The problem asks to graph the function $y = \cos(\frac{\pi}{6}x)$.

1. Problem Description

The problem asks to graph the function $y = \cos(\frac{\pi}{6}x)$.

2. Solution Steps

The general form of a cosine function is $y = A\cos(Bx - C) + D$, where:

- A is the amplitude

- B is related to the period (Period = $\frac{2\pi}{|B|}$)

- C is related to the phase shift (Phase shift = $\frac{C}{B}$)

- D is the vertical shift

In our case, $y = \cos(\frac{\pi}{6}x)$. Comparing with the general form:

$A = 1$

$B = \frac{\pi}{6}$

$C = 0$

$D = 0$

Amplitude: $A = 1$

Period: $\frac{2\pi}{|B|} = \frac{2\pi}{\frac{\pi}{6}} = 2\pi \cdot \frac{6}{\pi} = 12$

Phase shift: $\frac{C}{B} = \frac{0}{\frac{\pi}{6}} = 0$

Vertical shift: $D = 0$

The cosine function starts at its maximum value. In this case, the maximum value is

1. Since the period is 12, we can find key points for one cycle:

- x = 0: $y = \cos(\frac{\pi}{6}(0)) = \cos(0) = 1$

- x = $\frac{12}{4}$ = 3: $y = \cos(\frac{\pi}{6}(3)) = \cos(\frac{\pi}{2}) = 0$

- x = $\frac{12}{2}$ = 6: $y = \cos(\frac{\pi}{6}(6)) = \cos(\pi) = -1$

- x = $\frac{3 \cdot 12}{4}$ = 9: $y = \cos(\frac{\pi}{6}(9)) = \cos(\frac{3\pi}{2}) = 0$

- x = 12: $y = \cos(\frac{\pi}{6}(12)) = \cos(2\pi) = 1$

Since the x-axis of the provided graph is scaled in terms of $\pi$, we need to map these x-values accordingly.

- 0 remains 0

- 3 corresponds to $\frac{3}{\pi}\pi$

- 6 corresponds to $\frac{6}{\pi}\pi$

- 9 corresponds to $\frac{9}{\pi}\pi$

- 12 corresponds to $\frac{12}{\pi}\pi$

The problem asks to graph the function. I am unable to draw it in text format.

3. Final Answer

The function is a cosine function with amplitude 1, period 12, no phase shift, and no vertical shift.