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)   (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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 0, 50. Here's your graph:

Cheers,
Stan H.
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