6. If $f(x) = \sqrt{x - 2}$, find the domain of $f$. Step 1: 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. Step 2: Set up the inequality. In this function, $u = x - 2$. Therefore, we must have: $$ x - 2 \ge 0 $$ Step 3: Solve the inequality for $x$. Add 2 to both sides: $$ x \ge 2 $$ 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 $(\beta - 1)$. Step 1: Find the sum and product of the roots for the original equation. 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 $x^2 - 5x + 6 = 0$: $$ \alpha + \beta = -(-5)/1 = 5 $$ $$ \alpha \beta = 6/1 = 6 $$ Step 2: Find the sum of the new roots. Let the new roots be $r_1 = \alpha - 1$ and $r_2 = \beta - 1$. The sum of the new roots is: $$ r_1 + r_2 = (\alpha - 1) + (\beta - 1) $$ $$ r_1 + r_2 = \alpha + \beta - 2 $$ Substitute the value of $\alpha + \beta$: $$ r_1 + r_2 = 5 - 2 = 3 $$ Step 3: Find the product of the new roots. The product of the new roots is: $$ r_1 r_2 = (\alpha - 1)(\beta - 1) $$ $$ r_1 r_2 = \alpha \beta - \alpha - \beta + 1 $$ $$ r_1 r_2 = \alpha \beta - (\alpha + \beta) + 1 $$ Substitute the values of $\alpha \beta$ and $\alpha + \beta$: $$ r_1 r_2 = 6 - (5) + 1 $$ $$ r_1 r_2 = 6 - 5 + 1 $$ $$ r_1 r_2 = 2 $$ Step 4: Form the new quadratic equation. 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: $$ x^2 - (3)x + (2) = 0 $$ $$ x^2 - 3x + 2 = 0 $$ 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: Step 1: Identify the real and imaginary parts of the complex number. Let $z = -1 + \sqrt{3}i$. The real part is $a = -1$. The imaginary part is $b = \sqrt{3}$. Step 2: Determine the quadrant of the complex number. Since $a < 0$ and $b > 0$, the complex number lies in the second quadrant. Step 3: Calculate the reference angle. The reference angle $\alpha$ is given by $\tan \alpha = \left|\frac{b}{a}\right|$. $$ \tan \alpha = \left|\frac{\sqrt{3}}{-1}\right| $$ $$ \tan \alpha = \sqrt{3} $$ The angle whose tangent is $\sqrt{3}$ is $\frac{\pi}{3}$ radians (or $60^\circ$). $$ \alpha = \frac{\pi}{3} $$ Step 4: Calculate the argument. For a complex number in the second quadrant, the argument $\theta$ is $\pi - \alpha$. $$ \theta = \pi - \frac{\pi}{3} $$ $$ \theta = \frac{3\pi}{3} - \frac{\pi}{3} $$ $$ \theta = \frac{2\pi}{3} $$ The correct option is C. $\boxed{\text{C. } \frac{2\pi}{3}}$