Piecewise function worksheet with six problems requiring graphing on coordinate grids.

Let’s solve each piecewise function step by step. We’ll graph them one at a time, making sure to follow the rules for each part of the function.

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Problem 1:
f(x) =
{
-3x - 11 if x < -3
5 if -3 ≤ x ≤ 2
x + 1 if x > 2
}

Step 1: For x < -3 → use f(x) = -3x - 11
Pick x = -4: f(-4) = -3*(-4) - 11 = 12 - 11 = 1 → point (-4, 1)
Pick x = -5: f(-5) = -3*(-5) - 11 = 15 - 11 = 4 → point (-5, 4)
Draw a line with slope -3 starting from left, but stop BEFORE x = -3 (open circle at x=-3).

At x = -3: since it's not included in this piece, we don’t plot here yet — wait for next piece.

Step 2: For -3 ≤ x ≤ 2 → f(x) = 5 (horizontal line)
So from x = -3 to x = 2, y = 5.
Plot points: (-3, 5), (0, 5), (2, 5) → solid dots at both ends because inequalities include equality.

Step 3: For x > 2 → f(x) = x + 1
Pick x = 3: f(3) = 3 + 1 = 4 → point (3, 4)
Pick x = 4: f(4) = 5 → point (4, 5)
Start AFTER x=2 (open circle at x=2), draw line with slope 1.

Check continuity? At x=2: middle piece says f(2)=5; right piece at x=2 would be 3, but x>2 so not used. So jump at x=2.

Graph summary:
- Left: line going up to left, ending open at (-3, ?). Let’s compute limit as x→-3⁻: -3*(-3)-11 = 9-11 = -2 → so open circle at (-3, -2)
- Middle: horizontal line from (-3,5) to (2,5) — solid dots
- Right: line starting open at (2,3) [since 2+1=3] and going up-right

Wait — correction: when x approaches 2 from right, f(x)=x+1 → at x=2+, value is 3. But middle piece at x=2 is 5. So yes, jump.

But let me double-check calculation for left piece at x=-3:

f(x) = -3x -11 at x=-3: -3*(-3) -11 = 9 -11 = -2 → correct.

So graph has:
- Open circle at (-3, -2)
- Solid dot at (-3, 5) ← start of middle piece
- Solid dot at (2, 5) ← end of middle piece
- Open circle at (2, 3) ← start of right piece
- Then line through (3,4), (4,5)...

Okay, that’s Problem 1.

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Problem 2:
f(x) =
{
2x + 6 if x ≤ -1
x² if -1 < x ≤ 2
-3 if x > 2
}

Step 1: x ≤ -1 → f(x) = 2x + 6
At x = -1: 2*(-1)+6 = 4 → solid dot at (-1, 4)
At x = -2: 2*(-2)+6 = 2 → point (-2, 2)
At x = -3: 0 → point (-3, 0)
Draw line with slope 2, ending at (-1,4) with solid dot.

Step 2: -1 < x ≤ 2 → f(x) = x²
This is parabola. Start AFTER x=-1 (so open circle at x=-1), go to x=2 (solid dot).

At x=-1: not included → compute limit: (-1)^2 = 1 → open circle at (-1, 1)
At x=0: 0 → (0,0)
At x=1: 1 → (1,1)
At x=2: 4 → solid dot at (2,4)

Note: at x=-1, left piece gives 4, this piece gives 1 → big jump!

Step 3: x > 2 → f(x) = -3 (horizontal line)
Start after x=2 → open circle at (2, -3), then flat line to right.

At x=3: y=-3, etc.

Graph summary:
- Line from left to (-1,4) solid
- Parabola from just after (-1,1) to (2,4) solid
- Horizontal line at y=-3 starting open at (2,-3)

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Problem 3:
g(x) =
{
4 if x ≤ -3
(1/3)x if -3 < x < 3
-(x-3)² + 3 if x ≥ 3
}

Step 1: x ≤ -3 → g(x) = 4 → horizontal line
Solid dot at (-3,4), extend left.

Step 2: -3 < x < 3 → g(x) = (1/3)x → straight line through origin, slope 1/3
Open circles at x=-3 and x=3.

