Vertical and horizontal asymptotes
Problem 1.187 · medium
Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{4 x^{3} - 4 x^{2} + 4 x - 4}{x^{2} - x - 2} \).
- \[ x^{2} - x - 2 = \left(x - 2\right) \left(x + 1\right) \]Factor the denominator.✓ Proved
- The numerator is not zero at x = -1 or x = 2, so both are vertical asymptotes.
- The numerator has higher degree than the denominator, so f(x) grows without bound as x → ±∞: there is no horizontal asymptote.
Answer \( x = -1,\ x = 2;\ \text{no horizontal asymptote} \)
Lines: 1 proved, 2 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | Not checked | — | 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: 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 — 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 2026-10-04 — The solution correctly identifies the vertical asymptotes by checking that the numerator is non-zero at the roots of the denominator. It also correctly applies the degree comparison rule to conclude there is no horizontal asymptote.gpt-oss:20b: fail (misleading) 2026-10-04 — 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 2026-10-04 — The solution correctly identifies the vertical asymptotes by checking that the numerator is non-zero at the roots of the denominator. It also correctly applies the degree comparison rule to conclude there is no horizontal asymptote.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.