Math worksheet focused on identifying asymptotes in rational functions.

Let's solve each problem step by step to find the vertical, horizontal, and oblique (slant) asymptotes for each rational function.

---

General Rules for Asymptotes:

1. Vertical Asymptotes:
- Occur where the denominator is zero (and numerator is not zero).
- Solve $ \text{denominator} = 0 $, then check if numerator is non-zero at those points.

2. Horizontal Asymptotes:
- Compare degrees of numerator and denominator:
- If degree of numerator < degree of denominator → $ y = 0 $
- If degrees are equal → $ y = \frac{\text{leading coefficient of num}}{\text{leading coefficient of den}} $
- If degree of numerator > degree of denominator → No horizontal asymptote (but possibly oblique)

3. Oblique Asymptotes:
- Only when degree of numerator is exactly one more than denominator.
- Found by polynomial long division; the quotient (ignoring remainder) gives the oblique asymptote.

---

Now let’s go through each problem:

---

1. $ f(x) = \frac{5x}{x^2 - 27} $

- Vertical Asymptotes: Set denominator = 0
$ x^2 - 27 = 0 \Rightarrow x = \pm\sqrt{27} = \pm3\sqrt{3} $
Numerator ≠ 0 at these points → Vertical asymptotes at $ x = 3\sqrt{3},\ x = -3\sqrt{3} $

- Horizontal Asymptote: Degree of numerator = 1, denominator = 2 → numerator < denominator → $ y = 0 $

- Oblique: No (degree difference not 1)

✔ Answer:
- Vertical: $ x = \pm3\sqrt{3} $
- Horizontal: $ y = 0 $
- Oblique: None

---

2. $ f(x) = \frac{5x - 5}{x - 5} $

- Simplify: $ f(x) = \frac{5(x - 1)}{x - 5} $

- Vertical Asymptote: $ x - 5 = 0 \Rightarrow x = 5 $
Numerator ≠ 0 here → Vertical asymptote at $ x = 5 $

- Horizontal Asymptote: Degrees both = 1 → ratio of leading coefficients: $ \frac{5}{1} = 5 $ → $ y = 5 $

- Oblique: No (degrees same)

✔ Answer:
- Vertical: $ x = 5 $
- Horizontal: $ y = 5 $
- Oblique: None

---

3. $ f(x) = \frac{x^2 + 4}{x^2 + 6x + 8} $

Factor denominator: $ x^2 + 6x + 8 = (x+2)(x+4) $

- Vertical Asymptotes: $ x = -2, x = -4 $ (numerator ≠ 0 at these points)

- Horizontal Asymptote: Degrees both = 2 → $ y = \frac{1}{1} = 1 $

- Oblique: No (same degree)

✔ Answer:
- Vertical: $ x = -2,\ x = -4 $
- Horizontal: $ y = 1 $
- Oblique: None

---

4. $ f(x) = \frac{x^2 + 1}{x^2 - 6x + 6} $

- Vertical Asymptotes: Solve $ x^2 - 6x + 6 = 0 $
Discriminant: $ 36 - 24 = 12 $ → roots: $ x = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2\sqrt{3}}{2} = 3 \pm \sqrt{3} $
So vertical asymptotes at $ x = 3 + \sqrt{3},\ x = 3 - \sqrt{3} $

- Horizontal Asymptote: Degrees both = 2 → $ y = \frac{1}{1} = 1 $

- Oblique: No

✔ Answer:
- Vertical: $ x = 3 \pm \sqrt{3} $
- Horizontal: $ y = 1 $
- Oblique: None

---

5. $ f(x) = \frac{3x^2 + 4}{x^2 + 2x} $

Factor denominator: $ x(x + 2) $

- Vertical Asymptotes: $ x = 0,\ x = -2 $ (check numerator: at $ x=0 $, numerator = 4 ≠ 0; at $ x=-2 $, numerator = $ 3(4)+4=16≠0 $) → both valid

- Horizontal Asymptote: Degrees both = 2 → $ y = \frac{3}{1} = 3 $

- Oblique: No

✔ Answer:
- Vertical: $ x = 0,\ x = -2 $
- Horizontal: $ y = 3 $
- Oblique: None

---

6. $ f(x) = \frac{x^3 - 1}{x^2 - 2} $

- Vertical Asymptotes: $ x^2 - 2 = 0 \Rightarrow x = \pm\sqrt{2} $
Check numerator: at $ x = \pm\sqrt{2} $, $ x^3 - 1 \neq 0 $ → yes, vertical asymptotes at $ x = \pm\sqrt{2} $

- Horizontal Asymptote: Degree numerator = 3, denominator = 2 → num > den → No horizontal asymptote

