Question 980100
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x= (-2+-sqrt(2^2-(4)(-1)(2)))/(2)(-1)}}} -----> 


{{{x=(-2+-sqrt(4+8))/-2}}} ----->


{{{x=(-2 +-sqrt(12))/-2}}} ----->


{{{x=(-2+-2sqrt(3))/-2}}} ----->


{{{x=(-2(1 +-sqrt(3)))/-2}}} ----->


{{{x=1+-sqrt(3)}}} 


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+sqrt(3)}}} ----->


{{{x=1+1.732}}} ----->


{{{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( 400, 400, -1, 4, -1, 4, -x^2 + 2x + 2 ) }}}