This worksheet provides practice problems for graphing piecewise-defined functions on a coordinate plane.

Let’s solve each piecewise function step by step.

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Problem 1:

We are given:

\[
f(x) =
\begin{cases}
x - 3 & \text{if } x \leq -3 \\
3x + 2 & \text{if } -3 < x \leq 2 \\
2 & \text{if } x > 2
\end{cases}
\]

We need to graph this in three parts.

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Part 1: \( f(x) = x - 3 \) for \( x \leq -3 \)

This is a line with slope 1, y-intercept -3.

But we only draw it for \( x \leq -3 \).

Find key points:

- At \( x = -3 \): \( f(-3) = -3 - 3 = -6 \) → point (-3, -6)
- Since it's \( x \leq -3 \), include the endpoint at (-3, -6) — use a closed circle.
- Pick another point left of -3, say \( x = -4 \): \( f(-4) = -4 - 3 = -7 \) → (-4, -7)
- Draw a ray starting at (-3, -6) going down-left through (-4, -7), etc.

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Part 2: \( f(x) = 3x + 2 \) for \( -3 < x \leq 2 \)

Slope = 3, y-intercept = 2.

Domain: from just after -3 up to and including 2.

Check endpoints:

- At \( x = -3 \): NOT included (since it’s > -3). But let’s compute value anyway to know where to start:
\( f(-3) = 3*(-3) + 2 = -9 + 2 = -7 \) → so open circle at (-3, -7)
- At \( x = 2 \): INCLUDED → \( f(2) = 3*2 + 2 = 6 + 2 = 8 \) → closed circle at (2, 8)

Pick one more point in between, say \( x = 0 \): \( f(0) = 0 + 2 = 2 \) → (0, 2)

Draw line segment from open circle at (-3, -7) to closed circle at (2, 8), passing through (0, 2).

Note: There’s a jump at x = -3! The first part ends at (-3, -6), second starts near (-3, -7). That’s okay — piecewise functions can have jumps.

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Part 3: \( f(x) = 2 \) for \( x > 2 \)

Horizontal line at y = 2, but only for x > 2.

At x = 2: NOT included → open circle at (2, 2)

Then draw horizontal line to the right from (2, 2) — never touching x=2.

Wait — hold on! At x=2, the middle piece gives us y=8 (closed circle), and now this piece says for x>2, y=2. So at x=2, we already have a point at (2,8), and then immediately after x=2, it drops to y=2.

So:

- Open circle at (2, 2)
- Horizontal line to the right at y=2

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✔ Summary for Problem 1 Graph:

- Left of x=-3: line y=x-3 ending at (-3,-6) [closed]
- Between x=-3 and x=2: line y=3x+2 from open (-3,-7) to closed (2,8)
- Right of x=2: horizontal line y=2 starting with open circle at (2,2)

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Problem 2:

Given:

\[
f(x) =
\begin{cases}
3x - 2 & \text{if } x \leq -3 \\
-3 & \text{if } -3 < x \leq 2 \\
-3x - 2 & \text{if } x > 2
\end{cases}
\]

Again, three pieces.

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Part 1: \( f(x) = 3x - 2 \) for \( x \leq -3 \)

Line with slope 3, y-intercept -2.

Only for x ≤ -3.

At x = -3: \( f(-3) = 3*(-3) - 2 = -9 - 2 = -11 \) → closed circle at (-3, -11)

Another point: x = -4 → \( f(-4) = 3*(-4) - 2 = -12 - 2 = -14 \) → (-4, -14)

Draw ray from (-3, -11) going down-left through (-4, -14)

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Part 2: \( f(x) = -3 \) for \( -3 < x \leq 2 \)

Horizontal line at y = -3.

From just after x=-3 to x=2 (inclusive).

At x = -3: NOT included → open circle at (-3, -3)

At x = 2: INCLUDED → closed circle at (2, -3)

Draw horizontal line segment from open (-3, -3) to closed (2, -3)

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Part 3: \( f(x) = -3x - 2 \) for \( x > 2 \)

Line with slope -3, y-intercept -2.

Only for x > 2.

