f(x) = -16x^2 + 36
Where:
f(x) = height of the object
x = seconds after being dropped.
f(-1) = -16 (-1)^2 + 36
f(-1) = -16 (1) + 36
f(-1) = 20
-1 seconds after the object was dropped, the object was 20 ft above the ground.
This interpretation does not make sense, because seconds can't be negative.
f(0.5) = -16 (0.5)^2 + 36
f(0.5) = -16 (0.25) +36
f(0.5) = -4 + 36
f(0.5) = 32
0.5 seconds after the object was dropped, the object was 32 ft above the ground.
This interpretation makes sense in the context of the problem.
f(2) = -16 (2)^2 + 36
f(2) = -16 (4) +36
f(2) = -64+36
f(2) = -28
2 seconds after the object was dropped, the object was -28 ft above the ground.
This interpretation does not make sense in the context of the problem, because the height can't be negative.
Based on the observation, the domain of the function is real numbers in a <- x <-b , possible values of x where f(x) is true.
before the object is released x=0
next, calculate x when f(x)=0 ( after the object hits the ground)
0= -16x^2+36
16x^2 = 36
x^2 = 36/16
x^2 = 2.25
x = √2.25
x = 1.5
0 ≤ x ≤ 1.5