The particle is thrown from below the x-axis and intercepts the x-axis on it way up and down. When it intercepts the x-axis the y-coordinate is 0.
To determine the points where it intercepts the x-axis solve 9x - x^2 = 0
9x - x^2 = 0
=> x(9 - x) = 0
=> x = 0 and x = 9
The particle intercepts the x-axis at the points (0,0) and (9, 0)
The path of the particle is given by,
y = 9x - x^2 .....(1)
when the particle intercepts the x-axis the y cordinate of the particle must be zero. In orther words when the particle intercepts the x-axis the verticle distance of the particle from the x axis should be zero.
y = 0 .....(2)
Substitute this value in the equation (1)
0 = 9x - x^2
0 = x(9-x)
in order to satisfy this conditin ,
x = 0 or (9-x) = 0
x = 0 or x = 9
Hence we have obtains two points where the particle intercepts the x axis.
those two points are, (0,0) and (9,0)