Identification and graphing of piecewise functions with multiple examples and solutions.

To solve these problems, we need to look at how piecewise functions are defined and graphed. A piecewise function is like a rule book that changes depending on the value of $x$ (the horizontal position).

Here is the step-by-step breakdown for each section.

Part III: Identifying the Piecewise Function

In this section, we match the graphs to their algebraic expressions by looking at three things:
1. The Shape: Is it a straight line? A flat horizontal line? A curve (parabola)?
2. The Domain (x-values): Where does each piece start and stop?
3. The Endpoints: Are the circles open ($\circ$, meaning "not included") or closed ($\bullet$, meaning "included")?

1. Graph $f(x)$ (Top Left)
* Left piece ($x < -1$): This is a line going down. It passes through $(-3, 0)$ and $(-1, -1)$. The slope is $\frac{-1 - 0}{-1 - (-3)} = \frac{-1}{2}$. Using point-slope form, the equation is $y = -\frac{1}{2}x - \frac{3}{2}$. The circle at $x=-1$ is open.
* Middle piece ($-1 \le x \le 3$): This is a flat horizontal line at height $y=2$. The left endpoint is closed (solid dot implied by the previous open circle logic usually, but here let's check the solution provided in the image text which says $-1 \le x \le 3$). Wait, looking closely at the graph, there is an open circle at $(-1, -1)$ and the horizontal line starts at $x=-1$. Usually, one side must be closed. The solution key in the image defines the middle part as $-1 \le x \le 3$.
* Right piece ($x > 3$): This line goes down from $(3, 4)$ to $(5, 2)$. Slope is $\frac{2-4}{5-3} = -1$. Equation: $y - 2 = -1(x - 5) \Rightarrow y = -x + 7$. The circle at $x=3$ is open.

2. Graph $g(x)$ (Top Right)
* Left piece ($x \le 0$): Line passing through $(0, -2)$ and $(-2, 0)$. Slope is $-1$. Y-intercept is $-2$. Equation: $y = -x - 2$. The circle at $x=0$ is closed (solid).
* Right piece ($x > 0$): Line starting at $(0, 2)$ and going up through $(2, 4)$. Slope is $1$. Y-intercept would be $2$. Equation: $y = x + 2$. The circle at $x=0$ is open.

3. Graph $h(x)$ (Bottom Left)
* Left piece ($x < 3$): A diagonal line passing through origin $(0,0)$ and $(2,-2)$. Slope is $-1$. Equation: $y = -x$. The circle at $x=3$ is open.
* Middle point ($x = 3$): There is a single solid dot at $(3, 1)$. So, $h(3) = 1$.
* Right piece ($x > 3$): A horizontal line at height $y=3$. Starts with an open circle at $x=3$. Equation: $y = 3$.

4. Challenge Graph $f(x)$ (Bottom Right)
* Left piece ($x < -2$): Horizontal line at $y = -3$. Open circle at $x=-2$.
* Middle piece ($-2 \le x \le 2$): A parabola (U-shape) with vertex at $(0,0)$. It passes through $(2,4)$ and $(-2,4)$. The standard equation is $y = x^2$. The endpoints are solid dots (implied by continuity or standard notation if not specified otherwise, but the solution key uses $\le$).
* Right piece ($x > 2$): An absolute value shape pointing down. Vertex at $(3,1)$. Passes through $(2,0)$ and $(4,0)$. The slope on the left side of the V is $1$, right side is $-1$. The equation for an inverted V shifted right 3 and up 1 is $y = -|x - 3| + 1$. Let's check the solution key provided in the image text: it says $-|x - 4|$. Let's test that. If $x=2$, $-|2-4| = -2$. But the graph is at $y=0$ when $x=2$. Actually, looking at the graph, the vertex is at $(3,1)$. The point $(2,0)$ is on the graph. $-|2-3|+1 = 0$. Correct. The solution key in the image actually writes `-|x - 4|`? No, looking closer at the faint text, it looks like `-|x - 3| + 1` or similar. However, the provided text solution in the image says `-|x - 4|`. Let's re-read the graph. Ah, the vertex is at $x=3, y=1$. The line goes through $(4,0)$? No, it goes through $(5, -1)$. Let's look at the provided solution text in the image again. It says `-|x - 4|`. If $x=2$, $-|2-4| = -2$. That doesn't match the graph which is at $y=0$ at $x=2$. There might be a typo in the image's own solution key or my reading of the grid. Let's look at the graph very carefully.
* Vertex at $(3, 1)$.
* Passes through $(2, 0)$ and $(4, 0)$.
* Equation: $y = -|x - 3| + 1$.
* The text in the image seems to say `-|x - 4|`. This is likely incorrect based on the visual graph, OR the graph is drawn differently than I see. Let's look at the third crop. The vertex is clearly at $x=3$. The solution text is blurry. It might say `-|x - 3| + 1`. Regardless, the task is to identify the expression. The expressions are already written next to the graphs in the "SOLUTIONS" column. The user wants me to solve/verify them.

Part IV: Graphing Piecewise Functions

Here, we are given the equations and must draw the lines.

1. Graph $f(x)$ (Bottom Left)
$$ f(x) = \begin{cases} 4 & \text{if } x < 3 \\ -x + 3 & \text{if } x \ge 3 \end{cases} $$
* Step 1: Draw the line $y = 4$ for all $x$ values less than 3. This is a horizontal line at height 4. Since $x < 3$, put an open circle at $(3, 4)$.
* Step 2: Draw the line $y = -x + 3$ for $x \ge 3$.
* At $x = 3$, $y = -3 + 3 = 0$. Put a closed circle at $(3, 0)$.
* At $x = 4$, $y = -4 + 3 = -1$. Plot $(4, -1)$.
* Connect $(3,0)$ and $(4,-1)$ with a ray going downwards to the right.

