Vertical and horizontal asymptotes
Problem 1.386 · medium
Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{- 4 x - 8}{x^{2} - 2 x - 3} \).
- \[ x^{2} - 2 x - 3 = \left(x - 3\right) \left(x + 1\right) \]Factor the denominator.✓ Proved
- The numerator is not zero at x = -1 or x = 3, so both are vertical asymptotes.Reviewed
- \[ \lim_{x \to \infty}\left(\frac{- 4 x - 8}{x^{2} - 2 x - 3}\right) = 0 \]Divide through by the highest power of x in the denominator.✓ Proved
- \[ \lim_{x \to -\infty}\left(\frac{- 4 x - 8}{x^{2} - 2 x - 3}\right) = 0 \]The same as x → −∞.✓ Proved
Answer \( x = -1,\ x = 3;\ y = 0 \)
✓ Nihil obstat Lines: 3 proved, 1 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 vertical asymptotes by checking that the denominator is zero while the numerator is non-zero, and correctly identifies the horizontal asymptote by evaluating limits at infinity.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies vertical asymptotes by checking that the denominator is zero while the numerator is non-zero, and correctly identifies the horizontal asymptote by evaluating limits at infinity.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly identifies the vertical asymptotes by checking that the denominator is zero while the numerator is non-zero. It also correctly identifies the horizontal asymptote by evaluating the limit at infinity, which is 0 because the degree of the denominator is greater than the degree of the numerator.
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 2026-10-07 with SymPy 1.14.0.