Grade 11 Advanced Math worksheet on graphing rational functions, focusing on domain and asymptotes.

Let's solve each rational function step by step to find:

1. Domain — all real numbers except where the denominator is zero.
2. Vertical asymptotes — occur at values of $ x $ that make the denominator zero (but not canceled by the numerator).
3. Horizontal asymptotes — determined by comparing degrees of numerator and denominator.

We'll go through each function and match it with the correct answer choice (a–f).

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1. $ f(x) = \dfrac{x^2 - 2}{x^2 - 4} $

- Factor:
- Numerator: $ x^2 - 2 $ → doesn't factor nicely over integers.
- Denominator: $ x^2 - 4 = (x - 2)(x + 2) $
- Domain: Exclude values that make denominator zero:
- $ x \neq 2, -2 $
- So, $ D = \{x \mid x \neq 2, -2, x \in \mathbb{R}\} $
- Vertical asymptotes: $ x = 2 $, $ x = -2 $ (since neither cancels)
- Horizontal asymptote: Degrees are both 2 → ratio of leading coefficients: $ \frac{1}{1} = 1 $
- So, $ y = 1 $

✔ Answer: b.
$ D = \{x \mid x \neq 2, -2, x \in \mathbb{R}\};\quad x = 2, x = -2, y = 1 $

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2. $ h(x) = \dfrac{x^3 - 8}{x + 4} $

- Factor numerator: $ x^3 - 8 = (x - 2)(x^2 + 2x + 4) $
- Denominator: $ x + 4 $
- Domain: $ x \neq -4 $
- So, $ D = \{x \mid x \neq -4, x \in \mathbb{R}\} $
- Vertical asymptote: $ x = -4 $ (no cancellation)
- Horizontal asymptote: Degree of numerator (3) > degree of denominator (1) → No horizontal asymptote, but oblique (slant). But question only asks for vertical or horizontal → so no horizontal asymptote.

But look at choices: none say "no horizontal" directly. Let's check if any option matches domain $ x \ne -4 $ and vertical asymptote $ x = -4 $. Only c says that.

Wait — c says:
> c. $ D = \{x \mid x \neq -4, x \in \mathbb{R}\}; x = -4 $

That’s exactly what we have here — but no horizontal asymptote mentioned.

But c does not mention a horizontal asymptote, which is fine — if there is none, it's okay to omit it.

So this seems to be a candidate.

But let's double-check: Is there a horizontal asymptote?

- Degree numerator: 3
- Degree denominator: 1
- Since num degree > den degree → no horizontal asymptote, but slant asymptote exists.

So only vertical asymptote: $ x = -4 $

✔ Answer: c.
$ D = \{x \mid x \neq -4, x \in \mathbb{R}\}; x = -4 $

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3. $ f(x) = \dfrac{x(x - 1)(x + 2)^2}{(x + 3)(x - 4)} $

- Domain: Exclude values making denominator zero:
- $ x \neq -3, 4 $
- So, $ D = \{x \mid x \neq -3, 4, x \in \mathbb{R}\} $
- Vertical asymptotes: $ x = -3 $, $ x = 4 $ (no cancellation)
- Horizontal asymptote: Compare degrees
- Numerator: degree 4 (three factors: $ x, (x-1), (x+2)^2 $ → total degree 4)
- Denominator: degree 2
- Degree numerator > denominator → no horizontal asymptote

Now look at options: None of the options list both $ x = -3 $, $ x = 4 $, and no horizontal asymptote.

But wait — option a says:
> a. $ D = \{x \mid x \neq 4, -3, x \in \mathbb{R}\}; x = 4, x = -3 $

This matches our domain and vertical asymptotes — but does not mention a horizontal asymptote, which is acceptable since there isn’t one.

So a is correct.

✔ Answer: a.

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4. $ g(x) = \dfrac{x - 6}{(x + 3)(x + 5)} $

- Domain: $ x \neq -3, -5 $
- So $ D = \{x \mid x \neq -3, -5, x \in \mathbb{R}\} $
- Vertical asymptotes: $ x = -3 $, $ x = -5 $
- Horizontal asymptote: Num degree = 1, Den degree = 2 → as $ x \to \infty $, function → 0
- So $ y = 0 $

✔ Answer: d.
$ D = \{x \mid x \neq -3, -5, x \in \mathbb{R}\}; x = -3, x = -5, y = 0 $

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5. $ h(x) = \dfrac{(x - 1)(x + 1)}{(x - 2)^2(x + 4)^2} $

- Note: $ (x - 1)(x + 1) = x^2 - 1 $
- Domain: $ x \neq 2, -4 $
- So $ D = \{x \mid x \neq 2, -4, x \in \mathbb{R}\} $
- Vertical asymptotes: $ x = 2 $, $ x = -4 $ (both squared, but still asymptotes unless canceled — not canceled)
- Horizontal asymptote: Num degree = 2, Den degree = 4 → $ y = 0 $

So: $ y = 0 $

✔ Answer: f.
$ D = \{x \mid x \neq 2, -4, x \in \mathbb{R}\}; x = 2, x = -4, y = 0 $

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6. $ g(x) = \dfrac{(x - 4)(x + 2)}{(x + 1)(x - 3)} $

- Domain: $ x \neq -1, 3 $
- So $ D = \{x \mid x \neq -1, 3, x \in \mathbb{R}\} $
- Vertical asymptotes: $ x = -1 $, $ x = 3 $
- Horizontal asymptote: Both numerator and denominator degree 2 → ratio of leading coefficients: $ \frac{1}{1} = 1 $
- So $ y = 1 $

✔ Answer: e.
$ D = \{x \mid x \neq 3, -1, x \in \mathbb{R}\}; x = 3, x = -1, y = 1 $

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✔ Final Matching:

| Function | Answer |
|--------|--------|
| 1. $ f(x) = \dfrac{x^2 - 2}{x^2 - 4} $ | b |
| 2. $ h(x) = \dfrac{x^3 - 8}{x + 4} $ | c |
| 3. $ f(x) = \dfrac{x(x - 1)(x + 2)^2}{(x + 3)(x - 4)} $ | a |
| 4. $ g(x) = \dfrac{x - 6}{(x + 3)(x + 5)} $ | d |
| 5. $ h(x) = \dfrac{(x - 1)(x + 1)}{(x - 2)^2(x + 4)^2} $ | f |
| 6. $ g(x) = \dfrac{(x - 4)(x + 2)}{(x + 1)(x - 3)} $ | e |

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✔ Summary of Answers:

1. b
2. c
3. a
4. d
5. f
6. e

Let me know if you'd like a visual explanation or graphing tips!