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The function is $y = |x+3|$. This is an absolute value function, which means $y = x+3$ when $x+3 \ge 0$ (i.e., $x \ge -3$) and $y = -(x+3)$ when $x+3 < 0$ (i.e., $x < -3$).
We can define the function as: $$y = \begin{cases} x+3, & x \ge -3 \\ -(x+3), & x < -3 \end{cases}$$
The graph of $y = |x+3|$ is a V-shaped graph with the vertex at $(-3, 0)$. For $x \ge -3$, the graph is a straight line with slope 1. For $x < -3$, the graph is a straight line with slope -1.
We need to find the area enclosed by the curve $y = |x+3|$, the x-axis, and the lines $x = -6$ and $x = 0$. Since the function changes its form at $x = -3$, we need to split the integral into two parts: from $x = -6$ to $x = -3$ and from $x = -3$ to $x = 0$. The area is given by: $$A = \int_{-6}^{0} |x+3| \, dx = \int_{-6}^{-3} -(x+3) \, dx + \int_{-3}^{0} (x+3) \, dx$$
$$\int_{-6}^{-3} -(x+3) \, dx = -\int_{-6}^{-3} (x+3) \, dx = -\left[\frac{x^2}{2} + 3x\right]_{-6}^{-3}$$ $$= -\left[\left(\frac{(-3)^2}{2} + 3(-3)\right) - \left(\frac{(-6)^2}{2} + 3(-6)\right)\right] = -\left[\left(\frac{9}{2} - 9\right) - \left(\frac{36}{2} - 18\right)\right]$$ $$= -\left[\frac{9}{2} - 9 - 18 + 18\right] = -\left[\frac{9}{2} - 9\right] = -\left[\frac{9 - 18}{2}\right] = -\left[-\frac{9}{2}\right] = \frac{9}{2}$$
$$\int_{-3}^{0} (x+3) \, dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{0} = \left[\left(\frac{0^2}{2} + 3(0)\right) - \left(\frac{(-3)^2}{2} + 3(-3)\right)\right]$$ $$= \left[0 - \left(\frac{9}{2} - 9\right)\right] = -\left[\frac{9}{2} - 9\right] = -\left[\frac{9 - 18}{2}\right] = -\left[-\frac{9}{2}\right] = \frac{9}{2}$$
$$A = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9$$
Final Answer: 9
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