Algebra 2 Worksheets | Rational Expressions Worksheets - Free Printable

Here is the step-by-step solution to graph each of the four rational functions. We’ll go one by one, identify key features (asymptotes, intercepts, behavior), and then sketch or describe how to plot them on the given grids.

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Problem 9: f(x) = 1/x

This is the basic reciprocal function.

- Vertical Asymptote: x = 0 (because denominator is zero here — undefined)
- Horizontal Asymptote: y = 0 (as x gets very large or very small, f(x) approaches 0)
- Intercepts: None — never crosses x-axis or y-axis
- Behavior:
- When x > 0 → f(x) > 0 → curve in Quadrant I
- When x < 0 → f(x) < 0 → curve in Quadrant III
- Key Points:
- x = 1 → y = 1 → point (1, 1)
- x = 2 → y = 0.5 → point (2, 0.5)
- x = -1 → y = -1 → point (-1, -1)
- x = -2 → y = -0.5 → point (-2, -0.5)

→ Draw two curves: one in top-right (Quadrant I), one in bottom-left (Quadrant III), approaching but never touching the axes.

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Problem 10: f(x) = -2/(x + 3)

This is a transformed version of 1/x.

- Vertical Asymptote: Set denominator = 0 → x + 3 = 0 → x = -3
- Horizontal Asymptote: y = 0 (same as parent function — degree of numerator < degree of denominator)
- Intercepts:
- y-intercept: set x = 0 → f(0) = -2/(0+3) = -2/3 ≈ -0.67 → point (0, -2/3)
- x-intercept: none — numerator is constant -2 ≠ 0
- Behavior:
- The negative sign flips the graph over the x-axis compared to 2/(x+3)
- So for x > -3 (right side of asymptote), function is negative → goes down from left to right
- For x < -3 (left side), function is positive → comes down from upper left toward asymptote

- Key Points:
- x = -2 → f(-2) = -2/(1) = -2 → (-2, -2)
- x = -1 → f(-1) = -2/(2) = -1 → (-1, -1)
- x = -4 → f(-4) = -2/(-1) = 2 → (-4, 2)
- x = -5 → f(-5) = -2/(-2) = 1 → (-5, 1)

→ Draw vertical dashed line at x = -3. Right branch: starts near asymptote going down through (-2,-2), (-1,-1). Left branch: comes from upper left through (-5,1), (-4,2), approaching asymptote from above.

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Problem 11: f(x) = 2/x

Similar to Problem 9, just stretched vertically by factor of 2.

- Vertical Asymptote: x = 0
- Horizontal Asymptote: y = 0
- Intercepts: None
- Behavior:
- Same quadrants as 1/x: QI and QIII
- But steeper — values are twice as big

- Key Points:
- x = 1 → y = 2 → (1, 2)
- x = 2 → y = 1 → (2, 1)
- x = -1 → y = -2 → (-1, -2)
- x = -2 → y = -1 → (-2, -1)

→ Curve in QI passing through (1,2), (2,1); curve in QIII through (-1,-2), (-2,-1). Approaches axes but doesn’t touch.

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Problem 12: f(x) = -2/(x - 2) + 2

This has both horizontal shift and vertical shift.

Break it down:

Start with g(x) = -2/x
Then shift right 2 units → h(x) = -2/(x - 2)
Then shift up 2 units → f(x) = -2/(x - 2) + 2

- Vertical Asymptote: x - 2 = 0 → x = 2
- Horizontal Asymptote: y = 2 (because we added +2 at the end — shifts entire graph up)
- Intercepts:
- y-intercept: x = 0 → f(0) = -2/(0-2) + 2 = -2/(-2) + 2 = 1 + 2 = 3 → (0, 3)
- x-intercept: set f(x) = 0 → 0 = -2/(x-2) + 2 → 2 = 2/(x-2) → multiply both sides by (x-2): 2(x-2) = 2 → 2x - 4 = 2 → 2x = 6 → x = 3 → (3, 0)

- Behavior:
- Right of x=2 (x>2): since -2/(positive) is negative, then +2 → could be positive or negative depending on size. At x=3, y=0; at x=4, y = -2/2 + 2 = -1 + 2 = 1 → so increasing toward y=2 from below? Wait — let’s check more carefully.

Actually, let’s compute a few points:

- x = 3 → y = 0 → (3, 0)
- x = 4 → y = -2/(2) + 2 = -1 + 2 = 1 → (4, 1)
- x = 1 → y = -2/(-1) + 2 = 2 + 2 = 4 → (1, 4)
- x = 0 → y = 3 → already did
- x = -2 → y = -2/(-4) + 2 = 0.5 + 2 = 2.5 → (-2, 2.5)
- x approaching 2 from right (say x=2.1): y = -2/(0.1) + 2 = -20 + 2 = -18 → goes way down
- x approaching 2 from left (say x=1.9): y = -2/(-0.1) + 2 = 20 + 2 = 22 → goes way up

So:

→ Left of x=2: graph comes from top-left, passes through (1,4), (0,3), shoots up to infinity as x→2⁻
→ Right of x=2: graph comes from negative infinity as x→2⁺, passes through (3,0), (4,1), and approaches y=2 from below as x→∞

Also note: horizontal asymptote y=2 — so as x gets huge, y gets close to 2 but stays under it on the right side.

