8. Amy throws a ball so that when it is at its highest point, it passes through the centre of a
hoop. The path of the ball is modelled by the equation y=h+kx-x², where y is the
height in metres of the ball above the ground and x is the horizontal distance in metres
from the point at which the ball was thrown. The centre of the hoop is at the point where
x=2 and y = 5.
(a) Find the values of h and k.
(b) Find the value of x at which the ball hits the ground.

The values of h and k would be =2 and 2 respectively.

The value of x would be = 5.2

A horizontal distance is the distance that is be covered by a moving object through a horizontal pathway.

y = h + kx - 0.5x^2

When X = 2 and y = 5 is the centre of the hoop.

To calculate for y;

X max = -k/2× (-0.5)

X max = k = 2

y max = 5

Substitute the given variables into the equation and make h the subject of formula.

5 = (h + 2×2) - (0.5x^2)

5= (h + 4)- 1

h +4 = 5+1

h +4 = 6

h = 6-4

h = 2

Also from the graph, the value for x as the ball hits the ground would be = 5.2

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