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Given the equation [tex]$y=6 \sin \left(\frac{4 \pi}{3} x-4 \pi\right)+2$[/tex]:

The amplitude is: [tex]$\square$[/tex]

The period is: [tex]$\square$[/tex]

The horizontal shift is: [tex]$\square$[/tex]

The midline is: [tex]$y=$[/tex] [tex]$\square$[/tex]


Sagot :

Let's analyze the given sine function step-by-step:

The given equation is:
[tex]\[ y = 6 \sin \left(\frac{4 \pi}{3} x - 4 \pi\right) + 2 \][/tex]

1. Amplitude:
The amplitude of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is determined by the absolute value of the coefficient of the sine function (A). Here, the coefficient is [tex]\( 6 \)[/tex], so the amplitude is:
[tex]\[ \text{Amplitude} = 6 \][/tex]

2. Period:
The period of a sine function is given by [tex]\( \frac{2\pi}{|B|} \)[/tex], where B is the coefficient of [tex]\( x \)[/tex] inside the sine function. In this equation, [tex]\( B \)[/tex] is [tex]\( \frac{4\pi}{3} \)[/tex]. Therefore, the period is:
[tex]\[ \text{Period} = \frac{2\pi}{\frac{4\pi}{3}} = \frac{2\pi \times 3}{4\pi} = \frac{3}{2} = 1.5 \][/tex]

3. Horizontal Shift (Phase Shift):
The horizontal shift of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is found by solving the equation inside the sine function [tex]\( Bx + C = 0 \)[/tex] for [tex]\( x \)[/tex]. Here, we solve:
[tex]\[ \left(\frac{4\pi}{3}\right)x - 4\pi = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{4\pi}{3} x = 4\pi \][/tex]
[tex]\[ x = \frac{4\pi}{\frac{4\pi}{3}} \][/tex]
[tex]\[ x = 3 \][/tex]
So, the horizontal shift is:
[tex]\[ \text{Horizontal Shift} = 3 \][/tex]

4. Midline:
The midline of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is the vertical shift, which is given by the constant term [tex]\( D \)[/tex]. In this case, [tex]\( D = 2 \)[/tex]. Therefore, the midline is:
[tex]\[ \text{Midline} = y = 2 \][/tex]

In summary:
The amplitude is: [tex]\( 6 \)[/tex]
The period is: [tex]\( 1.5 \)[/tex]
The horizontal shift is: [tex]\( 3 \)[/tex]
The midline is: [tex]\( y = 2 \)[/tex]