How do you use transformation to graph the sin function and determine the amplitude and period of $y = sin x - 6$ $y=sinx-6$ ?

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Answer 1 · 2017-10-30T15:10:50

Amplitude = 1 and Period = $2 π$ $2pi$

Standard form of equation is
$y = sin (b x + c) + d$ $y=sin (bx +c) + d$

Where amplitude = a, Period = -(2pi)/b, phase shift = -c/b.

Given equation is $y = sin (x) - 6$ $y=sin(x) - 6$

$a = 1, b = 1, c = 0, d = - 6$ $a = 1, b = 1, c = 0, d = -6$

Therefore, $a m p l i t u d e = a = 1$ $amplitude = a = 1$

$P e r i o d = \frac{2 π}{b} = \frac{2 π}{1} = 2 π$ $Period = (2pi)/b = (2pi)/1 = 2pi$

Phase shift = $- \frac{c}{b} = 0$ $-c/b = 0$

Vertical shift $= d = - 6$ $= d = - 6$

How do you use transformation to graph the sin function and determine the amplitude and period of y=sinx−6y=sinx-6?