At x=-3: (1/3)*(-3) = -1 → open circle at (-3, -1)
At x=0: 0 → (0,0)
At x=3: (1/3)*3 = 1 → open circle at (3,1)

Step 3: x ≥ 3 → g(x) = -(x-3)² + 3 → downward parabola vertex at (3,3)
Solid dot at (3,3)
At x=4: -(1)² +3 = 2 → (4,2)
At x=5: -(2)² +3 = -1 → (5,-1)
At x=6: -(3)² +3 = -6 → (6,-6)

Note: at x=3, middle piece says open at (3,1), this piece says solid at (3,3) → jump.

Graph:
- Flat line at y=4 until x=-3 (solid)
- Line from (-3,-1) open to (3,1) open
- Parabola starting at (3,3) solid, curving down

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Problem 4:
f(x) =
{
x² + 6x + 7 if x < -1
-x + 3 if -1 ≤ x < 3
5 if x ≥ 3
}

Step 1: x < -1 → f(x) = x² + 6x + 7
Complete square or plug values.

Vertex at x = -b/(2a) = -6/2 = -3
f(-3) = 9 -18 +7 = -2 → vertex (-3,-2)
At x=-1: 1 -6 +7 = 2 → but x<-1, so open circle at (-1,2)
At x=-2: 4 -12 +7 = -1 → (-2,-1)
At x=-4: 16 -24 +7 = -1 → (-4,-1)

Parabola opening up, vertex (-3,-2), open at (-1,2)

Step 2: -1 ≤ x < 3 → f(x) = -x + 3
Line with slope -1
At x=-1: -(-1)+3 = 4 → solid dot at (-1,4)
At x=0: 3 → (0,3)
At x=2: 1 → (2,1)
At x=3: 0 → but x<3, so open circle at (3,0)

Step 3: x ≥ 3 → f(x) = 5 → horizontal line
Solid dot at (3,5), extend right.

Jumps:
- At x=-1: left piece → 2 (open), this piece → 4 (solid) → jump up
- At x=3: middle piece → 0 (open), this piece → 5 (solid) → big jump

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Problem 5:
f(x) =
{
2x + 9 if x ≤ -2
x² - 1 if -2 < x ≤ 1
-4 if x > 1
}

Step 1: x ≤ -2 → f(x) = 2x + 9
At x=-2: 2*(-2)+9 = 5 → solid dot at (-2,5)
At x=-3: 2*(-3)+9 = 3 → (-3,3)
At x=-4: 1 → (-4,1)

Step 2: -2 < x ≤ 1 → f(x) = x² - 1
Open at x=-2: (-2)^2 -1 = 4-1=3 → open circle at (-2,3)
At x=0: -1 → (0,-1)
At x=1: 1-1=0 → solid dot at (1,0)

Step 3: x > 1 → f(x) = -4
Open circle at (1,-4), then flat line.

Jumps:
- At x=-2: left piece → 5 (solid), this piece → 3 (open) → jump down
- At x=1: middle → 0 (solid), right → -4 (open) → jump down

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Problem 6:
f(x) =
{
-2x - 11 if x < -4
x² + 4x + 2 if -4 ≤ x < 0
-3 if x ≥ 0
}

Step 1: x < -4 → f(x) = -2x -11
At x=-4: -2*(-4)-11 = 8-11 = -3 → but x<-4, so open circle at (-4,-3)
At x=-5: -2*(-5)-11 = 10-11 = -1 → (-5,-1)
At x=-6: 12-11=1 → (-6,1)

Step 2: -4 ≤ x < 0 → f(x) = x² + 4x + 2
Complete square: (x+2)^2 - 4 + 2 = (x+2)^2 - 2 → vertex at (-2,-2)
At x=-4: 16 -16 +2 = 2 → solid dot at (-4,2)
At x=-2: 4 -8 +2 = -2 → (-2,-2)
At x=0: 0+0+2=2 → but x<0, so open circle at (0,2)

Step 3: x ≥ 0 → f(x) = -3
Solid dot at (0,-3), flat line right.

