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.

Hence,

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)**

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)**

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**Further Reading**