Overview
A sinusoidal formula contains changes that act on its output and changes that act on its input. Vertical scale and shift determine the range; the coefficient and shift inside the angle determine when the graph repeats and where its peaks occur. Keeping those roles separate lets you answer range, amplitude and location questions without drawing an entire graph.
Transform the output interval to find a range
Sine always takes values between −1 and 1. For y = 3sin(2x) + 1, multiplying those output bounds by 3 gives values from −3 to 3, and adding 1 shifts them to −2 through 4. Thus the range is [−2, 4].
The factor 2 inside sine changes how quickly the graph cycles as x changes; it does not enlarge or shrink the possible sine values. The distractor [−3, 3] ignores the vertical shift, while [1, 4] keeps only values above the midline. Treat the output interval as an object that undergoes the outside operations. This approach explains why the endpoints change as they do and avoids mixing a horizontal change with a vertical one.
Measure amplitude from the midline
A sinusoidal graph with maximum 2 and minimum −2 has total vertical span 4. Its amplitude is half that span, or 2, because amplitude measures the distance from the midline to a peak rather than from a trough to a peak. The midline is halfway between the extremes.
The answer 4 confuses full range width with amplitude. The answer 0 corresponds to the midline height, which describes position rather than size. In a shifted graph, these ideas remain distinct: a nonzero midline does not itself change amplitude. If a graph is supplied, use the labeled maximum and minimum instead of estimating pixel distances. Explain which distance you are measuring before calculating it; the vocabulary is part of the mathematical information, not merely a naming convention.
Find peak locations by solving for the inner angle
Consider y = 3cos(2x − π/3) − 1. Because the coefficient of cosine is positive, the graph reaches its maximum whenever cosine equals 1. Therefore its inner angle must be 2kπ for an integer k: 2x − π/3 = 2kπ. Solving gives x = π/6 + kπ.
The spacing between consecutive maxima is π, not 2π, because the input contains 2x. The horizontal shift is π/6; reading the visible π/3 as the shift without dividing by the coefficient would be incorrect. The outside factor 3 and subtraction of 1 affect peak height but do not change these peak locations. Deriving the locations directly from the cosine condition keeps all of those roles separate and prevents a memorized phase-shift rule from being applied to only part of the input.
A phase shift can be understood by factoring the coefficient inside the input: 2x − π/3 = 2(x − π/6). This confirms the shift derived by solving for a maximum. The algebraic and graphical interpretations agree, but solving the peak condition is often safer when the task asks for several locations. It gives a complete repeating family rather than only the first shifted peak.
Apply the interval and the requested aggregation
The problem asks for the sum of maximum locations in the open interval π < x < 4π. In x = π/6 + kπ, the allowed integers are k = 1, 2 and 3. The locations are 7π/6, 13π/6 and 19π/6. Their sum is 39π/6 = 13π/2. The option 7π/6 gives only the first peak rather than the requested sum.
For transfer, identify whether the target is a vertical feature, a horizontal location, or a count or sum of locations. Use the output bounds for range, half the vertical span for amplitude, and a trigonometric condition for peak timing. Then filter all solutions by the interval before performing the final count or sum. During practice, write the allowed integer indices explicitly. This makes it easier to catch omitted cycles, included endpoints and a correct first solution that was mistakenly treated as the complete answer.
Apply the method to a new case
A sinusoidal signal has maximum 7 and minimum −1. Consecutive maxima occur at times 2 and 8 seconds. Find its amplitude, midline and period. Amplitude is half the vertical range: (7 − (−1))/2 = 4. The midline is their average, (7 + (−1))/2 = 3. The period is the horizontal separation of consecutive matching maxima, 8 − 2 = 6 seconds. Reporting 8 as the amplitude uses the full range, while reporting 3 seconds as the period confuses the peak-to-trough interval with peak-to-peak. Check that the midline plus or minus the amplitude reproduces 7 and −1. A model is y = 3 + 4cos(π(t − 2)/3), whose angular coefficient gives period 2π/(π/3) = 6. Its phase shift changes the peak times without changing amplitude or period.