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question: an object was dropped off the top of a building. the function…

Question

question: an object was dropped off the top of a building. the function f(x)=-16x² + 144 represents the height of the object above the ground, in feet, x seconds after being dropped. find and interpret the given function values and determine an appropriate domain for the function. answer attempt 1 out of 2: f(-2)=80, meaning that -2 seconds after the object was dropped, the object was 80 feet above the ground. this interpretation in the context of the problem. f(1.5)=108, meaning that 1.5 seconds after the object was dropped, the object was 108 feet above the ground. this interpretation in the context of the problem. f(4)=-112, meaning that 4 seconds after the object was dropped, the object was 112 feet above the ground. this interpretation in the context of the problem. based on the observations above, it is clear that an appropriate domain for the function is

Explanation:

Step1: Calculate $f(-2)$

Substitute $x = - 2$ into $f(x)=-16x^{2}+144$. So $f(-2)=-16\times(-2)^{2}+144=-16\times4 + 144=-64 + 144=80$. But time $x=-2$ is not valid in the context of the object - dropping problem as time cannot be negative before the object is dropped.

Step2: Calculate $f(1.5)$

Substitute $x = 1.5$ into $f(x)=-16x^{2}+144$. Then $f(1.5)=-16\times(1.5)^{2}+144=-16\times2.25+144=-36 + 144 = 108$. This represents the height of the object 1.5 seconds after it is dropped.

Step3: Calculate $f(4)$

Substitute $x = 4$ into $f(x)=-16x^{2}+144$. We get $f(4)=-16\times4^{2}+144=-16\times16+144=-256+144=-112$. A negative height is not valid in the real - world context as it means the object is below the ground level which is not possible for the height function before it hits the ground. The object hits the ground when $f(x)=0$. Set $-16x^{2}+144 = 0$, then $16x^{2}=144$, $x^{2}=9$, $x = 3$ (we take the positive value since time $x\geq0$).

Step4: Determine the domain

Since time $x$ represents the time after the object is dropped and the object hits the ground at $x = 3$ seconds, the domain of the function is $0\leq x\leq3$.

Answer:

The domain of the function $f(x)=-16x^{2}+144$ is $0\leq x\leq3$.