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Problem: Graphing Piecewise Functions

The task is to graph the given piecewise functions on the provided coordinate grids. Below, I will solve each problem step by step and explain the process.

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Problem 1:

$$
f(x) =
\begin{cases}
-x & \text{if } x \leq 2 \\
x & \text{if } x > 2
\end{cases}
$$

#### Steps:
1. Identify the two pieces of the function:
- For \( x \leq 2 \): \( f(x) = -x \)
- For \( x > 2 \): \( f(x) = x \)

2. Graph \( f(x) = -x \) for \( x \leq 2 \):
- This is a straight line with a slope of \(-1\) and a y-intercept of \(0\).
- Plot points such as \((0, 0)\), \((1, -1)\), and \((2, -2)\).
- Since the condition is \( x \leq 2 \), include the point \((2, -2)\) (closed circle).

3. Graph \( f(x) = x \) for \( x > 2 \):
- This is a straight line with a slope of \(1\) and a y-intercept of \(0\).
- Plot points such as \((3, 3)\), \((4, 4)\), and \((5, 5)\).
- Since the condition is \( x > 2 \), do not include the point \((2, 2)\) (open circle).

4. Combine the graphs:
- Draw the line \( f(x) = -x \) for \( x \leq 2 \) with a closed circle at \((2, -2)\).
- Draw the line \( f(x) = x \) for \( x > 2 \) with an open circle at \((2, 2)\).

#### Final Graph:
- A V-shaped graph with the vertex at \((2, -2)\), where the left side is \( f(x) = -x \) and the right side is \( f(x) = x \).

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Problem 2:

$$
f(x) =
\begin{cases}
2 & \text{if } x > -3 \\
-5 & \text{if } x < -3
\end{cases}
$$

#### Steps:
1. Identify the two pieces of the function:
- For \( x > -3 \): \( f(x) = 2 \) (a horizontal line at \( y = 2 \)).
- For \( x < -3 \): \( f(x) = -5 \) (a horizontal line at \( y = -5 \)).

2. Graph \( f(x) = 2 \) for \( x > -3 \):
- Draw a horizontal line at \( y = 2 \).
- Since the condition is \( x > -3 \), do not include the point \((-3, 2)\) (open circle).

3. Graph \( f(x) = -5 \) for \( x < -3 \):
- Draw a horizontal line at \( y = -5 \).
- Since the condition is \( x < -3 \), do not include the point \((-3, -5)\) (open circle).

4. Combine the graphs:
- Draw the horizontal line \( y = 2 \) for \( x > -3 \) with an open circle at \((-3, 2)\).
- Draw the horizontal line \( y = -5 \) for \( x < -3 \) with an open circle at \((-3, -5)\).

#### Final Graph:
- Two horizontal lines: \( y = 2 \) for \( x > -3 \) and \( y = -5 \) for \( x < -3 \), with open circles at \((-3, 2)\) and \((-3, -5)\).

---

Problem 3:

$$
f(x) =
\begin{cases}
-1 & \text{if } x \leq -2 \\
2 & \text{if } x > -2
\end{cases}
$$

#### Steps:
1. Identify the two pieces of the function:
- For \( x \leq -2 \): \( f(x) = -1 \) (a horizontal line at \( y = -1 \)).
- For \( x > -2 \): \( f(x) = 2 \) (a horizontal line at \( y = 2 \)).

2. Graph \( f(x) = -1 \) for \( x \leq -2 \):
- Draw a horizontal line at \( y = -1 \).
- Since the condition is \( x \leq -2 \), include the point \((-2, -1)\) (closed circle).

3. Graph \( f(x) = 2 \) for \( x > -2 \):
- Draw a horizontal line at \( y = 2 \).
- Since the condition is \( x > -2 \), do not include the point \((-2, 2)\) (open circle).

4. Combine the graphs:
- Draw the horizontal line \( y = -1 \) for \( x \leq -2 \) with a closed circle at \((-2, -1)\).
- Draw the horizontal line \( y = 2 \) for \( x > -2 \) with an open circle at \((-2, 2)\).

#### Final Graph:
- Two horizontal lines: \( y = -1 \) for \( x \leq -2 \) and \( y = 2 \) for \( x > -2 \), with a closed circle at \((-2, -1)\) and an open circle at \((-2, 2)\).

---

Problem 4:

$$
f(x) =
\begin{cases}
-1 & \text{if } x \leq -1 \\
1 & \text{if } -1 < x < 1 \\
x & \text{if } x > 1
\end{cases}
$$

#### Steps:
1. Identify the three pieces of the function:
- For \( x \leq -1 \): \( f(x) = -1 \) (a horizontal line at \( y = -1 \)).
- For \( -1 < x < 1 \): \( f(x) = 1 \) (a horizontal line at \( y = 1 \)).
- For \( x > 1 \): \( f(x) = x \) (a straight line with slope \(1\) and y-intercept \(0\)).

