OK, try this or a version thereof...
>> y=bsxfun(@plus,randn(10),linspace(0,10,10).');
>> yCrs=6;
>> del=[zeros(1,10); diff(sign(y-yCrs))==2];
>> [r,c]=ind2sub(size(del),find(del)); r=r.';
>> for i=1:10
xCrs(i)=interp1(y(r(i)-1:r(i),i),r(i)-1:r(i),yCrs);
end
>> xCrs=[min(xCrs) max(xCrs)]
xCrs =
5.6484 7.4861
Above gives--
>> plot(y)
>> line(xlim,[yCrs yCrs],'linestyle',':','color','k')
>> line([xCrs(1) xCrs(1)],ylim,'linestyle',':','color','k')
>> line([xCrs(2) xCrs(2)],ylim,'linestyle',':','color','k')
ADDENDUM Depending upon just how "jaggedy" the individual lines can be, may need to do the above bounding for both first and last crossing(s) if, besides there being a sign change in the slope the observation can cross the threshold and then go back below again. That'd essentially be find with the 'first' and again with 'last' option to set the range. Then you've got the issue that will have to interpolate both sections as well since the inverse interpolation will not work with the double-valued independent variable when you turn x, y on their heads as dependent, independent variables instead. My sample data had slope reservals, but didn't actually cross below the chosen threshold; admit I didn't think of it specifically then.