- Oblique Asymptote: Degree diff = 1 → YES
Perform polynomial division:
Divide $ x^3 - 1 $ by $ x^2 - 2 $

$$
\begin{array}{r|rr}
x^2 - 2 & x^3 + 0x^2 + 0x - 1 \\
\hline
& x \\
& x^3 - 2x \\
\hline
& 0 + 2x - 1 \\
\end{array}
$$

Quotient: $ x $, remainder $ 2x - 1 $ → so $ f(x) = x + \frac{2x - 1}{x^2 - 2} $

→ Oblique asymptote: $ y = x $

✔ Answer:
- Vertical: $ x = \pm\sqrt{2} $
- Horizontal: None
- Oblique: $ y = x $

---

7. $ f(x) = \frac{6x^2 + x + 12}{2x^2 - 5x - 2} $

- Vertical Asymptotes: Solve $ 2x^2 - 5x - 2 = 0 $
Discriminant: $ 25 + 16 = 41 $ → roots: $ x = \frac{5 \pm \sqrt{41}}{4} $

These are real and distinct → vertical asymptotes at $ x = \frac{5 \pm \sqrt{41}}{4} $

- Horizontal Asymptote: Degrees both = 2 → $ y = \frac{6}{2} = 3 $

- Oblique: No

✔ Answer:
- Vertical: $ x = \frac{5 \pm \sqrt{41}}{4} $
- Horizontal: $ y = 3 $
- Oblique: None

---

8. $ f(x) = \frac{x^5}{x^2 - 1} $

- Vertical Asymptotes: $ x^2 - 1 = 0 \Rightarrow x = \pm1 $

At $ x = \pm1 $, numerator $ x^5 = \pm1 \ne 0 $ → both are vertical asymptotes

- Horizontal Asymptote: Num degree = 5, den = 2 → num > den → None

- Oblique Asymptote? Degree difference = 3 → No oblique (only if difference is exactly 1)

So no oblique asymptote.

✔ Answer:
- Vertical: $ x = -1,\ x = 1 $
- Horizontal: None
- Oblique: None

> Note: Even though degree is higher, it's not exactly 1 more → no oblique.

---

9. $ f(x) = \frac{-2x^2 + 1}{2x^2 + 4x^2} $

Wait — denominator: $ 2x^2 + 4x^2 = 6x^2 $? That seems like a typo.

But looking closely: $ 2x^2 + 4x^2 $? That's just $ 6x^2 $. But likely meant to be $ 2x^2 + 4x $ or $ 2x^2 + 4x + 2 $? But as written:
$$
f(x) = \frac{-2x^2 + 1}{2x^2 + 4x^2} = \frac{-2x^2 + 1}{6x^2}
$$

So we proceed with that.

- Vertical Asymptotes: Denominator $ 6x^2 = 0 \Rightarrow x = 0 $
But numerator at $ x=0 $: $ -2(0)^2 + 1 = 1 \ne 0 $ → so vertical asymptote at $ x = 0 $

- Horizontal Asymptote: Degrees both = 2 → $ y = \frac{-2}{6} = -\frac{1}{3} $

- Oblique: No

✔ Answer:
- Vertical: $ x = 0 $
- Horizontal: $ y = -\frac{1}{3} $
- Oblique: None

> ⚠️ Possible typo in problem — perhaps meant $ 2x^2 + 4x $ or similar? But based on given expression, this is correct.

---

10. $ f(x) = \frac{6x^2 + 2}{x^4} $

- Vertical Asymptotes: $ x^4 = 0 \Rightarrow x = 0 $
Numerator at $ x=0 $: $ 6(0)^2 + 2 = 2 \ne 0 $ → vertical asymptote at $ x = 0 $

- Horizontal Asymptote: Num degree = 2, den = 4 → num < den → $ y = 0 $

- Oblique: No (degree difference not 1)

✔ Answer:
- Vertical: $ x = 0 $
- Horizontal: $ y = 0 $
- Oblique: None

---

## ✔ Final Answers Summary:

| Problem | Vertical Asymptotes | Horizontal Asymptote | Oblique Asymptote |
|--------|----------------------|------------------------|---------------------|
| 1 | $ x = \pm3\sqrt{3} $ | $ y = 0 $ | None |
| 2 | $ x = 5 $ | $ y = 5 $ | None |
| 3 | $ x = -2, -4 $ | $ y = 1 $ | None |
| 4 | $ x = 3 \pm \sqrt{3} $ | $ y = 1 $ | None |
| 5 | $ x = 0, -2 $ | $ y = 3 $ | None |
| 6 | $ x = \pm\sqrt{2} $ | None | $ y = x $ |
| 7 | $ x = \frac{5 \pm \sqrt{41}}{4} $ | $ y = 3 $ | None |
| 8 | $ x = \pm1 $ | None | None |
| 9 | $ x = 0 $ | $ y = -\frac{1}{3} $ | None |
| 10 | $ x = 0 $ | $ y = 0 $ | None |

---

Let me know if you'd like a graph sketch or explanation for any specific one!