Determine the amplitude, period, and phase shift of the following trigonometric equation.

[tex]\[ y = 8 - \frac{1}{2} \sin \left( \pi x + \frac{\pi}{4} \right) \][/tex]

Amplitude: [tex]\(\square\)[/tex]

Period: [tex]\(\square\)[/tex]

Phase Shift:
A. no phase shift
B. shifted to the right
C. shifted to the left



Answer :

To determine the amplitude, period, and phase shift of the trigonometric equation:

[tex]\[ y = 8 - \frac{1}{2} \sin \left(\pi x + \frac{\pi}{4}\right) \][/tex]

we proceed as follows:

1. Amplitude:
The amplitude of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is given by [tex]\(|A|\)[/tex]. In this equation, the coefficient of the sine function (without the negative sign and divided by the constant) is [tex]\(-\frac{1}{2}\)[/tex].
Thus, the amplitude is:
[tex]\[ \text{Amplitude} = \left| -\frac{1}{2} \right| = \frac{1}{2} = 0.5 \][/tex]

2. Period:
The period of the sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is given by [tex]\( \frac{2\pi}{|B|} \)[/tex].
Here, the coefficient [tex]\( B \)[/tex] of [tex]\( x \)[/tex] inside the sine function is [tex]\(\pi\)[/tex].
Thus, the period is:
[tex]\[ \text{Period} = \frac{2\pi}{\pi} = 2 \][/tex]

3. Phase Shift:
The phase shift of the sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is determined by the formula [tex]\( -\frac{C}{B} \)[/tex].
Here, [tex]\( C \)[/tex] is [tex]\(\frac{\pi}{4}\)[/tex] and [tex]\( B \)[/tex] is [tex]\(\pi\)[/tex].
Thus, the phase shift is:
[tex]\[ \text{Phase Shift} = -\frac{\frac{\pi}{4}}{\pi} = -\frac{1}{4} = -0.25 \][/tex]

Since the phase shift is negative, it means the graph of the sine function is shifted to the right by 0.25 units.

So, we summarize our answers:
- Amplitude: [tex]\( 0.5 \)[/tex]
- Phase Shift: shifted to the right

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