How do you graph $y = sin 2 (x - \frac{3 π}{4})$ $y=sin2(x-(3pi)/4)$ ?

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Answer 1 · 2018-03-23T11:15:19

See below.

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

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

$A m p l i t u d e = | A | = 1$ $Amplitude =| A| = 1$

$P e r i o d = P = \frac{2 π}{| B |} = \frac{2 π}{2} = π$ $"Period = P = (2pi) / |B| = (2pi) / 2 = pi$

$Phase Shift = (- \frac{C}{B}) = \frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}$ $"Phase Shift " = (-C/B) = ((3pi)/2) / 2 = (3pi)/4$

$Vertical Shift = D = 0$ $"Vertical Shift " = D = 0$

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

How do you graph y=sin2(x-(3pi)/4)y=sin2(x−3π4)y=sin2(x-(3pi)/4)?