math
posted by lyne .
function rule for tis
points 1 2 3 4
lines 0 1 3 6
its similar to an earlir one i posted but i still don't quite understand
thanks

it deals with the number of points that a certain number of lines drawn have.
BTW, I think you have your headings backwards,
the first set of numbers should be lines,
the second set of numbers are the number of intersection points those lines have
e.g.
1 line has no intersections points
2 lines have one intersection points
3 lines have 3 intersection points
4 lines have 6 intersection points
etc.
can you see a pattern here? 
yeaa i understand and can see the patterns i just don't know how to find the function the way you did...

I don't know what grade level you are in, so I can't guess how sophisticated the method should be that I describe.
suppose we double your second line from
0 1 3 6 ... to
0 2 6 12 ...
notice that those are like
1x0, 2x1, 3x2, 4x3 .... or n(n1)
but we multiplied our original numbers by 2, so let's divide our formula by 2 to get back to the originals.
so the function rule would be
number of points = n(n1)/2 , where n is the number of lines drawn.
testing:
If I had 4 lines drawn, n=6
and no. of intersection points is (6x5)/2 = 15
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