To find the Taylor polynomial of $f(x) = \sqrt{1 + \frac{x}{2}}$ of order 2 about the origin, we need to calculate the derivatives of $f(x)$ and evaluate them at $x = 0$.

1. Find $f(0)$: $f(0) = \sqrt{1 + \frac{0}{2}} = \sqrt{1} = 1$

2. Find $f'(x)$: $f'(x) = \frac{d}{dx}\left(1 + \frac{x}{2}\right)^{1/2} = \frac{1}{2}(1 + \frac{x}{2})^{-1/2} \cdot \frac{1}{2} = \frac{1}{4}(1 + \frac{x}{2})^{-1/2}$ $f'(0) = \frac{1}{4}(1 + 0)^{-1/2} = \frac{1}{4}$

3. Find $f''(x)$: $f''(x) = \frac{d}{dx}\left(\frac{1}{4}(1 + \frac{x}{2})^{-1/2}\right) = \frac{1}{4} \cdot \left(-\frac{1}{2}\right)(1 + \frac{x}{2})^{-3/2} \cdot \frac{1}{2}$ $f''(x) = -\frac{1}{16}(1 + \frac{x}{2})^{-3/2}$ $f''(0) = -\frac{1}{16}(1 + 0)^{-3/2} = -\frac{1}{16}$

4. Construct the Taylor polynomial: $P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2$ $P_2(x) = 1 + \frac{1}{4}x - \frac{1}{16}\frac{x^2}{2}$ $P_2(x) = 1 + \frac{1}{4}x - \frac{1}{32}x^2$

The Taylor polynomial of order 2 is $P_2(x) = 1 + \frac{1}{4}x - \frac{1}{32}x^2$.