Question 205704
Chuck is standing atop a high platform and fires a rock up into the air with his slingshot. While it is in flight, the rock's distance about the ground is a quadratic function of time. 1 second after firing it, the rock is 68 meters high, at 2 seconds it's 96 meters, and at 3 seconds it's 114 meters above ground. Use this information to answer the following questions.
a. Find and write the quadratic equation for the function
Because a "quadratic equation" is:
f(t) = at^2 + bt + c
We have three unknowns: a, b, and c
Therefore, we need three equations.  We get this from:
1 second after firing it, the rock is 68 meters high, 
at 2 seconds it's 96 meters, and 
at 3 seconds it's 114 meters above ground.
.
Our three equations are:
68 = a(1)^2 + b(1) + c
96 = a(2)^2 + b(2) + c
114 = a(3)^2 + b(3) + c
.
Or, rewritten:
68 = a + b + c
96 = 4a + 2b + c
114 = 9a + 3b + c
.
Solve the above system of equations to find a, b and c.  Then put it back into:
f(t) = at^2 + bt + c
This will be your "quadratic equation" describing the the height (meters) based on time (sec).  
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b. What is the highest point the rock will be above ground?
find the vertex of the quadratic found in parta
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c. How high is the platform from which Chuck fires the rock? (Hint: Time = 0)
Set t=0 and solve.  It will be 'c'.
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d. When will the rock hit the ground?
Set f(x)=0 and solve for t
0 = at^2 + bt + c
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e. On its way down, instead of hitting the ground, the rock goes into a nearly empty well. It splashes into the water (hits bottom) exactly 10 seconds after Chuck fired it. How deep is the well? 
Add 10 the answer from part d and plug it into your quadratic equation to find f(x), it will be a negative number (because it is below the ground).  The absolute value will be the depth of the well.

Write back if you get stuck.