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Vertical and horizontal asymptotes

Problem 1.389 · medium

Find all vertical and horizontal asymptotes of \( \displaystyle f(x) = \frac{3 x^{2} - 21 x + 36}{x^{2} + 2 x - 3} \).

\[ x^{2} + 2 x - 3 = \left(x - 1\right) \left(x + 3\right) \]
Factor the denominator.✓ Proved
The numerator is not zero at x = -3 or x = 1, so both are vertical asymptotes.
Reviewed
\[ \lim_{x \to \infty}\left(\frac{3 x^{2} - 21 x + 36}{x^{2} + 2 x - 3}\right) = 3 \]
Divide through by the highest power of x in the denominator.✓ Proved
\[ \lim_{x \to -\infty}\left(\frac{3 x^{2} - 21 x + 36}{x^{2} + 2 x - 3}\right) = 3 \]
The same as x → −∞.✓ Proved

Answer \( x = -3,\ x = 1;\ y = 3 \)

✓ 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	claude-sonnet-5-5	claude-sonnet-5-5: Numerator at x=-3 is 27+63+36=126 and at x=1 is 3-21+36=18; both are nonzero while the denominator (x-1)(x+3) vanishes there, so both are vertical asymptotes.
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: pass
qwen3.6:27b-mlx: fail (error) — The claim that the numerator is not zero at x = 1 is false; f(1) = 0/0, so x = 1 is a removable discontinuity, not a vertical asymptote. Only x = -3 is a vertical asymptote.

Senior review claude-sonnet-5-5, 2026-10-07: pass — The numerator 3(x-3)(x-4) vanishes only at 3 and 4, so both x=-3 and x=1 are vertical asymptotes. The horizontal asymptote y=3 is correct for both ends.

qwen3.6:27b-mlx: dismiss — The numerator at x=1 is 3-21+36=18, not 0 (it factors as 3(x-3)(x-4)), so f(1) is 18/0 and x=1 is a genuine vertical asymptote, not a hole.

Every verdict on record (5)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The claim that the numerator is not zero at x = 1 is false; f(1) = 0/0, so x = 1 is a removable discontinuity, not a vertical asymptote. Only x = -3 is a vertical asymptote.
claude-sonnet-5-5: pass 2026-10-07 — The numerator 3(x-3)(x-4) vanishes only at 3 and 4, so both x=-3 and x=1 are vertical asymptotes. The horizontal asymptote y=3 is correct for both ends.
gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution incorrectly claims the numerator is not zero at x = 1. Since f(1) = 0/0, x = 1 is a removable discontinuity (hole), not a vertical asymptote. Only x = -3 is a vertical asymptote.

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.