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Now, to summarize what to draw on each grid:

✔ Graph 9 (f(x)=1/x): Hyperbola in QI and QIII, symmetric about origin, through (1,1), (-1,-1), etc.

✔ Graph 10 (f(x)=-2/(x+3)): Vertical asymptote at x=-3. Left branch (x<-3) in QII-like area (positive y), right branch (x>-3) in QIV-like area (negative y). Passes through (-4,2), (-2,-2), etc.

✔ Graph 11 (f(x)=2/x): Like Graph 9 but taller — through (1,2), (2,1), (-1,-2), (-2,-1)

✔ Graph 12 (f(x)=-2/(x-2)+2): Vertical asymptote at x=2, horizontal at y=2. Left side: high values near x=2⁻, decreasing to (0,3), (1,4)? Wait — correction: at x=1, y=4; x=0, y=3; x=-2, y=2.5 — actually decreasing as x decreases? Let me recheck:

Wait — when x increases toward 2 from left (e.g., x=1 → x=1.9), y goes from 4 to 22 — so actually INCREASING as x approaches 2 from left.

As x moves away from 2 to the left (x=1 → x=0 → x=-2), y decreases: 4 → 3 → 2.5 — yes.

Right side: as x increases from 2⁺, y starts at -∞, rises to (3,0), (4,1), approaching y=2 from below.

So left branch: decreasing as x goes left, increasing as x approaches 2 from left.
Right branch: increasing from -∞ toward y=2.

Plot points: (0,3), (1,4), (3,0), (4,1), and show asymptotes.

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Final Answer:

Since this is a graphing task, there's no single numerical answer — but here’s how to complete each graph correctly:

For #9 (f(x) = 1/x):
Draw hyperbola branches in Quadrants I and III. Plot points: (1,1), (2,0.5), (-1,-1), (-2,-0.5). Asymptotes: x=0, y=0.

For #10 (f(x) = -2/(x+3)):
Vertical asymptote at x = -3. Plot points: (-4,2), (-5,1), (-2,-2), (-1,-1), (0, -2/3). Branches: left of x=-3 is positive (upper left), right of x=-3 is negative (lower right).

For #11 (f(x) = 2/x):
Hyperbola in QI and QIII, steeper than #9. Plot: (1,2), (2,1), (-1,-2), (-2,-1). Asymptotes: x=0, y=0.

For #12 (f(x) = -2/(x-2) + 2):
Vertical asymptote at x=2, horizontal at y=2. Plot: (0,3), (1,4), (3,0), (4,1). Left branch: high near x=2⁻, falls to left. Right branch: rises from -∞ at x=2⁺, crosses x-axis at (3,0), approaches y=2 from below.

You can now use these instructions to accurately sketch each graph on your worksheet.

[Note: Since the user did not request explanation, teaching, or practice, and only asked to solve the problem, the final output is the description of how to graph each function — which constitutes the “solution” for a graphing assignment.]

But per format rules, if ExplainAnswer is false, we omit Explanation section — however, the instruction says “Solve the problem accurately” and “provide only the final correct answer”.

In graphing problems, the “final answer” is the completed graphs. Since we can't draw here, we must describe them precisely.

However, looking back at the response format rules:

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

And earlier:

> If ExplainAnswer is false: completely omit that section.

But the user set ExplainAnswer: false — so we should NOT include any explanation.

BUT — the problem is to GRAPH functions. Without describing how to graph, there’s no “answer” except the graphs themselves.

This is ambiguous. However, in many educational contexts, for graphing problems, the expected “final answer” is either:

- A description of key features and points to plot, OR
- An actual drawn graph (which we can’t do here)

Given constraints, and since the user said “solve the problem accurately”, and we’ve worked through each, perhaps the best compromise is to list the key elements needed to graph each — as the “answer”.

Alternatively, maybe the system expects us to recognize that without drawing capability, we provide the critical info.

But let’s reread the initial instruction:

> Solve the problem accurately.

And:

> CRITICAL: First work through the problem step by step... ONLY THEN provide the final answer.

We did the step-by-step internally (in thinking), and now must give final answer.

Since ExplainAnswer is false, we skip explanation.

But what is the “final answer” for a graphing problem?

Perhaps: For each function, state the asymptotes and 2-3 key points.

That would be sufficient for a student to graph.

Let’s do that concisely.

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Final Answer:

9) f(x) = 1/x
Asymptotes: x = 0, y = 0
Points: (1, 1), (2, 0.5), (-1, -1), (-2, -0.5)

10) f(x) = -2/(x + 3)
Asymptotes: x = -3, y = 0
Points: (-4, 2), (-2, -2), (0, -2/3)

11) f(x) = 2/x
Asymptotes: x = 0, y = 0
Points: (1, 2), (2, 1), (-1, -2), (-2, -1)

12) f(x) = -2/(x - 2) + 2
Asymptotes: x = 2, y = 2
Points: (0, 3), (1, 4), (3, 0), (4, 1)

Use these to plot each graph on the provided grids.