Test whether the relation y = -sqrt(-5 - 6x - x2) is a function or not? If it is a function, determine its domain and range.

Step 1: Test whether the relation is a function. A relation is a function if for every input $x$ in its domain, there is exactly one output $y$. The given relation is $y = -\sqrt{-5 - 6x - x^2}$. For any valid value of $x$, the expression under the square root, $-5 - 6x - x^2$, will evaluate to a single non-negative number. The square root of this number, $\sqrt{-5 - 6x - x^2}$, will yield a single non-negative value (the principal square root). The negative sign in front, $-\sqrt{-5 - 6x - x^2}$, ensures that for each valid $x$, there is only one unique $y$ value. Therefore, the relation is a function.

Step 2: Determine the domain of the function. For the expression under the square root to be defined in real numbers, it must be non-negative: $-5 - 6x - x^2 \ge 0$ Multiply the inequality by $-1$ and reverse the inequality sign: $x^2 + 6x + 5 \le 0$ Factor the quadratic expression: $(x+1)(x+5) \le 0$ The roots of the quadratic equation $x^2 + 6x + 5 = 0$ are $x = -1$ and $x = -5$. Since the parabola $y = x^2 + 6x + 5$ opens upwards, the expression $(x+1)(x+5)$ is less than or equal to zero between its roots. Thus, the domain is: $-5 \le x \le -1$ The domain is $\boxed{[-5, -1]}$.

Step 3: Determine the range of the function. The function is $y = -\sqrt{-5 - 6x - x^2}$. Let's complete the square for the expression under the square root: $-x^2 - 6x - 5 = -(x^2 + 6x + 5)$ $= -(x^2 + 6x + 9 - 9 + 5)$ $= -((x+3)^2 - 4)$ $= 4 - (x+3)^2$ So, the function can be written as: $y = -\sqrt{4 - (x+3)^2}$ From the domain, we know that $-5 \le x \le -1$. Let's find the range of $(x+3)^2$: $-5+3 \le x+3 \le -1+3$ $-2 \le x+3 \le 2$ When we square this interval, the minimum value of $(x+3)^2$ is $0$ (when $x+3=0$, i.e., $x=-3$) and the maximum value is $4$ (when $x+3=-2$ or $x+3=2$). So, $0 \le (x+3)^2 \le 4$.

Now, consider the expression under the square root, $4 - (x+3)^2$: Since $0 \le (x+3)^2 \le 4$, we have: $4 - 4 \le 4 - (x+3)^2 \le 4 - 0$ $0 \le 4 - (x+3)^2 \le 4$ This means the value inside the square root ranges from $0$ to $4$. Next, consider $\sqrt{4 - (x+3)^2}$: $\sqrt{0} \le \sqrt{4 - (x+3)^2} \le \sqrt{4}$ $0 \le \sqrt{4 - (x+3)^2} \le 2$ Finally, consider $y = -\sqrt{4 - (x+3)^2}$: Multiplying by $-1$ reverses the inequality signs: $-2 \le -\sqrt{4 - (x+3)^2} \le 0$ So, the range of the function is: $-2 \le y \le 0$ The range is $\boxed{[-2, 0]}$.