Verified answer

A piecewise defined function has more than one definition where one holds for one piece but may not hold for the whole domain of the function. And thus the other part of the domain will be left out to the other part of the function.

If you look at you coordinate grid, using the usual convention where the horizontal axis is the x-axis, and the vertical axis the y-axis, you should notice that the path of the ball has a positive slope - rising along the y-axis - in the x interval 2<=x<=9. This is where you should have your first definition which apply only for these interval.

The slope of the graph is now negative in the interval 9<x<=15.

Now use your knowledge of y=mx+c to find the equations of the two lines;

Lets suppose that your first interval is a<=x<=b, and the second interval b<x<=c.

Moreover, p be the positive gradient and n the negative gradient. This mean your two functions over the two intervals would be y=px+c and y=nx+c. Then letting f(x)=3, your piecewise function will be like this:

f(x)= {px+c,a<=x<=b;

nx+c,b<x<=c;

This says use px+c if a<=x<=b, and nx+c if b<x<=c.

The value of the c in the second equation is where the ball hits the other side of the pool table.

The x and y values are the coordinates of the ball, location, on the pool table.

I hope this helps you understand and to solve the rest of the problem.