Vertical and horizontal asymptotes
Problem 1.275 · medium
Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{3 x^{3} - 3 x^{2} + 3 x - 3}{x^{2} - 5 x + 6} \).
- \[ x^{2} - 5 x + 6 = \left(x - 3\right) \left(x - 2\right) \]Factor the denominator.✓ Proved
- The numerator is not zero at x = 2 or x = 3, 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 \( x = 2,\ x = 3;\ \text{no horizontal asymptote} \)
✓ 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 | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 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: passqwen3.6:27b-mlx: pass — The solution correctly identifies the vertical asymptotes by checking that the numerator is non-zero at the roots of the denominator. It also correctly applies the degree comparison rule to conclude there is no horizontal asymptote.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the vertical asymptotes by checking that the numerator is non-zero at the roots of the denominator. It also correctly applies the degree comparison rule to conclude there is no horizontal asymptote.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — 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.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.