2. Graph \( f(x) = -1 \) for \( x \leq -1 \):
- Draw a horizontal line at \( y = -1 \).
- Since the condition is \( x \leq -1 \), include the point \((-1, -1)\) (closed circle).

3. Graph \( f(x) = 1 \) for \( -1 < x < 1 \):
- Draw a horizontal line at \( y = 1 \).
- Since the condition is \( -1 < x < 1 \), do not include the points \((-1, 1)\) and \((1, 1)\) (open circles).

4. Graph \( f(x) = x \) for \( x > 1 \):
- Draw a straight line with slope \(1\) and y-intercept \(0\).
- Since the condition is \( x > 1 \), do not include the point \((1, 1)\) (open circle).

5. Combine the graphs:
- Draw the horizontal line \( y = -1 \) for \( x \leq -1 \) with a closed circle at \((-1, -1)\).
- Draw the horizontal line \( y = 1 \) for \( -1 < x < 1 \) with open circles at \((-1, 1)\) and \((1, 1)\).
- Draw the line \( f(x) = x \) for \( x > 1 \) with an open circle at \((1, 1)\).

#### Final Graph:
- A combination of a horizontal line at \( y = -1 \) for \( x \leq -1 \), a horizontal line at \( y = 1 \) for \( -1 < x < 1 \), and a diagonal line \( f(x) = x \) for \( x > 1 \), with appropriate closed and open circles.

---

Problem 5:

$$
f(x) =
\begin{cases}
-x + 2 & \text{if } x \leq 0 \\
\frac{1}{2}x + 3 & \text{if } x > 0
\end{cases}
$$

#### Steps:
1. Identify the two pieces of the function:
- For \( x \leq 0 \): \( f(x) = -x + 2 \) (a line with slope \(-1\) and y-intercept \(2\)).
- For \( x > 0 \): \( f(x) = \frac{1}{2}x + 3 \) (a line with slope \(\frac{1}{2}\) and y-intercept \(3\)).

2. Graph \( f(x) = -x + 2 \) for \( x \leq 0 \):
- Plot points such as \((0, 2)\) and \((-2, 4)\).
- Since the condition is \( x \leq 0 \), include the point \((0, 2)\) (closed circle).

3. Graph \( f(x) = \frac{1}{2}x + 3 \) for \( x > 0 \):
- Plot points such as \((0, 3)\) and \((2, 4)\).
- Since the condition is \( x > 0 \), do not include the point \((0, 3)\) (open circle).

4. Combine the graphs:
- Draw the line \( f(x) = -x + 2 \) for \( x \leq 0 \) with a closed circle at \((0, 2)\).
- Draw the line \( f(x) = \frac{1}{2}x + 3 \) for \( x > 0 \) with an open circle at \((0, 3)\).

#### Final Graph:
- A piecewise graph with a line \( f(x) = -x + 2 \) for \( x \leq 0 \) and a line \( f(x) = \frac{1}{2}x + 3 \) for \( x > 0 \), with a closed circle at \((0, 2)\) and an open circle at \((0, 3)\).

---

Problem 6:

$$
f(x) =
\begin{cases}
x + 2 & \text{if } x \leq 2 \\
-\frac{1}{2}x + 4 & \text{if } x > 2
\end{cases}
$$

#### Steps:
1. Identify the two pieces of the function:
- For \( x \leq 2 \): \( f(x) = x + 2 \) (a line with slope \(1\) and y-intercept \(2\)).
- For \( x > 2 \): \( f(x) = -\frac{1}{2}x + 4 \) (a line with slope \(-\frac{1}{2}\) and y-intercept \(4\)).

2. Graph \( f(x) = x + 2 \) for \( x \leq 2 \):
- Plot points such as \((0, 2)\) and \((2, 4)\).
- Since the condition is \( x \leq 2 \), include the point \((2, 4)\) (closed circle).

3. Graph \( f(x) = -\frac{1}{2}x + 4 \) for \( x > 2 \):
- Plot points such as \((2, 3)\) and \((4, 2)\).
- Since the condition is \( x > 2 \), do not include the point \((2, 3)\) (open circle).

4. Combine the graphs:
- Draw the line \( f(x) = x + 2 \) for \( x \leq 2 \) with a closed circle at \((2, 4)\).
- Draw the line \( f(x) = -\frac{1}{2}x + 4 \) for \( x > 2 \) with an open circle at \((2, 3)\).

#### Final Graph:
- A piecewise graph with a line \( f(x) = x + 2 \) for \( x \leq 2 \) and a line \( f(x) = -\frac{1}{2}x + 4 \) for \( x > 2 \), with a closed circle at \((2, 4)\) and an open circle at \((2, 3)\).

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Final Answers:

The graphs for each problem are described above. The final boxed answer is:

\[
\boxed{\text{See detailed explanations above for each graph.}}
\]