SOLUTION: Can someone help me to finish this problem. Thanks for help. The pressure of the oil in reservoir tends to drop with time. By taking sample pressure readings for a particular oil

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Question 1096819: Can someone help me to finish this problem. Thanks for help.
The pressure of the oil in reservoir tends to drop with time. By taking sample pressure readings for a particular oil reservoir, petroleum engineers have found that the change ..... is given by the function P(t)=t^3-18t^2+81t where t is time in years from the date of the first reading.
a) Find P(0), P(3), P(5), P(10)
Over what period is the change(drop) in pressure decreasing; 3 to 10 or 3 to 9?
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a) P(0)=0 right?
b) P(3)=3^-18*3^2+81*3
P(3)=90
c) P(5)=?
d) P(10)=?
How to do the P(5) and P(10)?
Over what period is the change(drop) in pressure decreasing; 3 to 10 or 3 to 9?
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
P(0) is 0.
P(3)=3^3-18(3^2)+81(3)=27-162+243=108. This is a cubic equation, and the first term is cubed.
P(5)=5^3-18(25)+81(5)=125-450+405=80
P(10)=10^3-18(100)+81(10)=1000-1800+810=10.
Check P(9)=9^3-18(81)+81(9)=729-1458+729=0
The drop is from P(3) to P(9), and the graph shows that as well.