At x = 2: NOT included → compute value: \( f(2) = -3*(2) - 2 = -6 - 2 = -8 \) → open circle at (2, -8)

Another point: x = 3 → \( f(3) = -9 - 2 = -11 \) → (3, -11)

Draw ray starting at open (2, -8) going down-right through (3, -11)

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✔ Summary for Problem 2 Graph:

- Left of x=-3: line y=3x-2 ending at (-3,-11) [closed]
- Between x=-3 and x=2: horizontal line y=-3 from open (-3,-3) to closed (2,-3)
- Right of x=2: line y=-3x-2 starting with open circle at (2,-8) going down-right

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Now, since the user did not ask for explanation or teaching or practice, and only wants the final answer — but note: the task is to “graph” the functions. Since we can’t draw here, we must describe the graphs clearly as the final answer? Wait — re-read instructions.

The user said: “Solve the problem accurately.” And the problem is to graph. But in text-based response, we can't produce actual images. However, looking back at the rules:

> Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]

But if ExplainAnswer is false, we skip explanation. In this case, ExplainAnswer is false, TeachTopic is false, GeneratePractice is false.

So we must provide ONLY the final answer — but what is the "final answer" for a graphing problem?

Actually, perhaps the expected output is to describe the key points and behavior so that someone could plot it — OR maybe since it’s a worksheet, the “answer” is the completed graph. But we can’t upload images.

Wait — rereading the initial instruction:

> You are an educational assistant helping a student solve homework problems.

And the user uploaded an image containing a task — which is to graph two piecewise functions.

Since we cannot draw, and the options are set to false for explanation/teach/practice, perhaps the best we can do is give a textual description of how to graph them — but the rule says: “If an option is false, completely omit that section.”

And “Final Answer: [Provide only the final correct answer]”

Hmm. Maybe for graphing problems, the “final answer” is the set of key coordinates and behaviors.

Alternatively, perhaps the system expects us to realize that without drawing capability, we should still provide the critical information needed to construct the graph — even if briefly.

But the strict rule is: if ExplainAnswer is false, omit explanation. So no step-by-step.

However, the problem is inherently graphical. Let me check similar cases.

Perhaps the intended interpretation is: since the student needs to graph, and we’re solving it, the “answer” is the description of the graph components.

But to comply strictly:

I think the safest path is to provide, under Final Answer, a concise summary of each piece’s domain, formula, and key points — enough to plot.

But let’s see the exact wording: “Solve the problem accurately.” The problem is to graph. So solving means providing the graph. Since we can’t, we describe it minimally.

Alternatively, maybe the platform expects LaTeX or something — but unlikely.

Another thought: perhaps for such worksheets, the “answer” is considered to be the plotted graph, and since we can’t do that, we state that — but that violates “solve accurately”.

I think the best compromise is to list the key features for each piece in bullet form under Final Answer, as that constitutes the solution.

Let me try:

──────────────────────────────────────

Final Answer:

For Problem 1:

- For \( x \leq -3 \): Line \( y = x - 3 \); plot closed circle at (-3, -6), and go through (-4, -7), etc.
- For \( -3 < x \leq 2 \): Line \( y = 3x + 2 \); open circle at (-3, -7), closed circle at (2, 8), pass through (0, 2).
- For \( x > 2 \): Horizontal line \( y = 2 \); open circle at (2, 2), extend right.

For Problem 2:

- For \( x \leq -3 \): Line \( y = 3x - 2 \); closed circle at (-3, -11), go through (-4, -14).
- For \( -3 < x \leq 2 \): Horizontal line \( y = -3 \); open circle at (-3, -3), closed circle at (2, -3).
- For \( x > 2 \): Line \( y = -3x - 2 \); open circle at (2, -8), go through (3, -11).

──────────────────────────────────────

But wait — the rule says “only the final correct answer”, and this is descriptive. Is there a better way?

Perhaps the expected answer is simply to recognize that the graphs consist of those segments, and since it’s a worksheet, the student would draw based on this.

I think this is acceptable.

To make it even more minimal:

Final Answer:

Problem 1 Graph:
- Left: y = x - 3, x ≤ -3 → ends at (-3, -6) ●
- Middle: y = 3x + 2, -3 < x ≤ 2 → from (-3, -7) ○ to (2, 8) ●
- Right: y = 2, x > 2 → starts at (2, 2) ○, horizontal right

Problem 2 Graph:
- Left: y = 3x - 2, x ≤ -3 → ends at (-3, -11) ●
- Middle: y = -3, -3 < x ≤ 2 → from (-3, -3) ○ to (2, -3) ●
- Right: y = -3x - 2, x > 2 → starts at (2, -8) ○, goes down-right

Yes, this is concise and contains all necessary info to graph.

I’ll go with that.