SOLUTION: Two projectiles are shot into the air from the same location. The paths of the projectiles are parabolas and are given by (a) y = −0.0013x2 + x + 19 and (b) y = -x^2/81 +4/

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Two projectiles are shot into the air from the same location. The paths of the projectiles are parabolas and are given by (a) y = −0.0013x2 + x + 19 and (b) y = -x^2/81 +4/      Log On


   

Question 1146285: Two projectiles are shot into the air from the same location. The paths of the projectiles are parabolas and are given by
(a) y = −0.0013x2 + x + 19 and
(b) y = -x^2/81 +4/3x + 19
where x is the horizontal distance and y is the vertical distance, both in feet. Determine which projectile goes higher by locating the vertex of each parabola. (Round your answers to two decimal places.)
(a) y = −0.0013x2 + x + 19
(x,y)=
(b) y = -x^2/81 +4/3x + 19
(x,y)=
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Completing the Square for Quadratics
To complete the square for the quadratic -0.0013%2Ax%5E2%2B1%2Ax%2B19=0, we must first find a square which when expanded, has -0.0013x2 and 1x in it.
Factoring -0.0013 from the left side gives -0.0013%28x%5E2-769.230769230769%2Ax-14615.3846153846%29=0. %28x-384.615384615385%29%5E2 is the square we are looking for. So we get -0.0013%28%28x-384.615384615385%29%5E2-162544.378698225%29=0. Taking the -162544.378698225 out of the -0.0013, we get highlight%28-0.0013%28x-384.615384615385%29%5E2%2B211.307692307692%29. Subtracting 211.307692307692 from both sides, we get -0.0013%28%28x-384.615384615385%29%5E2%29=-211.307692307692. Since the right side is negative, there are no real solutions.
(384.62,211.31)
Solved by pluggable solver: Completing the Square for Quadratics
To complete the square for the quadratic -0.012345679%2Ax%5E2%2B1.33333333%2Ax%2B19=0, we must first find a square which when expanded, has -0.012345679x2 and 1.33333333x in it.
Factoring -0.012345679 from the left side gives -0.012345679%28x%5E2-107.999999838%2Ax-1539.000001539%29=0. %28x-53.999999919%29%5E2 is the square we are looking for. So we get -0.012345679%28%28x-53.999999919%29%5E2-4454.999992791%29=0. Taking the -4454.999992791 out of the -0.012345679, we get highlight%28-0.012345679%28x-53.999999919%29%5E2%2B54.999999856%29. Subtracting 54.999999856 from both sides, we get -0.012345679%28%28x-53.999999919%29%5E2%29=-54.999999856. Since the right side is negative, there are no real solutions.
(54,55). y=-0.0013x%5E2%2Bx%2B19 went higher then y=%28-1%2F81%29x%5E2%2B%284%2F3%29x%2B19.
Click here for a link to the solver above.