The function given in the image is:

$f(x) = -\frac{1}{2} |x + 3| + 4$

Let's analyze the function step by step:

1. Basic Structure: The basic form of the absolute value function is $f(x) = a |x - h| + k$, where $(h, k)$ represents the vertex of the graph. In this case, the function has been altered to $f(x) = -\frac{1}{2}|x + 3| + 4$.

2. Identify the vertex: The expression $x + 3$ indicates a shift to the left by 3 units, and the "+4" means a vertical shift up by 4 units. Therefore, the vertex of the graph is at $(-3, 4)$.

3. Slope analysis: The coefficient $-\frac{1}{2}$ affects the steepness and direction of the graph. Since it is negative, the graph opens downward, and the slope of each "arm" of the V-shaped graph will be $-\frac{1}{2}$.

4. Drawing the graph:

   • Start at the vertex $(-3, 4)$.
   • From the vertex, as you move to the right (positive direction), the slope is $-\frac{1}{2}$. This means for every 2 units moved horizontally, you move down 1 unit vertically.
   • Similarly, as you move to the left from the vertex, you apply the slope $-\frac{1}{2}$.

Key Points:

• Vertex: $(-3, 4)$
• Slope to the right of the vertex: $-\frac{1}{2}$
• Slope to the left of the vertex: $-\frac{1}{2}$

Plotting specific points:

• At $x = -3$, $f(-3) = 4$. This is the vertex.
• At $x = -1$, $f(-1) = -\frac{1}{2} \times 2 + 4 = 3$. So, $(-1, 3)$ is a point on the graph.
• At $x = -5$, $f(-5) = -\frac{1}{2} \times 2 + 4 = 3$. So, $(-5, 3)$ is also a point on the graph.
• At $x = 0$, $f(0) = -\frac{1}{2} \times 3 + 4 = 2.5$.

Sketch:

The graph is a "V" shape with the vertex at $(-3, 4)$ opening downward with arms having a slope of $-\frac{1}{2}$.

Would you like a more detailed sketch or further explanation?

Relative Questions:

1. How does changing the coefficient of $x$ inside the absolute value affect the graph's shape?
2. What happens to the graph if the absolute value expression is positive instead of negative?
3. How would the graph shift if the "+4" in the equation was changed to "-4"?
4. What effect does altering the slope to $-1$ instead of $-\frac{1}{2}$ have on the graph?
5. How do we derive points from the graph if $f(x) = |x|$ without any transformations?

Tip: Always start by finding the vertex and direction of the slope when sketching absolute value functions.