Part 1The function graphed is of the form y equals a sine bxy=asinbx or y equals a cosine bxy=acosbx, where b 0. Determine the equation of the graph. Part 1
-66xy00piπ2 pi2π Edit coordinates (0,0) A coordinate system has a horizontal x-axis labeled from 0 to 2 pi in increments of StartFraction pi Over 2 EndFraction and a vertical y-axis labeled from negative 6 to 6 in increments of 1. Within 1 period, a repeating curve starts at the origin and rises to a maximum at (StartFraction pi Over 4 EndFraction, 4), falls to a minimum at (StartFraction 3 pi Over 4 EndFraction, negative 4), then rises to the x-axis at pi. Part 1yequals=[input]enter your response here (Type an expression using x as the variable.)

Question

Part 1The function graphed is of the form y equals a sine bxy=asinbx or y equals a cosine bxy=acosbx​, where b​ > 0. Determine the equation of the graph. Part 1
-66xy00piπ2 pi2π Edit coordinates (0,0) A coordinate system has a horizontal x-axis labeled from 0 to 2 pi in increments of StartFraction pi Over 2 EndFraction and a vertical y-axis labeled from negative 6 to 6 in increments of 1. Within 1 period, a repeating curve starts at the origin and rises to a maximum at (StartFraction pi Over 4 EndFraction, 4), falls to a minimum at (StartFraction 3 pi Over 4 EndFraction, negative 4), then rises to the x-axis at pi. Part 1yequals=[input]enter your response here ​(Type an expression using x as the​ variable.)

BlackTom AI · Accepted Answer

First, I’ll restate what the problem asks and what is given, so we’re examining the same setup.
- The function graphed has the form y = a sin(bx) or y = a cos(bx) with b > 0.
- In one period, the graph starts at the origin (0,0), rises to a maximum of 4 at x = π/4, then falls to a minimum of −4 at x = 3π/4, and returns to y = 0 at x = π.
- The requested answer is an expression in x, presumably in the form a sin(bx) or a cos(bx).

Now, I’ll analyze the behavior of the graph to determine a and b (and whether sine or cosine is appropriate).
- The graph starts at the origin (0,0). A sine function y = a sin(bx) naturally has value 0 at x = 0, since sin(0) = 0. In contrast, a cosine function y = a cos(bx) would start at y = a, not 0, unless a = 0 or a phase shift is involved. Since we’re not given any phase shift and the graph begins at the origin, sine is the more natural base form here.
- The graph rises from 0 to a maximum of 4 when x = π/4. For y = a sin(bx), the first peak after x = 0 occurs at x = π/(2b). Setting π/(2b) = π/4 gives b = 2.
- With b = 2, the amplitude a must be 4 to match the maximum value y = 4 and minimum y = −4, so a = 4. The function y = 4 sin(2x) then has range [−4, 4] and periods of length π, which aligns with the graph completing a full cycle by x = π.
- Let me verify key points using y = 4 sin(2x): at x = 0, sin(0) = 0 → y = 0; at x = π/4, sin(π/2) = 1 → y = 4 (maximum); at x = 3π/4, sin(3π/2) = −1 → y = −4 (minimum); at x = π, sin(2π) = 0 → y = 0. This exactly matches the described points.

Next, I’ll consider the alternative options conceptually to show why they don’t fit as well.
- If we tried a cosine form y = a cos(bx), we’d have y(0) = a, which would be nonzero unless a = 0. Since the graph starts at 0, a cosine form would require a phase shift to align with the origin, which the problem statement does not indicate. Therefore, y = a cos(bx) is unlikely to describe this graph without additional shifts.
- Trying different amplitudes or frequencies would misplace either the peak value (4) or the timing of the peak and zero crossings. The given peak occurs at π/4 and the zero crossing at π, which uniquely pins down a = 4 and b = 2 for the sine form as shown.
- A negative amplitude or a different sign for b would flip the graph’s orientation or shift the peak locations, contradicting the observed sequence of points.

Finally, consolidating the deductions: the graph’s features—origin start, rising to a maximum of 4 at π/4, falling to −4 at 3π/4, and returning to 0 at π—are all exactly matched by y = 4 sin(2x). Therefore, among the standard sine/cosine forms without phase shifts, the expression that fits is 4 sin(2x).

Homework:Section 6.3 Homework

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