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Vertical and horizontal asymptotes

Problem 1.586 · medium

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

\[ x^{2} + 2 x - 8 = \left(x - 2\right) \left(x + 4\right) \]
Factor the denominator.✓ Proved
The numerator is not zero at x = -4 or x = 2, 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 \( x = -4,\ x = 2;\ \text{no horizontal asymptote} \)

✓ 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 script, run by sympy 1.14.0	claude-sonnet-5-5 script, run by sympy 1.14.0: script failed: SyntaxError: invalid syntax (`checks/1.586/line2-111857c7.py`)
3	Reviewed	claude-sonnet-5-5 script, run by sympy 1.14.0	claude-sonnet-5-5 script, run by sympy 1.14.0: script failed: SyntaxError: invalid syntax (`checks/1.586/line3-cf15842e.py`)
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: pass
qwen3.6:27b-mlx: pass

Senior review claude-sonnet-5-5, 2026-10-13: pass — The numerator is 5 at x=2 and 119 at x=-4, so both are vertical asymptotes. The limits at ±∞ are -∞ and +∞, so there is no horizontal asymptote, and the stated answer is correct.

gpt-oss:20b: dismiss — A slant asymptote does not conflict with the claim: when the numerator's degree exceeds the denominator's by one, there is an oblique asymptote and no horizontal one, so 'no horizontal asymptote' is correct.

Every verdict on record (5)

gpt-oss:20b: pass 2026-10-13
qwen3.6:27b-mlx: pass 2026-10-13
claude-sonnet-5-5: pass 2026-10-13 — The numerator is 5 at x=2 and 119 at x=-4, so both are vertical asymptotes. The limits at ±∞ are -∞ and +∞, so there is no horizontal asymptote, and the stated answer is correct.
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 denominator is zero while the numerator is non-zero at those points. 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.