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the function h(t)=-16t2+144 represents the height, h(t), in feet, of an object from the ground at t second after its dropped. A realistic domain for this function is

Answer :

The domain is the limits of the function. Since time doesnt go negative, start with 0. At time 0, the height is

-16(0)^2 + 144 = 144ft

Then, solve for t to find the upper limit for t, which is when the height is zero (since you're dropping the object).

-16t^2 + 144 = 0
-16t^2 = -144
t^2 = 9
t = sqrt(9)
t = 3

The domain is 0 to 3 seconds.

Following are the calculation on the domain:

Given:

[tex]\bold{h(t)=-16t\times 2+144 }[/tex]

To find:

The domain of the function.

Solution:

The domain refers to the function's boundaries. It begins with 0 because time does not go backward. The altitude is 0 at time 0.

[tex]\bold{-16(0)^2 + 144 = 144ft}[/tex]

Then calculate for t to determine the upper limit for t, however when the height is zero (because the object is being dropped).

[tex]-16t^2 + 144 = 0\\\\-16t^2 = -144\\\\t^2 = \frac{144}{16}\\\\t^2 = 9\\\\t = \sqrt{9}\\\\t = 3[/tex]

0 to 3 seconds is the domain.

Therefore, the function domain is "0 to 3".

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