The compulsory math problem [trigonometric function] in senior two needs steps ~
1, the maximum value of y is sin(2x+π/6)= 1, 2x+π/6=π/2+2kπ(k belongs to z), and x=π/6+kπ(k belongs to z).
2. Firstly, π/6 is shifted to the left, contracted twice on the abscissa to get y=sin(2x+π/6), stretched twice on the ordinate, and moved up by two units to get y=2sin(2x+π/6)+2.