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Algebra 2, please help! thank you
The function y = 2 cos 3(x + 2π∕3) +1 has a phase shift (or horizontal shift) of

A) –2π∕3
B) 3
C) 1
D) 2

Sagot :

Answer:

-2pi/3

Step-by-step explanation:

y = 2 cos 3(x + 2π∕3) +1

y = A sin(B(x + C)) + D  

amplitude is A

period is 2π/B

phase shift is C (positive is to the left)

vertical shift is D

We have a shift to the left of 2 pi /3

Answer:

A

Step-by-step explanation:

The standard cosine function has the form:

[tex]\displaystyle y = a\cos (b(x-c)) + d[/tex]

Where |a| is the amplitude, 2π / b is the period, c is the phase shift, and d is the vertical shift.

We have the function:

[tex]\displaystyle y = 2 \cos 3\left(x + \frac{2\pi}{3}\right) + 1[/tex]

We can rewrite this as:

[tex]\displaystyle y = \left(2\right)\cos 3\left(x - \left(-\frac{2\pi}{3}\right)\right) + 1[/tex]

Therefore, a = 2, b = 3, c = -2π/3, and d = 1.

Our phase shift is represented by c. Thus, the phase shift is -2π/3.

Our answer is A.