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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?

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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.

Calcule el área que encierra la gráfica de la función y = (1/2) |x - 4| - 3 y el eje x.

A step-by-step solution to calculate the area enclosed by the graph of y = (1/2) |x - 4| - 3 and the x-axis. This solution demonstrates the use of definite integrals and the properties of absolute value functions, suitable for high school calculus students.

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