Jumps:
- At x=-4: left → -3 (open), this → 2 (solid) → jump up
- At x=0: middle → 2 (open), this → -3 (solid) → jump down

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Now, since the user didn't ask for graphs drawn, but to "solve" — and given the context, I believe they want the key points and behavior described, or perhaps verification of understanding.

But looking back at the original request: “Solve the problem accurately.” And the image shows blank graphs — likely expecting the student to graph them.

However, since I can’t draw, and the instruction says “provide only the final correct answer” after reasoning, and no specific question was asked beyond “solve”, I think the best approach is to summarize the critical points for each function that define the graph — especially the transition points and whether they are open/closed.

Alternatively, maybe the task is to evaluate or describe — but without explicit questions, I'll assume the goal is to correctly identify the pieces and their domains, and perhaps compute some key values.

But to match the format, and since all options are false, I must provide ONLY the Final Answer.

Given that, and since there are six functions, perhaps the expected answer is to list the graphs’ key features or confirm correctness.

Wait — re-reading the user message: “You are an educational assistant helping a student solve homework problems.” and “Solve the problem accurately.”

The “problem” is the entire worksheet — which is to graph these piecewise functions.

Since I cannot produce images, and the instructions say to avoid describing the image, I think the intended output is to provide the correct graphical descriptions or key coordinates.

But the Final Answer section should contain only the final correct answer — which for graphing problems might be the set of plotted points or the description.

However, to comply strictly, and since no specific question was posed (like “what is f(-3)?”), I will instead provide a concise summary of each function’s graph structure as the “answer”.

But let me check if any of the functions have errors or need correction — actually, all seem correctly defined.

Perhaps the student is supposed to graph them, and my role is to verify or guide — but again, per rules, I must give Final Answer after reasoning.

I think the safest Final Answer is to state that each piecewise function has been analyzed and its graph can be constructed using the provided pieces with attention to domain boundaries and open/closed circles.

But that’s vague.

Alternative: since the worksheet is titled “More Piecewise Functions!” and has blank grids, the “solution” is the completed graphs. Since I can’t draw, I’ll list for each function the critical points with their status (open/closed) and the shape.

Let me do that concisely.

Final Answer:

For each piecewise function, the graph consists of the following segments with specified endpoints:

1. f(x):
- Line y = -3x - 11 for x < -3: open circle at (-3, -2)
- Horizontal line y = 5 for -3 ≤ x ≤ 2: solid dots at (-3,5) and (2,5)
- Line y = x + 1 for x > 2: open circle at (2,3)

2. f(x):
- Line y = 2x + 6 for x ≤ -1: solid dot at (-1,4)
- Parabola y = x² for -1 < x ≤ 2: open circle at (-1,1), solid dot at (2,4)
- Horizontal line y = -3 for x > 2: open circle at (2,-3)

3. g(x):
- Horizontal line y = 4 for x ≤ -3: solid dot at (-3,4)
- Line y = (1/3)x for -3 < x < 3: open circles at (-3,-1) and (3,1)
- Parabola y = -(x-3)² + 3 for x ≥ 3: solid dot at (3,3)

4. f(x):
- Parabola y = x² + 6x + 7 for x < -1: open circle at (-1,2)
- Line y = -x + 3 for -1 ≤ x < 3: solid dot at (-1,4), open circle at (3,0)
- Horizontal line y = 5 for x ≥ 3: solid dot at (3,5)

5. f(x):
- Line y = 2x + 9 for x ≤ -2: solid dot at (-2,5)
- Parabola y = x² - 1 for -2 < x ≤ 1: open circle at (-2,3), solid dot at (1,0)
- Horizontal line y = -4 for x > 1: open circle at (1,-4)

6. f(x):
- Line y = -2x - 11 for x < -4: open circle at (-4,-3)
- Parabola y = x² + 4x + 2 for -4 ≤ x < 0: solid dot at (-4,2), open circle at (0,2)
- Horizontal line y = -3 for x ≥ 0: solid dot at (0,-3)

These points and shapes define the complete graphs for each piecewise function.