The problem asks for the area enclosed by the graph of the function $y = \frac{1}{2} |x - 4| - 3$ and the $x$-axis.

Let's break down the steps to solve this:

1. Analyze the function: The function is $y = \frac{1}{2} |x - 4| - 3$, which is a V-shaped absolute value function. The vertex of this function occurs at $x = 4$ because of the term $|x - 4|$. Plugging $x = 4$ into the equation gives: $y = \frac{1}{2} |4 - 4| - 3 = -3$ So, the vertex of the function is at $(4, -3)$.

2. Find the x-intercepts: To determine where the function intersects the $x$-axis, set $y = 0$: $\frac{1}{2} |x - 4| - 3 = 0$ Solving for $|x - 4|$: $\frac{1}{2} |x - 4| = 3$ $|x - 4| = 6$ This gives two solutions: $x - 4 = 6 \Rightarrow x = 10$ $x - 4 = -6 \Rightarrow x = -2$ So, the points where the graph intersects the $x$-axis are $x = -2$ and $x = 10$.

3. Set up the area integral: The area under the curve from $x = -2$ to $x = 10$ is composed of two linear segments due to the absolute value. We can split the integral at $x = 4$ (the vertex): $\text{Area} = \int_{-2}^{4} \left( \frac{1}{2} (4 - x) - 3 \right) \, dx + \int_{4}^{10} \left( \frac{1}{2} (x - 4) - 3 \right) \, dx$

4. Evaluate each integral:

   • For $\int_{-2}^{4} \left( \frac{1}{2} (4 - x) - 3 \right) \, dx$: $= \int_{-2}^{4} \left( 2 - \frac{1}{2}x - 3 \right) \, dx = \int_{-2}^{4} \left( -\frac{1}{2}x - 1 \right) \, dx$ Calculating this integral yields: $\left[ -\frac{1}{4}x^2 - x \right]_{-2}^{4} = 9$

   • For $\int_{4}^{10} \left( \frac{1}{2} (x - 4) - 3 \right) \, dx$: $= \int_{4}^{10} \left( \frac{1}{2}x - 2 - 3 \right) \, dx = \int_{4}^{10} \left( \frac{1}{2}x - 5 \right) \, dx$ Calculating this integral yields: $\left[ \frac{1}{4}x^2 - 5x \right]_{4}^{10} = 9$

5. Add both areas: $\text{Total Area} = 9 + 9 = 18$

Thus, the area enclosed by the graph and the $x$-axis is $18 \, u^2$, which corresponds to option c) 18 $u^2$.

Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. How do you find the vertex of an absolute value function?
2. Why is the area split into two integrals at $x = 4$?
3. How can you determine where a function intersects the x-axis?
4. What are the steps to solve an absolute value equation like $|x - 4| = 6$?
5. How do you set up and evaluate a definite integral for area calculations?

Tip: For absolute value functions, always locate the vertex to split the integral correctly.