SOLUTION: We are working with quadratic functions and their graphs. The question is: If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then

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Question 23273: We are working with quadratic functions and their graphs. The question is: If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 64t. Graph this function for 0 < or equal to t < or equal to 4. What is the maximum height reached by the ball?
Answer by stanbon(48546) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B64x%2B0+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2864%29%5E2-4%2A-16%2A0=4096.

Discriminant d=4096 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-64%2B-sqrt%28+4096+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2864%29%2Bsqrt%28+4096+%29%29%2F2%5C-16+=+-0
x%5B2%5D+=+%28-%2864%29-sqrt%28+4096+%29%29%2F2%5C-16+=+4

Quadratic expression -16x%5E2%2B64x%2B0 can be factored:
-16x%5E2%2B64x%2B0+=+-16%28x--0%29%2A%28x-4%29
Again, the answer is: -0, 4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B64%2Ax%2B0+%29

Your quadratic has a=-16, b=64, c=0
The max or min of this quadratic is
found at t=-b/2a =64/32 = 2
Put that value of "t" into the original
equation to find the correspondins "s"
value.
Cheers,
stan H.