I forgot the question number, but on the radio-wave graphing problem the guidelines specify full credit only for y=3sin(2x) or y= -3sin(2x). However, any translation of the former that takes some point on the graph of 3sin(2x) to the origin will also satisfy the criteria given in the problem. So the general solution really is
(y - k) = 3 sin[2(x-h)], where k = 3sin(2h), i.e. y = 3 sin[2(x - h)] + 3 sin(2h) for any real h.
Here the -3 cases are covered by translations of the +3 cases, for example by adding pi to h.
It seems to me that some student somewhere must have drawn a correct graph with range 0 to 6, and should receive credit.