SOLUTION: A ball was thrown and followed a pth descibed by y= -0.02xto the second power+ x. What was the maximum height(in feet) of the thrown ball?

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Question 26501: A ball was thrown and followed a pth descibed by y= -0.02xto the second power+ x. What was the maximum height(in feet) of the thrown ball?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y=-0.02x^2+x
There are many ways to find the maximum height depending
on the math course you are taking.
Yours is a quadratic equation with a=-0.02 and b=1
The max point can be found at x=-b/2a =-1(1/-0.04)=25
Then height=y=-0.02(25)^2+25
y=12.5
Thought you might like to see this curve: the following.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -0.02x%5E2%2B1x%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=%281%29%5E2-4%2A-0.02%2A0=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+1+%29%29%2F2%5C-0.02+=+0
x%5B2%5D+=+%28-%281%29-sqrt%28+1+%29%29%2F2%5C-0.02+=+50

Quadratic expression -0.02x%5E2%2B1x%2B0 can be factored:
-0.02x%5E2%2B1x%2B0+=+-0.02%28x-0%29%2A%28x-50%29
Again, the answer is: 0, 50. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-0.02%2Ax%5E2%2B1%2Ax%2B0+%29

Cheers,
Stan H.