Answer:
Part A: Vertex of function is at (0.5, 4.5)
h(t) = -6*(t - 0.5)^2 + 4.5
Part B: We want the y-intercept to be (0, 3) because Holly starts at 3 meters above the water surface and at time t =0 seconds.
Solving for when she enters the water, we do not need to use t = 1.5 seconds, because it is unclear if h< 0 at that time.
h(t ) =0 = -6*(t - 0.5)^2 + 4.5
-4.5 = -6*(t - 0.5)^2
4.5/6 = (t - 0.5)^2
0.866025 = t - 0.5
t = 0.5 + 0.866025 = 1.366025 seconds
So at 1.366 seconds, Holly is at the surface of the water.
Step-by-step explanation:
We know that for a quadratic function that models a trajectory path, that the vertex is the maximum point.
With a maximum at h = 4.5 meters at time t = 0.5 seconds
Vertex = (0.5, 4.5)
x-intercept at (1.5, 0)
Parabolic equation: h(t) = a *(t - h)^2 + k
h(t) = a*(t - 0.5)^2 + 4.5
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We need h(t) = 3 at t = 0
3 = a*(0 - 0.5)^2 + 4.5
-1.5 = a * (-0.5)^2
-1.5 = a*0.25
a = -1.5/0.25
a = -6
so: h(t) = -6*(t - 0.5)^2 + 4.5
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With t = 1.5 at h < 0
0 = a*(1.5 - 0.5)^2 + 4.5
0 = a*1 + 4.5
a = -4.5 ? this part we really do not know if t = 1.5 seconds is when h = 0
it says she enters the water, it could mean h(t) < 0 here.
