SOLUTION: A projectile follows a parabolic path whose height in meters, is given by the function f(x) = -x^2 +2x +2. Find the mazimum horizontal distance that the projectile may cover.

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Question 980100: A projectile follows a parabolic path whose height in meters, is given by the function f(x) = -x^2 +2x +2.
Find the mazimum horizontal distance that the projectile may cover.
Answer by josh_jordan(263) About Me  (Show Source):
You can put this solution on YOUR website!
To find the maximum horizontal distance, which represents x in our quadratic equation, we will need to use the quadratic formula to find x.

ax^2 + bx + c -----> -x^2 + 2x + 2

a = -1
b = 2
c = 2

quadratic formula: x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x=+%28-2%2B-sqrt%282%5E2-%284%29%28-1%29%282%29%29%29%2F%282%29%28-1%29 ----->

x=%28-2%2B-sqrt%284%2B8%29%29%2F-2 ----->

x=%28-2+%2B-sqrt%2812%29%29%2F-2 ----->

x=%28-2%2B-2sqrt%283%29%29%2F-2 ----->

x=%28-2%281+%2B-sqrt%283%29%29%29%2F-2 ----->

x=1%2B-sqrt%283%29

We will now disregard the - sign, because if we subtract the square root of 3 from 1, we will obtain a negative number. A projectile that has not yet moved will start at the point (0,0) on a graph (unless we are told the projectile starts off at a different height, in which case our y coordinate may be higher than 0), so our x coordinate cannot be a negative number. We only want to find the distance between the original x coordinate (0) and the x coordinate where the projectile lands.

x=1%2Bsqrt%283%29 ----->

x=1%2B1.732 ----->

x=2.732

Therefore, the maximum horizontal distance that the projectile may cover is approximately 2.732 meters.

On a graph, the parabola will look like the following (note that the x coordinate where the projectile lands is 2.732)

+graph%28+400%2C+400%2C+-1%2C+4%2C+-1%2C+4%2C+-x%5E2+%2B+2x+%2B+2+%29+