2. Graph $g(x)$ (Bottom Right)
$$ g(x) = \begin{cases} 2x & \text{if } x < -3 \\ |x| & \text{if } -3 \le x < 3 \\ 5 & \text{if } x \ge 3 \end{cases} $$
* Step 1: Draw $y = 2x$ for $x < -3$.
* At $x = -3$, $y = 2(-3) = -6$. Put an open circle at $(-3, -6)$.
* Pick another point like $x = -4$, $y = -8$. Draw a line going down to the left from $(-3, -6)$.
* Step 2: Draw $y = |x|$ for $-3 \le x < 3$.
* This is a V-shape centered at $(0,0)$.
* At $x = -3$, $y = |-3| = 3$. Put a closed circle at $(-3, 3)$.
* At $x = 0$, $y = 0$. Vertex at $(0,0)$.
* At $x = 3$, $y = |3| = 3$. Put an open circle at $(3, 3)$.
* Connect $(-3,3)$ to $(0,0)$ and $(0,0)$ to $(3,3)$.
* Step 3: Draw $y = 5$ for $x \ge 3$.
* This is a horizontal line at height 5.
* At $x = 3$, $y = 5$. Put a closed circle at $(3, 5)$.
* Draw a ray to the right from $(3, 5)$.

*(Note: The graph shown in the bottom right of the original image seems to correspond to a different problem or has errors compared to the equation provided next to it. For example, the graph shows a V-shape at the bottom with vertex at $(0,0)$ going up to $(-3,3)$ and $(3,3)$, which matches the middle part $|x|$. But the left part in the graph stops at $(-3,3)$ with an open circle, whereas the equation $2x$ would be way down at $y=-6$. The right part in the graph is a horizontal line at $y=5$ starting at $x=3$. The graph in the image actually matches the equation if we ignore the left-most part's position or if the grid scale is different. Wait, looking at the bottom-right graph in the image: The left ray comes from top-left? No, it's cut off. The V shape is between -3 and 3. The right ray is at y=5. The left part of the graph in the image shows a line segment ending at an open circle at roughly (-3, 3)? If the equation is $2x$, the point should be $(-3, -6)$. The graph in the image appears to have a mistake or corresponds to a different equation like $-x$ for the left part. However, based on the text equation provided: $2x$ for $x<-3$, the graph should go deep down.)*

Let's re-examine the bottom-right graph in the image provided.
The graph shows:
- A V-shape from $x=-3$ to $x=3$. Vertices at $(-3,3), (0,0), (3,3)$. This matches $|x|$.
- A horizontal line to the right from $x=3$ at height $y=5$. This matches $5$ if $x \ge 3$.
- To the left of $x=-3$, there is a line segment coming down from the top left? Or going up? It looks like it ends at an open circle at $(-3, 3)$. If the equation was $y = -x$ for $x < -3$, then at $x=-3, y=3$. That would match the visual. But the text says $2x$.
Conclusion: The graph printed in the worksheet for the bottom-right problem seems to contain an error regarding the left-hand piece ($2x$ vs what is drawn), or the grid lines are misleading. However, the task is to "Graph Piecewise functions". I will provide the description of how the graph *should* look based on the equations provided.

Final Answer:

III. Identifying the Piecewise Function

The expressions describing the graphs are:

1. $f(x)$ (Top Left):
$$ f(x) = \begin{cases} -\frac{1}{2}x - \frac{3}{2} & \text{if } x < -1 \\ 2 & \text{if } -1 \le x \le 3 \\ -x + 7 & \text{if } x > 3 \end{cases} $$

2. $g(x)$ (Top Right):
$$ g(x) = \begin{cases} x + 2 & \text{if } x > 0 \\ -x - 2 & \text{if } x \le 0 \end{cases} $$

3. $h(x)$ (Bottom Left):
$$ h(x) = \begin{cases} 3 & \text{if } x > 3 \\ 1 & \text{if } x = 3 \\ -x & \text{if } x < 3 \end{cases} $$

4. Challenge $f(x)$ (Bottom Right):
$$ f(x) = \begin{cases} -3 & \text{if } x < -2 \\ x^2 & \text{if } -2 \le x \le 2 \\ -|x - 3| + 1 & \text{if } x > 2 \end{cases} $$
*(Note: Based on the visual graph, the rightmost piece is an inverted absolute value with vertex at $(3,1)$. The text in the image solution key is slightly ambiguous but likely intends this or a similar shift.)*

IV. Graphing Piecewise Functions

1. Graph for $f(x) = \begin{cases} 4 & \text{if } x < 3 \\ -x + 3 & \text{if } x \ge 3 \end{cases}$
* Draw a horizontal line at $y = 4$ extending to the left from $x = 3$. Place an open circle at $(3, 4)$.
* Draw a line starting with a closed circle at $(3, 0)$ and passing through $(4, -1)$, extending downwards to the right.

2. Graph for $g(x) = \begin{cases} 2x & \text{if } x < -3 \\ |x| & \text{if } -3 \le x < 3 \\ 5 & \text{if } x \ge 3 \end{cases}$
* Left Part: Draw the line $y = 2x$. Place an open circle at $(-3, -6)$. Draw a ray going down and to the left from this point.
* Middle Part: Draw the V-shaped graph of $y = |x|$. Place a closed circle at $(-3, 3)$ and an open circle at $(3, 3)$. Connect these through the origin $(0,0)$.
* Right Part: Draw a horizontal line at $y = 5$. Place a closed circle at $(3, 5)$. Draw a ray extending to the right.