How do you find the amplitude and period of #y = 3sin2x-(pi/2)#?

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Answer 1 · 2018-07-25T14:04:12

As below. graph{3 sin (2x) - pi/2 [-10, 10, -5, 5]}

Standard form of sine function is #y = A sin (Bx - C) + D#

Given: #y = 3 sin 2x - pi/2#

#A = 3, B = 2, C = 0, D = pi/2#

Amplitude #= |A| = 3#

Period #= (2pi) / |B| = (2pi) / 2 = pi#

Phase Shift #= -C / B = 0#

Vertical Shift #= D = -(pi/2)#