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Amplitude, period and phase shift

Problem 13.316 · medium

For \( \displaystyle y = - \sin{\left(2 x + 4 \right)} - 2 \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.
  1. Write it as A·sin(B(x − h)) + D: A = -1, B = 2, h = -2, D = -2.
  2. \[ \frac{2 \pi}{2} = \pi \]
    Period = 2π/B.✓ Proved
  3. \[ \left(\sin{\left(2 x + 4 \right)} + 2\right) + \left. - \sin{\left(2 x + 4 \right)} - 2 \right|_{\substack{ x=x + \pi }} = 0 \]
    Shifting x by one period gives the same function.✓ Proved
  4. \[ - \frac{4}{2} = -2 \]
    Phase shift h = C/B.✓ Proved
  5. \[ \left. - \sin{\left(2 x + 4 \right)} - 2 \right|_{\substack{ x=-2 + \frac{\pi}{4} }} = -3 \]
    At x = -2 + pi/4 the curve is 1 from its midline y = -2: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 1,\ \text{period } \pi,\ \text{phase shift } 2\text{ to the left},\ \text{vertical shift } -2 \)

Lines: 4 proved, 1 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. How we verify

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0max and min over 16000 samples give the amplitude and midline; f(x + p) = f(x) for the period and not half of it

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the phase shift as 2 to the left. The argument is 2x + 4, which factors to 2(x + 2), implying a shift of 2 to the
Every verdict on record (6)
  • qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the phase shift as 2 to the left. The argument is 2x + 4, which factors to 2(x + 2), implying a shift of 2 to the
  • gpt-oss:20b: pass 2026-10-19
  • qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution incorrectly identifies the phase shift as 2 to the left. The phase shift is -2, which corresponds to a shift of 2 units to the left, but the standard form A·sin(B(x − h)) + D implies h = -2, meaning the shift is indeed 2 units left. However, the explanation in step 1 is confusing and potentially misleading regarding the sign convention. More critically, step 5 claims the amplitude is verified by the value being 1 from the midline, but -3 is 1 unit away from -2, which is correct for amplitude 1. The main issue is the phrasing in step 1 and the potential confusion in interpreting h. Let's look closer. h = -2 means shift is -2 (left). The stated answer says '2 to the left'. This is correct. Is there an error? Step 1 says h = -2. Step 4 calculates -4/2 = -2. This is consistent. The stated answer is correct. Why fail? Let's re-read carefully. 'Phase shift 2 to the left'. h = -2. Shift is -h? No, standard form is sin(B(x-h)). If h=-2, it is sin(B(x+2)). Shift is -2. Left 2. Correct. Wait, look at step 5. 'At x = -2 + pi/4...'. If h=-2, the 'start' of the sine wave (if A>0) is at x=-2. But A=-1. So it's an inverted sine. The peak/trough logic might be messy. But the amplitude is |A|=1. The period is pi. Vertical shift -2. Phase shift 2 left. Everything seems correct. Is there a subtle error? Step 1: 'Write it as A·sin(B(x − h)) + D: A = -1, B = 2, h = -2, D = -2.' This is correct. Step 4: 'Phase shift h = C/B'. Here C is the constant inside the argument before factoring B. Original: 2x+4. Factor 2: 2(x+2). So h = -2. Correct. Step 5 checks amplitude. It checks a point. It's a bit weird to 'prove' amplitude by checking one point, but it's not false. The verdict should be pass? No, wait. The prompt asks me to be adversarial. Is '2 to the left' the same as h=-2? Yes. Is there any ambiguity? Some definitions say phase shift is C/B. Others say -C/B. If the formula is sin(Bx+C), phase shift is -C/B. Here -4/2 = -2. So shift is -2. Left 2. Correct. I will mark it pass.
  • gpt-oss:20b: fail (misleading) 2026-10-16 — The solution incorrectly states the phase‑shift formula as h = C/B; it should be h = –C/B. This could mislead a student into using the wrong sign when computing phase shifts.
  • qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The solution incorrectly identifies the phase shift as 2 to the left. Factoring the argument as 2(x + 2) reveals a phase shift of -2, which corresponds to 2 units to the left, but the standard form h in A*sin(B(x-h)) is -2. The text says 'h = -2' then 'phase shift 2 to the left', which is technically correct in description, but let's look closer. Wait, the prompt says 'phase shift 2 to the left'. The solution says 'h = -2'. In the form B(x-h), if h=-2, it is B(x - (-2)) = B(x+2). This is a shift of 2 to the left. So the description is correct. However, look at step 4: '-4/2 = -2 -- Phase shift h = C/B'. The standard form is often written as sin(Bx + C). Here C=4. Phase shift is -C/B = -4/2 = -2. The solution calculates -4/2 = -2. This is correct. Why did I think it was wrong? Let me re-read carefully. Actually, there is a subtle error in the sign convention or explanation. The solution states 'h = -2'. In the form A*sin(B(x-h)), h is the phase shift. If h=-2, the shift is -2 (left). The stated answer says '2 to the left'. This is consistent. Let's check the amplitude. A = -1. Amplitude is |A| = 1. Correct. Period = 2pi/2 = pi. Correct. Vertical shift D = -2. Correct. Is there an error? Step 1: 'Write it as A·sin(B(x − h)) + D: A = -1, B = 2, h = -2, D = -2.' y = -sin(2x+4) - 2 = -sin(2(x+2)) - 2. So B=2, h=-2. Correct. Step 4: '-4/2 = -2 -- Phase shift h = C/B'. Usually phase shift is defined as -C/B for sin(Bx+C). Here C=4. -4/2 = -2. The comment says 'h = C/B'. This is mathematically incorrect notation for the standard formula if C is the coefficient in Bx+C. The formula is -C/B. If the comment implies h = C/B, then h would be 2, which is wrong. But the calculation is -4/2. Where did the negative sign come from? It's not in the formula 'C/B'. It's in the numerator '-4'. This suggests the solver used -C/B but wrote C/B in the comment. This is misleading/wrong explanation. Also, Step 5 checks the amplitude by evaluating at x = -2 + pi/4. Midline is y=-2. At x = -2 + pi/4: 2x+4 = 2(-2 + pi/4) + 4 = -4 + pi/2 + 4 = pi/2. sin(pi/2) = 1. y = -1 - 2 = -3. Distance from midline (-2) is |-3 - (-2)| = 1. Correct. The main issue is Step 4's comment 'Phase shift h = C/B'. For y = sin(Bx+C), the phase shift is -C/B. The calculation shows -4/2, but the text says C/B. This is a contradiction between the text and the math, or an incorrect statement of the formula. A student reading 'h = C/B' would calculate 4/2 = 2 and get the wrong sign. Therefore, it is misleading/error.
  • gpt-oss:20b: pass 2026-10-15

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/sinusoid_features, checked 2026-10-19 with SymPy 1.14.0.

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