SOLUTION: Amy throws a ball so that when it is at its highest point, it passes through a hoop. The path of the ball is modelled by the equation y = h + kx - 0.5x^2 where y is the height of

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Amy throws a ball so that when it is at its highest point, it passes through a hoop. The path of the ball is modelled by the equation y = h + kx - 0.5x^2 where y is the height of       Log On


   

Question 1094800: Amy throws a ball so that when it is at its highest point, it passes through a hoop. The path of the ball is modelled by the equation y = h + kx - 0.5x^2
where y is the height of the ball above the ground and x is the horizontal distance from the point at which the ball was thrown. The centre of the Hoop is at the point where x = 2 and y = 5
Find the values of h and k and find the value of x for when the ball hits the ground.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The x-value of the maximum is:
+x%5Bmax%5D+=+-k%2F%282%2A%28-.5%29%29+
+x%5Bmax%5D+=+k+
-----------------
+y%5Bmax%5D+=+-.5%2Ak%5E2+%2B+k%2Ak+%2B+h+
+y%5Bmax%5D+=+.5%2Ak%5E2+%2B+h+
-------------------------------
The maximum point is at ( 2,5 )
+x%5Bmax%5D+=+2+
+k+=+2+
and
+y%5Bmax%5D+=+5+
+.5k%5E2+%2B+h+=+5+
+.5%2A2%5E2+%2B+h+=+5+
+h+=+5+-+2+
+h+=+3+
---------------------
The ball hits the ground at +y+=+0+
+-.5x%5E2+%2B+2x+%2B+3+=+0+
Use quadratic formula
+x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A%28-.5%29%2A3+%29%29%2F%282%2A%28-.5%29%29+
+x+=+%28+-2+-+sqrt%28+4+%2B+6+%29+%29+%2F+%28-1%29+
+x+=+2+%2B+sqrt%2810%29+
+x+=+2+%2B+3.1623+
+x+=+5.1623%0D%0ASo%2C+the+equation+is%3A%0D%0A%7B%7B%7B+y+=+-.5x%5E2+%2B+2x+%2B+3+
Here's the plot:
+graph%28+400%2C+400%2C+-2%2C+6%2C+-1%2C+6%2C+-.5x%5E2+%2B+2x+%2B+3+%29+
Looks close to answers