The given equation is:

$y = 2|x+1| - 10$

This is a V-shaped graph because of the absolute value function. The equation represents two linear functions that join at the vertex, which occurs where the expression inside the absolute value equals zero, i.e., $x = -1$.

Step 1: Break the equation into two linear pieces based on the absolute value function:

• When $x + 1 \geq 0$ (i.e., $x \geq -1$), the absolute value disappears: $y = 2(x+1) - 10 = 2x + 2 - 10 = 2x - 8$

• When $x + 1 < 0$ (i.e., $x < -1$), the absolute value becomes negative: $y = 2(-(x+1)) - 10 = 2(-x-1) - 10 = -2x - 2 - 10 = -2x - 12$

Thus, the piecewise function becomes:

-2x - 12 & \text{for } x < -1 \\ 2x - 8 & \text{for } x \geq -1 \end{cases}$$ ### Step 2: Identify the third quadrant region In the third quadrant, both $$x$$ and $$y$$ are negative. We need to find the area under the graph in this quadrant, specifically for the interval $$x = -6$$ to $$x = -1$$ (since $$x = -1$$ is the vertex, it divides the two linear sections). For $$x = -6$$ to $$x = -1$$, the equation of the line is $$y = -2x - 12$$. ### Step 3: Find the area in the third quadrant The area under the graph from $$x = -6$$ to $$x = -1$$ can be found by integrating the linear function $$y = -2x - 12$$ over this interval. $$\text{Area} = \int_{-6}^{-1} (-2x - 12) \, dx$$ Let's calculate this integral. $$\int (-2x - 12) \, dx = -x^2 - 12x$$ Now, evaluate this from $$x = -6$$ to $$x = -1$$: $$\left[ -x^2 - 12x \right]_{-6}^{-1} = \left( -(-1)^2 - 12(-1) \right) - \left( -(-6)^2 - 12(-6) \right)$$ $$= \left( -1 + 12 \right) - \left( -36 + 72 \right)$$ $$= 11 - 36$$ $$= -25$$ Since the area is positive, the area in the third quadrant is $$25$$ square units. ### Conclusion The area bounded by the graph in the third quadrant from $$x = -6$$ to $$x = -1$$ is **25 square units**. Would you like more details or have any questions about the calculation? --- Here are 5 related questions to expand your understanding: 1. How does the vertex of the absolute value function affect the shape of the graph? 2. What would happen to the area if the graph had a different slope? 3. Can you find the area of this graph in the second quadrant for another interval? 4. How would the graph change if the absolute value term was multiplied by a different coefficient? 5. Can you calculate the point where the graph crosses the x-axis? **Tip**: For absolute value functions, the vertex is key in dividing the function into its piecewise linear parts. Always check the domain for where each linear part applies!