which of the following functions illustrates a phase shift? a. y = -2 - cos(x - π) b. y = 3 cos4x c. y = 1 +…

Answer:

A. $y=-2 - \cos(x-\pi)$

Explanation:

Step1: Recall phase - shift formula

The general form of a trigonometric function is $y = A\sin(Bx - C)+D$ or $y=A\cos(Bx - C)+D$, and the phase - shift is given by $\frac{C}{B}$.

Step2: Analyze option A

For $y=-2-\cos(x - \pi)$, comparing with $y = A\cos(Bx - C)+D$ where $A=-1$, $B = 1$, $C=\pi$, $D=-2$. The phase - shift is $\frac{C}{B}=\pi$.

Step3: Analyze option B

For $y = 3\cos(4x)$, comparing with $y = A\cos(Bx - C)+D$, $C = 0$. So, there is no phase - shift.

Step4: Analyze option C

For $y=1+\sin(x)$, comparing with $y = A\sin(Bx - C)+D$, $C = 0$. So, there is no phase - shift.

Step5: Analyze option D

For $y=\tan(2x)$, comparing with $y = A\tan(Bx - C)+D$, $C = 0$. So, there is no phase - shift.