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Vertical and horizontal asymptotes

Problem 1.585 · medium

Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{2 x^{3} - 8 x^{2} + 2 x - 8}{x^{2} + 2 x - 3} \).

\[ x^{2} + 2 x - 3 = \left(x - 1\right) \left(x + 3\right) \]
Factor the denominator.✓ Proved
The numerator is not zero at x = -3 or x = 1, so both are vertical asymptotes.
Reviewed
The numerator has higher degree than the denominator, so f(x) grows without bound as x → ±∞: there is no horizontal asymptote.
Reviewed

Answer \( x = -3,\ x = 1;\ \text{no horizontal asymptote} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	claude-sonnet-5-5	claude-sonnet-5-5: N(1) = 2-8+2-8 = -12 and N(-3) = -54-72-6-8 = -140, both nonzero. The denominator vanishes at both points, so there is no cancellation and both are vertical asymptotes.
3	Reviewed	claude-sonnet-5-5	claude-sonnet-5-5: Degree 3 over degree 2 means f ~ 2x as x → ±∞, which is unbounded, so there is no horizontal asymptote. A slant asymptote exists but is not asked for and does not contradict the statement.
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	f probed just beside each vertical asymptote and at x = ±1e9

Reviewers

gpt-oss:20b: fail (misleading) — The solution incorrectly states that there is no horizontal asymptote. Since the numerator’s degree (3) exceeds the denominator’s degree (2), the function has an oblique (slant) asymptote, not a horizontal one.
qwen3.6:27b-mlx: pass

Senior review claude-sonnet-5-5, 2026-10-13: pass — The numerator equals 2(x^2+1)(x-4), which is -12 at x=1 and -140 at x=-3. It is nonzero at both roots of the denominator, so both are vertical asymptotes. The degree comparison correctly rules out a horizontal asymptote.

gpt-oss:20b: dismiss — A slant asymptote does not contradict the claim: the degree of the numerator exceeds that of the denominator, so f is unbounded and has no horizontal asymptote, which is exactly what the solution says, and the question asks only for vertical and horizontal asymptotes.

Every verdict on record (5)

gpt-oss:20b: fail (misleading) 2026-10-13 — The solution incorrectly states that there is no horizontal asymptote. Since the numerator’s degree (3) exceeds the denominator’s degree (2), the function has an oblique (slant) asymptote, not a horizontal one.
qwen3.6:27b-mlx: pass 2026-10-13
claude-sonnet-5-5: pass 2026-10-13 — The numerator equals 2(x^2+1)(x-4), which is -12 at x=1 and -140 at x=-3. It is nonzero at both roots of the denominator, so both are vertical asymptotes. The degree comparison correctly rules out a horizontal asymptote.
gpt-oss:20b: fail (misleading) 2026-10-13 — The solution incorrectly states that there is no horizontal asymptote. Since the numerator’s degree (3) exceeds the denominator’s degree (2), the function has an oblique (slant) asymptote, not a horizontal one. The conclusion about horizontal asymptotes is false and would mislead a student.
qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the vertical asymptotes by checking that the numerator is non-zero at the roots of the denominator. It also correctly concludes there is no horizontal asymptote because the degree of the numerator exceeds the degree of the denominator.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/asymptotes, checked 2026-10-13 with SymPy 1.14.0.