Examples of horizontal and slanted asymptotes in rational functions with graphical representations.

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Since I can't view images directly, I’ll guide you through solving the types of problems shown in your image based on typical content found in such worksheets. Let's go through each type of asymptote and then walk through examples similar to what’s likely in your image.

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🔹 1. Horizontal Asymptotes

Rules for horizontal asymptotes depend on the degrees of the numerator and denominator:

Let:
- $ f(x) = \frac{P(x)}{Q(x)} $
- Degree of $ P(x) $ = $ n $
- Degree of $ Q(x) $ = $ m $

| Case | Horizontal Asymptote |
|------|------------------------|
| $ n < m $ | $ y = 0 $ |
| $ n = m $ | $ y = \frac{\text{leading coefficient of } P}{\text{leading coefficient of } Q} $ |
| $ n > m $ | No horizontal asymptote (but may have slant asymptote) |

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🔹 2. Vertical Asymptotes

- Occur where the denominator is zero and the numerator is not zero.
- Solve $ Q(x) = 0 $, then check if those values make numerator zero (if yes, it might be a hole, not a vertical asymptote).

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🔹 3. Slant (Oblique) Asymptotes

- Occur when degree of numerator is exactly one more than degree of denominator ($ n = m + 1 $).
- Found by polynomial long division: divide $ P(x) $ by $ Q(x) $, and the quotient (ignoring remainder) is the slant asymptote.

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Now let's analyze each problem as they appear in the image (based on common formats):

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## ✔ Problem 1: $ f(x) = \frac{x^2 - 4}{x - 2} $

Step 1: Simplify

Factor numerator:
$ x^2 - 4 = (x - 2)(x + 2) $

So,
$$
f(x) = \frac{(x - 2)(x + 2)}{x - 2}
$$

Cancel $ x - 2 $ (but note: hole at $ x = 2 $)
→ $ f(x) = x + 2 $, $ x \neq 2 $

Vertical Asymptote?

- Denominator zero at $ x = 2 $, but numerator also zero → not a vertical asymptote, it's a hole.
- So: No vertical asymptote (but a hole at $ x = 2 $)

Horizontal Asymptote?

- Degree of numerator: 2
- Degree of denominator: 1 → $ n > m $
- So no horizontal asymptote

Slant Asymptote?

- Since $ n = m + 1 $, yes
- Perform division: $ \frac{x^2 - 4}{x - 2} $
- Long division: $ x + 2 $ with remainder 0
- So slant asymptote: $ y = x + 2 $

Wait! But this function simplifies to $ x + 2 $, so the graph is a line with a hole at $ x = 2 $ — no asymptotes.

But since the original expression has a removable discontinuity, we say:
- No vertical asymptote
- No horizontal asymptote
- No slant asymptote (because the function reduces to a polynomial)

> 🚫 Conclusion: This function has no asymptotes, just a hole.

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## ✔ Problem 2: $ f(x) = \frac{2x^2 + 1}{x^2 - 1} $

Step 1: Vertical Asymptotes

Set denominator = 0:
$ x^2 - 1 = 0 $ → $ x = \pm 1 $

Check numerator at $ x = 1 $: $ 2(1)^2 + 1 = 3 \neq 0 $
At $ x = -1 $: $ 2(-1)^2 + 1 = 3 \neq 0 $

✔ So vertical asymptotes at $ x = 1 $ and $ x = -1 $

Horizontal Asymptote?

- Degrees: both 2 → $ n = m $
- Ratio of leading coefficients: $ \frac{2}{1} = 2 $
- So horizontal asymptote: $ y = 2 $

Slant Asymptote?

- $ n = m $ → no slant asymptote

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## ✔ Problem 3: $ f(x) = \frac{x^2 - 1}{x - 1} $

Step 1: Simplify

Numerator: $ x^2 - 1 = (x - 1)(x + 1) $

So,
$$
f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1, \quad x \neq 1
$$

Again, removable discontinuity at $ x = 1 $

- Vertical asymptote? ✘ (hole at $ x = 1 $)
- Horizontal asymptote? Degree num = 2, den = 1 → $ n > m $ → no horizontal
- Slant asymptote? Yes, since $ n = m + 1 $

Do long division:
$$
\frac{x^2 - 1}{x - 1} = x + 1 \quad \text{(remainder 0)}
$$

So slant asymptote: $ y = x + 1 $

But again, the function simplifies to a line with a hole → no asymptotes.

Wait — contradiction?

Actually, since the function simplifies to a line, there is no asymptote, even though the degree condition suggests a slant asymptote.

💡 Important: If the rational function simplifies to a polynomial, then there are no asymptotes, only possible holes.

So in this case:
- Hole at $ x = 1 $
- No asymptotes

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## ✔ Final Summary Table (Likely What You Need to Fill In)

| Problem | Function | Vertical Asymptotes | Horizontal Asymptote | Slant Asymptote | Graph Notes |
|--------|---------|---------------------|-----------------------|------------------|-------------|
| 1 | $ f(x) = \frac{x^2 - 4}{x - 2} $ | None (hole at $ x = 2 $) | None | None | Line $ y = x + 2 $ with hole at $ x = 2 $ |
| 2 | $ f(x) = \frac{2x^2 + 1}{x^2 - 1} $ | $ x = \pm 1 $ | $ y = 2 $ | None | Two vertical asymptotes, approaches $ y = 2 $ |
| 3 | $ f(x) = \frac{x^2 - 1}{x - 1} $ | None (hole at $ x = 1 $) | None | None | Line $ y = x + 1 $ with hole at $ x = 1 $ |

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## 📈 Graphing Tips

For each function:
- Draw asymptotes as dashed lines
- Plot key points
- Indicate holes with open circles
- For rational functions that simplify to polynomials, draw the polynomial but omit the point where undefined

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## ✔ Final Answer: How to Complete the Worksheet

For each box:

Example 1: $ f(x) = \frac{x^2 - 4}{x - 2} $

- Vertical Asymptote: None
- Horizontal Asymptote: None
- Slant Asymptote: None
- Graph: Draw $ y = x + 2 $, but leave a hole at $ x = 2 $

Example 2: $ f(x) = \frac{2x^2 + 1}{x^2 - 1} $

- Vertical Asymptotes: $ x = -1 $, $ x = 1 $
- Horizontal Asymptote: $ y = 2 $
- Slant Asymptote: None
- Graph: Curve approaching $ x = \pm1 $ vertically, and $ y = 2 $ horizontally

Example 3: $ f(x) = \frac{x^2 - 1}{x - 1} $

- Vertical Asymptote: None (hole)
- Horizontal Asymptote: None
- Slant Asymptote: None
- Graph: Line $ y = x + 1 $ with hole at $ x = 1 $

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## 🔍 Key Takeaway

Always factor first! Look for common factors — they indicate holes, not asymptotes.

Only irreducible denominators give asymptotes.

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If you can describe the exact expressions in your image or clarify which parts are unclear, I can tailor the explanation further!