Jesse - you could compare each element to its neighbour and see if there is a switch in sign i.e. from positive to negative or negative to positive. The easiest way to do this, without a for loop, is to use arrayfun and just check to see if each neighbour is different. Something like
idcs = arrayfun(@(k)sign(y1(k))~=sign(y1(k+1)),1:length(y1)-1);
We use sign to check the sign of the number, and if the sign between two neighbouring elements is different then
sign(y1(k))~=sign(y1(k+1)
is true (or 1). Note how we use the anonymous function
@(k)sign(y1(k))~=sign(y1(k+1))
to take as an input k which will be an index into y1. The resulting logical array, idcs, will have ones where there is a difference in sign which is the zero crossing. Use find to find those indices as
find(idcs==1)
which you can then use to draw your X or circle at the zero crossing. Try the above and see what happens!
