Identify the restriction for the function.

For a square root function $f(x) = \sqrt{u}$, the expression under the square root, $u$, must be non-negative (greater than or equal to zero) for the function to produce real numbers.

In this function, $u = x - 2$. Therefore, we must have:

Add 2 to both sides:

The domain of $f(x)$ is all real numbers $x$ such that $x \ge 2$.

The correct option is A.
$\boxed{\text{A. } x \ge 2}$

7. If $\alpha$ and $\beta$ are roots of $x^2 - 5x + 6 = 0$, find an equation whose roots are $(\alpha - 1)$ and $(\beta - 1)$.

For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.
For $x^2 - 5x + 6 = 0$:

Let the new roots be $r_1 = \alpha - 1$ and $r_2 = \beta - 1$.
The sum of the new roots is:

Substitute the value of $\alpha + \beta$:

The product of the new roots is:

Substitute the values of $\alpha \beta$ and $\alpha + \beta$:

A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1 + r_2)x + (r_1 r_2) = 0$.
Substitute the sum and product of the new roots:

The correct option is A.
$\boxed{\text{A. } x^2 - 3x + 2 = 0}$

8. The argument of the complex number $-1 + \sqrt{3}i$ is:

Let $z = -1 + \sqrt{3}i$.
The real part is $a = -1$.
The imaginary part is $b = \sqrt{3}$.

Since $a < 0$ and $b > 0$, the complex number lies in the second quadrant.

The reference angle $\alpha$ is given by $\tan \alpha = \left|\frac{b}{a}\right|$.

The angle whose tangent is $\sqrt{3}$ is $\frac{\pi}{3}$ radians (or $60^\circ$).

For a complex number in the second quadrant, the argument $\theta$ is $\pi - \alpha$.

The correct option is C.
$\boxed{\text{C. } \frac{2\pi}{3}}$