To write the piecewise function represented by the graph, we need to define each linear piece separately.

Step 1: First Linear Piece (From $x = -4$ to $x = -2$)

The first segment goes through the points $(-4, -6)$ and $(-2, -2)$. We will determine the equation of the line using the slope-intercept form $y = mx + b$.

Find the slope $m$:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-6)}{-2 - (-4)} = \frac{-2 + 6}{-2 + 4} = \frac{4}{2} = 2$

So, the slope $m = 2$.

Use point-slope form to find the equation:

Using point $(-4, -6)$ and the slope $m = 2$, the equation of the line is: $y - y_1 = m(x - x_1)$ $y - (-6) = 2(x - (-4))$ $y + 6 = 2(x + 4)$ $y = 2x + 8 - 6$ $y = 2x + 2$

Thus, the equation for the first piece is $y = 2x + 2$, valid for $x \in [-4, -2]$.

Step 2: Second Linear Piece (From $x = -2$ to $x = 2$)

The second segment goes through the points $(-2, -2)$ and $(2, 0)$.

Find the slope $m$:

$m = \frac{0 - (-2)}{2 - (-2)} = \frac{0 + 2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$

So, the slope $m = \frac{1}{2}$.

Use point-slope form to find the equation:

Using point $(-2, -2)$ and the slope $m = \frac{1}{2}$, the equation of the line is: $y - (-2) = \frac{1}{2}(x - (-2))$ $y + 2 = \frac{1}{2}(x + 2)$ $y + 2 = \frac{1}{2}x + 1$ $y = \frac{1}{2}x + 1 - 2$ $y = \frac{1}{2}x - 1$

Thus, the equation for the second piece is $y = \frac{1}{2}x - 1$, valid for $x \in (-2, 2]$.

Piecewise Function:

Now, we can write the piecewise function:

\begin{cases} 2x + 2 & \text{if } -4 \leq x \leq -2 \\ \frac{1}{2}x - 1 & \text{if } -2 < x \leq 2 \end{cases}$$ Would you like further details on any part of this? Here are some related questions to deepen your understanding: 1. How do you find the slope between two points? 2. What is the point-slope form of a line, and when is it used? 3. Can a piecewise function be continuous? Why or why not? 4. How would you graph this piecewise function? 5. What is the domain of this piecewise function? **Tip:** When working with piecewise functions, always check if the pieces are connected at the boundary points!