How do you find the amplitude, period, phase shift given $y = sin (2 π x - π)$ $y=sin(2pix-pi)$ ?

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Answer 1 · 2018-01-08T11:07:58

Amplitude is $1$ $1$ , period is $1$ $1$ and phase shift is $\frac{1}{2}$ $1/2$

The sinusidal function is $y = A sin (B x + C) + D$ $y= A sin(Bx+C)+D$ where $A$ $A$ is

amplitude , period is $\frac{2 π}{| B |}$ $(2pi)/|B|$ , phase shift is $- \frac{C}{| B |}$ $-C/|B|$ , vertical

shift is $D; y = sin (2 π \cdot x - π) or y = 1 \cdot sin (2 π \cdot x - π) + 0$ $D ; y= sin(2pi*x-pi) or y= 1*sin(2pi*x-pi)+0$

$:. A=1, B=2pi , C = -pi ,D=0$

Therefore amplitude is $1$ , period is $(2pi)/|B|= (2pi)/(2pi)=1$

phase shift is $-C/|B|= -(-pi)/(2pi)=1/2$

graph{sin(2pix-pi) [-10, 10, -5, 5]} [Ans]

How do you find the amplitude, period, phase shift given y=sin(2pix-pi)y=sin(2πx−π)y=sin(2pix-pi)?