How do you graph $y = 3 sin (\frac{1}{3} x + \frac{π}{2}) - 2$ $y=3sin(1/3x+ pi/2)-2$ ?

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How do you graph $y = 3 sin (\frac{1}{3} x + \frac{π}{2}) - 2$ $y=3sin(1/3x+ pi/2)-2$ ?

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Answer 1 · 2018-04-01T05:33:03

As below.

Explanation:

$y = A sin (B x - C) + D$ $y = A sin (Bx - C) + D$

Given equation is $y = 3 sin ((\frac{x}{3}) + (\frac{π}{2})) - 2$ $y = 3 sin((x/3) + (pi/2)) - 2$

$A m p l i t u d e = | A | = 3$ $Amplitude = |A| = 3$

$period = \frac{2 π}{| B |} = \frac{2 π}{\frac{1}{3}} = 6 π$ $"period " = (2pi) / |B| = (2pi) / (1/3) = 6pi$

$Phase Shift = \frac{C}{B} = \frac{- \frac{π}{2}}{\frac{1}{3}} = - \frac{3 π}{2}, \frac{3 π}{2} to the left$ $"Phase Shift " = C / B = (-pi/2) / (1/3) = -(3pi)/2, " " (3pi)/2 " to the left"$

$Vertical Shift = D = - 2$ $"Vertical Shift " = D = -2$

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

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How do you graph $y = 3 sin (\frac{1}{3} x + \frac{π}{2}) - 2$ $y=3sin(1/3x+ pi/2)-2$ ?

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