How do you graph $y = 3 sin (2 x - π) - 1$ $y=3sin(2x-pi)-1$ ?

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Answer 1 · 2018-03-19T08:29:47

As below.

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

Given : $y = 3 sin (2 x - π) - 1$ $y = 3 sin (2x - pi) - 1$

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

$Period = P = \frac{2 π}{| B |} = \frac{2 π}{2} = π$ $"Period " = P = (2pi) / |B| = (2pi) / 2 = pi$

$Phase Shift = P = \frac{- C}{B} = \frac{π}{2}$ $"Phase Shift" = P = (-C)/B = pi / 2$ $to the right$ $"to the right"$

$Vertical Shift = D = - 1$ $"Vertical Shift " = D = -1$

graph{3 sin(2x-pi) - 1 [-10, 10, -5, 5]}

How do you graph y=3sin(2x−π)−1y=3sin(2x-pi)-1?