1. The problem is to analyze the function $y = |x+3| - 4$. 2. The absolute value function $|x|$ outputs the distance of $x$ from zero, always non-negative. 3. The function $y = |x+3| - 4$ shifts the basic absolute value function $|x|$ left by 3 units and down by 4 units. 4. To find the vertex, set the inside of the absolute value to zero: $x+3=0 \Rightarrow x=-3$. 5. At $x=-3$, $y = |0| - 4 = -4$, so the vertex is $(-3, -4)$. 6. For $x \geq -3$, $y = (x+3) - 4 = x - 1$ (linear increasing). 7. For $x < -3$, $y = -(x+3) - 4 = -x - 7$ (linear decreasing). 8. The graph has a V shape with vertex at $(-3, -4)$, decreasing to the left and increasing to the right. 9. The y-intercept is found by setting $x=0$: $y = |0+3| - 4 = 3 - 4 = -1$. 10. The x-intercepts are found by solving $|x+3| - 4 = 0 \Rightarrow |x+3| = 4$. 11. This gives $x+3 = 4$ or $x+3 = -4$, so $x=1$ or $x=-7$. 12. Therefore, the x-intercepts are at $(1,0)$ and $(-7,0)$. Final answer: The vertex is at $(-3, -4)$, the function decreases linearly for $x < -3$ and increases linearly for $x \geq -3$, with x-intercepts at $x=1$ and $x=-7$, and y-intercept at $y=-1$.