The function shown in the image is:

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

This is a transformed absolute value function. The basic shape of the absolute value function $f(x) = |x|$ is a "V" shape, with the vertex at the origin $(0, 0)$ and the slope of the two arms being 1 and -1.

Let's break down the transformations applied to $v(x)$:

1. Horizontal Shift:

The expression $|x + 2|$ indicates a horizontal shift. Since the "+2" is inside the absolute value, it shifts the graph 2 units to the left. So the vertex, instead of being at $(0, 0)$, moves to $(-2, 0)$.

2. Vertical Stretch/Compression:

The coefficient $\frac{1}{3}$ outside the absolute value affects the slope of the lines. Normally, the slope is 1 for the right arm and -1 for the left arm. But multiplying by $\frac{1}{3}$ compresses the graph vertically, making the slopes $\frac{1}{3}$ and $-\frac{1}{3}$. This makes the graph less steep.

3. Vertical Shift:

The "-3" outside the absolute value indicates a vertical shift down by 3 units. So the vertex, which was at $(-2, 0)$, moves down to $(-2, -3)$.

Resulting Graph:

• The vertex of the graph is at $(-2, -3)$.
• The slopes of the arms are $\frac{1}{3}$ and $-\frac{1}{3}$, making the graph less steep than the standard absolute value function.

The graph in the image matches this description, with the "V" shape shifted left and down, and the slopes of the lines appear to be less steep due to the vertical compression.

Do you want more details or have any questions about this transformation?

Here are 5 related questions to expand your understanding:

1. How would the graph change if the coefficient were $\frac{2}{3}$ instead of $\frac{1}{3}$?
2. What would happen if we changed $|x + 2|$ to $|x - 2|$?
3. How does a negative sign in front of the absolute value function, like $-|x + 2|$, affect the graph?
4. What transformations would shift the graph up by 4 units?
5. Can you graph $v(x) = |x + 2| - 3$ without the $\frac{1}{3}$ factor? How would the shape differ?

Tip: When transforming graphs, pay close attention to the placement of constants inside and outside the function (e.g., inside the absolute value vs. outside), as they indicate horizontal vs. vertical changes!