How do you graph #y=3sin(1/3x+ pi/2)-2#?

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Answer 1 · 2018-04-01T05:33:03

As below.

#y = A sin (Bx - C) + D#

Given equation is #y = 3 sin((x/3) + (pi/2)) - 2#

#Amplitude = |A| = 3#

#"period " = (2pi) / |B| = (2pi) / (1/3) = 6pi#

#"Phase Shift " = C / B = (-pi/2) / (1/3) = -(3pi)/2, " " (3pi)/2 " to the left"#

#"Vertical Shift " = D = -2#

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