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
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:
\(\bold{h(t)=-16t\times 2+144 }\)
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.
\(\bold{-16(0)^2 + 144 = 144ft}\)
Then calculate for t to determine the upper limit for t, however when the height is zero (because the object is being dropped).
\(-16t^2 + 144 = 0\\\\-16t^2 = -144\\\\t^2 = \frac{144}{16}\\\\t^2 = 9\\\\t = \sqrt{9}\\\\t = 3\)
0 to 3 seconds is the domain.
Therefore, the function domain is "0 to 3".
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