Hi John.

Another mathematical definition of abs(x) is sqr(x^2)

I just tried it out, and it appears to work fine in the negative domain, basically mirroring your function about x = 0.

So what you will type into the equation line is:

.15*sin(sqr(x^2)^(1+sqr(x^2)/100)*pi/3)+.33

Another idea: you might consider swapping sin for cos or (1-cos(...)), so you avoid the cusp at x = 0 (although I'm not sure what your application is).