Amplitude, period and phase shift
Problem 13.319 · medium
For \( \displaystyle y = - 2 \sin{\left(3 x + 6 \right)} \), find the amplitude, the period, the phase shift (with its direction) and the vertical shift.
- Write it as A·sin(B(x − h)) + D: A = -2, B = 3, h = -2, D = 0.
- Period = 2π/B: 2*pi/3.
- \[ 2 \sin{\left(3 x + 6 \right)} + \left. - 2 \sin{\left(3 x + 6 \right)} \right|_{\substack{ x=x + \frac{2 \pi}{3} }} = 0 \]Shifting x by one period gives the same function.✓ Proved
- \[ - \frac{6}{3} = -2 \]Phase shift h = C/B.✓ Proved
- \[ \left. - 2 \sin{\left(3 x + 6 \right)} \right|_{\substack{ x=-2 + \frac{\pi}{6} }} = -2 \]At x = -2 + pi/6 the curve is 2 from its midline y = 0: that is the amplitude.✓ Proved
Answer \( \text{amplitude } 2,\ \text{period } \frac{2 \pi}{3},\ \text{phase shift } 2\text{ to the left},\ \text{vertical shift } 0 \)
Lines: 3 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | Not checked | — | 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 |
| 5 | ✓ 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 | max 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: fail (error) — Sentence 3 incorrectly states that adding the shifted function to the original yields zero; it should assert equality of the function with its shift, not zero. This misstates the period test.qwen3.6:27b-mlx: pass — The solution correctly identifies the parameters A, B, h, and D from the standard form, leading to the correct amplitude, period, phase shift, and vertical shift. The algebraic checks provided are consistent with the stated results.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the parameters A, B, h, and D from the standard form, leading to the correct amplitude, period, phase shift, and vertical shift. The algebraic checks provided are consistent with the stated results.gpt-oss:20b: fail (error) 2026-10-19 — Sentence 3 incorrectly states that adding the shifted function to the original yields zero; it should assert equality of the function with its shift, not zero. This misstates the period test.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution incorrectly identifies the phase shift as 2 to the left. The argument is 3x + 6, so the phase shift is -6/3 = -2, which corresponds to a shift of 2 units to the left. However, the solution states 'h = -2' in step 1 but then claims the shift is '2 to the left' in the final answer, which is consistent, BUT step 4 calculates -6/3 = -2 and labels it 'Phase shift h = C/B'. The standard form y = A sin(B(x - h)) + D implies h is the shift. If h = -2, the shift is -2 (left). The final answer says '2 to the left', which is correct. Wait, let me re-read carefully. Step 1 says h = -2. Step 4 confirms h = -2. The final answer says 'phase shift 2 to the left'. This is actually correct. Is there an error? Let's check the amplitude. A = -2. Amplitude is |A| = 2. Correct. Period 2pi/3. Correct. Vertical shift 0. Correct. Phase shift: 3x+6 = 3(x+2). So shift is -2, i.e., 2 units left. The solution seems correct. Why would it fail? Ah, look at step 1: 'Write it as A·sin(B(x − h)) + D: A = -2, B = 3, h = -2, D = 0.' If the form is B(x-h), then 3(x - (-2)) = 3(x+2) = 3x+6. This is correct. The final answer is correct. Is there a subtle error? Maybe the sign of the amplitude? Amplitude is always positive. The solution says 'amplitude 2'. Correct. Maybe the direction? '2 to the left'. Correct. Let me look for a 'misleading' or 'style' issue. Step 5 checks the amplitude by evaluating at x = -2 + pi/6. 3(-2 + pi/6) + 6 = -6 + pi/2 + 6 = pi/2. sin(pi/2) = 1. -2(1) = -2. The value is -2. The midline is 0. The distance is |-2 - 0| = 2. This is correct. The solution appears sound. Wait, the prompt asks me to find what is wrong. Let me re-read the 'Stated answer' vs 'Solution'. Stated answer: 'phase shift 2 to the left'. Solution Step 1: 'h = -2'. In the form B(x-h), h is the shift. A shift of -2 is 2 to the left. This is consistent. Is it possible the system considers 'h' to be the value inside the parenthesis without the sign flip? No, standard convention is x-h. If h=-2, shift is left. I will mark it pass if it is truly correct. But usually these adversarial tasks have a trap. Let's look at Step 1 again. 'A = -2'. Amplitude is |A|. The solution doesn't explicitly state |A|, it just lists A=-2. Then in the final answer it says amplitude 2. This is a slight disconnect but not an error. Let's look at Step 4. '-6/3 = -2 -- Phase shift h = C/B'. This is a standard formula. It is correct. Let's look at Step 5. 'Subs(..., x, -2 + pi/6) = -2'. It claims 'that is the amplitude'. The value is -2. The amplitude is 2. The sentence says 'the curve is 2 from its midline... that is the amplitude'. This is slightly imprecise language (the value is -2, the distance is 2), but the conclusion is correct. Is there a bigger error? What if the question implies the form A sin(Bx + C)? Then phase shift is -C/B = -6/3 = -2. Left 2. Everything seems correct. I will verify the 'direction' part. '2 to the left'. Correct. I'll bet on 'pass' unless I'm missing a very subtle convention. Actually, looking at Step 1, it defines the form as `A·sin(B(x − h))`. If `h = -2`, then `x - (-2) = x + 2`. The term is `3(x+2)`. This matches `3x+6`. So `h=-2` is correct. The shift is `h`. A shift of `-2` is `2` units left. The final answer says `2 to the left`. This is correct. I will mark it pass.gpt-oss:20b: fail (error) 2026-10-16 — Sentence 3 incorrectly asserts that adding the shifted function equals zero; it should state equality of the original and shifted functions. Sentence 5 evaluates the function at a point but mislabels the resulting value as the amplitude; the amplitude is the absolute value 2, not the negative value obtained.qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the parameters A, B, h, and D from the standard form, leading to the correct amplitude, period, phase shift, and vertical shift. The unchecked equations serve as valid verifications of these properties.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.