What is the amplitude, period, phase shift and vertical displacement of $y = sin x - 1$ $y=sinx-1$ ?

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Answer 1 · 2017-12-17T03:08:20

Amplitude $= 1$ $= 1$
Period $= 2 π$ $= 2pi$
Phase shift $= 0$ $= 0$
Vertical Displacement $= - 1$ $= -1$

Consider this skeletal equation:

$y = a \cdot sin (b x - c) + d$ $y = a*sin(bx - c) + d$

From $y = sin (x) - 1$ $y = sin(x) - 1$ , we now that

$a = 1$ $a = 1$

$b = 1$ $b = 1$

$c = 0$ $c = 0$

$d = - 1$ $d = -1$

The a value is basically the amplitude , which is $1$ $1$ here.

Since

$period = \frac{2 π}{b}$ $"period" = (2pi) / b$

and the b value from the equation is $1$ $1$ , you have

$period = \frac{2 π}{1} \Rightarrow period = 2 π$ $"period" = (2pi) / 1 => "period" = 2pi$

^ (use $2 π$ $2pi$ if the equation is cos, sin, csc, or sec; use $π$ $pi$ only if the equation is tan, or cot)

Since the c value is $0$ $0$ , there is no phase shift (left or right).

Finally, the d value is $- 1$ $-1$ , which means the vertical displacement is $- 1$ $-1$ (the graph shifts down 1).

What is the amplitude, period, phase shift and vertical displacement of y=sinx−1y=sinx-1?