Question 147844: Jake Johnson dives off the high dive at a pool. His distance from the surface of the water varies quadratically with the number of seconds that have passed since he left the board. His distances at times of 1, 2, and 3 seconds since he left the board are 24, 18, and 2 meters above the water, respectively.
How do you write the particular equation expressing distance in terms of time?
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Jake Johnson dives off the high dive at a pool. His distance from the surface of the water varies quadratically with the number of seconds that have passed since he left the board. His distances at times of 1, 2, and 3 seconds since he left the board are 24, 18, and 2 meters above the water, respectively.
How do you write the particular equation expressing distance in terms of time?
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You have three points: (1,24), (2,18), (3,2)
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Form of the equation: y = ax^2 + bx + c
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Substitute each of the ordered pairs into the form to get three
equations with variables a, b, and c.
(1,24): a + b + c = 24
(2,18): 4a +2b + c = 18
(3,2): 9a +3b + c = 2
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Solve the system of equations be any method to get:
a = -5
b = 9
c = 20
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EQUATION:
y = -5x^2 + 9x + 20
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Cheers,
Stan H.
Answer by josmiceli(19441)  (Show Source):
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