To determine the amplitude, period, and phase shift of the given trigonometric equation [tex]\( y = \frac{-1}{3} \sin (x-2) \)[/tex], let's analyze each component one by one.
### Amplitude:
The amplitude of a sine function [tex]\( y = A \sin(Bx + C) \)[/tex] is given by the absolute value of the coefficient [tex]\( A \)[/tex]. In this equation, we have:
[tex]\[ A = \frac{-1}{3} \][/tex]
Thus, the amplitude is:
[tex]\[ \text{Amplitude} = \left| \frac{-1}{3} \right| = \frac{1}{3} \approx 0.3333 \][/tex]
### Period:
The general form for the period of a sine function [tex]\( y = A \sin(Bx + C) \)[/tex] is [tex]\( \frac{2\pi}{|B|} \)[/tex]. In this equation, the coefficient of [tex]\( x \)[/tex] is 1, so:
[tex]\[ B = 1 \][/tex]
Thus, the period is:
[tex]\[ \text{Period} = \frac{2\pi}{1} = 2\pi \approx 6.2832 \][/tex]
### Phase Shift:
The phase shift is determined by the horizontal shift of the sine function, which is given by [tex]\( \frac{-C}{B} \)[/tex] where [tex]\( y = A \sin(Bx + C) \)[/tex]. In this equation:
[tex]\[ B = 1 \][/tex]
[tex]\[ C = -2 \][/tex] (since [tex]\( x - 2 \)[/tex] can be written as [tex]\( x + (-2) \)[/tex])
Thus, the phase shift is:
[tex]\[ \text{Phase Shift} = \frac{-(-2)}{1} = 2 \][/tex]
Since the phase shift is positive, the function is shifted to the right.
In summary, we have:
- Amplitude: [tex]\( \frac{1}{3} \)[/tex]
- Period: [tex]\( 2\pi \)[/tex]
- Phase Shift: The function is shifted to the right by 2 units.
### Final Answer:
Amplitude: [tex]\( \frac{1}{3} \)[/tex]
Phase Shift: shifted to the right
