter, where [tex]\( x \)[/tex] is the time in seconds. The context of the problem involves the packet being dropped and falling until it reaches the ground level (height of 0 feet).
1. Understanding the Function:
- The function [tex]\( f(x) \)[/tex] is a quadratic function that opens downwards, as indicated by the negative coefficient of [tex]\( x^2 \)[/tex].
- The maximum height occurs at the vertex of the parabola. For a function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], the vertex [tex]\( x \)[/tex]-coordinate is given by [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, there's no [tex]\( x \)[/tex] term (i.e., [tex]\( b = 0 \)[/tex]).
- So, the maximum height occurs at [tex]\( x = 0 \)[/tex] seconds: [tex]\( f(0) = -16(0)^2 + 14400 = 14400 \)[/tex] feet.
2. Finding When the Packet Reaches the Ground:
- The packet reaches the ground when [tex]\( f(x) = 0 \)[/tex]:
[tex]\[
-16x^2 + 14400 = 0
\][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
-16x^2 = -14400 \implies x^2 = \frac{14400}{16} = 900 \implies x = \sqrt{900} = 30
\][/tex]
- Thus, the packet hits the ground 30 seconds after it is dropped.
3. Determining the Domain:
- The packet is in the air from the moment it is dropped until it hits the ground. Therefore, the viable values of [tex]\( x \)[/tex] are from 0 seconds to 30 seconds.
Hence, the domain of [tex]\( f(x) \)[/tex], based on the context of the function modeling the height of the food packet as it falls, is [tex]\( 0 \leq x \leq 30 \)[/tex].
