SOLUTION: Hi there I need to calculate out how far and high a boy named Jose throws in a ball in a football match, Y stands for the height in meters and X for the diatance in meters from whe

Question 750622: Hi there I need to calculate out how far and high a boy named Jose throws in a ball in a football match, Y stands for the height in meters and X for the diatance in meters from where jose is standing.
How do I calculate Y=-0,04x^2+0,6x+2
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Y=-0.04x^2+0.6x+2
.
Maximum is at vertex:
x-value of vertex is "axis of symmetry":
x = -b/(2a)
x = -0.6/(2*(-0.04))
x = -0.6/(-0.08)
x = 0.6/(0.08)
x = 7.5
Max height:
Y=-0.04(7.5)^2+0.6(7.5)+2
Y= 4.25 meters (max height)
.
Max distance, we set Y to zero and solve for x:
0=-0.04x^2+0.6x+2
0=0.04x^2-0.6x-2
0=0.02x^2-0.3x-1
applying the "quadratic formula" yields:
x = {17.81, -2.81}
throw out the negative solution (extraneous) leaving
x = 17.81 meters (max distance)
.
details of "quadratic formula":

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=0.17 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 17.8077640640442, -2.80776406404415. Here's your graph:

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