The function \( f(x) = 2 \sin(x + \frac{\pi}{6}) \) can be analyzed for phase shift by examining its argument.
In the standard form of a sine function, \( f(x) = a \sin(bx - c) \), the phase shift is given by \( \frac{c}{b} \).
For the function \( f(x) = 2 \sin(x + \frac{\pi}{6}) \), we can rewrite it as:
\[ f(x) = 2 \sin\left(1 \cdot x + \frac{\pi}{6}\right) = 2 \sin\left(1 \cdot x - \left(-\frac{\pi}{6}\right)\right) \]
Here, \( b = 1 \) and \( c = -\frac{\pi}{6} \).
To find the phase shift, we calculate:
\[ \text{Phase shift} = \frac{c}{b} = \frac{-\frac{\pi}{6}}{1} = -\frac{\pi}{6} \]
A negative phase shift means the graph is shifted to the left by \( \frac{\pi}{6} \).
So, the correct phase shift for the function \( f(x) = 2 \sin(x + \frac{\pi}{6}) \) is \( \frac{\pi}{6} \) units to the left.