SPUI wrote in message <7a23eh$2bc$1@client2.news.psi.net>... >No. Use y1=A*sin(BX + C) + D Although the original question was for the 86, I tried both ways on my 89 and both work. y1(x) = A*sin(B*X+C)*D and A*sin(B*X+C)+D -> y1(x) both work. Tom Lake ICQ # 25589135