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

1. Problem Description

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

2. Solution Steps

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

$A$ is the amplitude.

The period is $\frac{2\pi}{|B|}$.

The phase shift is $\frac{C}{B}$.

$D$ is the vertical shift.

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

$A = 1$

$B = 1$

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

$D = 0$

The amplitude is $1$.

The period is $\frac{2\pi}{|1|} = 2\pi$.

The phase shift is $\frac{\frac{\pi}{6}}{1} = \frac{\pi}{6}$.

The vertical shift is $0$.

The function $y = \cos(x - \frac{\pi}{6})$ is a cosine function shifted to the right by $\frac{\pi}{6}$.

The standard cosine function $y = \cos(x)$ starts at $(0, 1)$, reaches $( \frac{\pi}{2}, 0)$, $(\pi, -1)$, $(\frac{3\pi}{2}, 0)$ and ends at $(2\pi, 1)$.

Since the phase shift is $\frac{\pi}{6}$, we shift these points to the right by $\frac{\pi}{6}$.

The key points of the graph are:

$(\frac{\pi}{6}, 1)$

$(\frac{\pi}{2} + \frac{\pi}{6}, 0) = (\frac{3\pi}{6} + \frac{\pi}{6}, 0) = (\frac{4\pi}{6}, 0) = (\frac{2\pi}{3}, 0)$

$(\pi + \frac{\pi}{6}, -1) = (\frac{6\pi}{6} + \frac{\pi}{6}, -1) = (\frac{7\pi}{6}, -1)$

$(\frac{3\pi}{2} + \frac{\pi}{6}, 0) = (\frac{9\pi}{6} + \frac{\pi}{6}, 0) = (\frac{10\pi}{6}, 0) = (\frac{5\pi}{3}, 0)$

$(2\pi + \frac{\pi}{6}, 1) = (\frac{12\pi}{6} + \frac{\pi}{6}, 1) = (\frac{13\pi}{6}, 1)$

3. Final Answer

The graph of $y = \cos(x - \frac{\pi}{6})$ is a cosine curve with amplitude 1, period $2\pi$, and phase shift $\frac{\pi}{6}$ to the right. The key points for one period are $(\frac{\pi}{6}, 1)$, $(\frac{2\pi}{3}, 0)$, $(\frac{7\pi}{6}, -1)$, $(\frac{5\pi}{3}, 0)$, and $(\frac{13\pi}{6}, 1)$.