Vertical and horizontal asymptotes
Problem 1.185 · medium
Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{- 4 x^{3} + 12 x^{2} - 4 x + 12}{x^{2} + 2 x - 8} \).
- \[ x^{2} + 2 x - 8 = \left(x - 2\right) \left(x + 4\right) \]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 | 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 concludes there is no horizontal asymptote because the degree of the numerator exceeds the degree of the denominator.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — 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 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the vertical asymptotes by checking that the denominator is zero and the numerator is non-zero at those points. It also correctly concludes there is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.