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5. Graph Y=3sin(pi )/(2)(x+1)-1

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5. Graph y=3sin(pi )/(2)(x+1)-1

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Answer

Here's how to graph the function y = 3sin[π/2(x+1)] - 1:1. **Identify the key features of the sine function:*** **Amplitude:** The amplitude is 3, meaning the graph oscillates 3 units above and below the midline.* **Period:** The period is found using the formula 2π/|B|, where B is the coefficient of x within the sine function. In this case, B = π/2. So, the period is 2π / (π/2) = 4. This means the graph completes one full cycle every 4 units along the x-axis.* **Phase Shift (Horizontal Shift):** The phase shift is determined by the value inside the parentheses with x. Here, we have (x+1), which indicates a horizontal shift of 1 unit to the *left*.* **Vertical Shift:** The vertical shift is -1, meaning the midline of the graph is shifted 1 unit *down* from the x-axis. The midline is y = -1.2. **Determine key points for one cycle:**It's helpful to find the x-values for the maximum, minimum, and midline points of one cycle. We can start with the standard sine function and apply the transformations:* **Standard Sine:** Key points are typically at x = 0, π/2, π, 3π/2, and 2π.* **Adjust for Period:** Divide the standard x-values by B (π/2): 0, 1, 2, 3, and 4.* **Adjust for Phase Shift:** Subtract 1 from each x-value: -1, 0, 1, 2, and 3.Now we have the x-values for one cycle of the transformed sine function.3. **Calculate the corresponding y-values:**Substitute the x-values we found into the equation y = 3sin[π/2(x+1)] - 1:* x = -1: y = 3sin[π/2(-1+1)] - 1 = 3sin(0) - 1 = -1* x = 0: y = 3sin[π/2(0+1)] - 1 = 3sin(π/2) - 1 = 3(1) - 1 = 2* x = 1: y = 3sin[π/2(1+1)] - 1 = 3sin(π) - 1 = 3(0) - 1 = -1* x = 2: y = 3sin[π/2(2+1)] - 1 = 3sin(3π/2) - 1 = 3(-1) - 1 = -4* x = 3: y = 3sin[π/2(3+1)] - 1 = 3sin(2π) - 1 = 3(0) - 1 = -14. **Plot the points and sketch the graph:**Plot the points (-1, -1), (0, 2), (1, -1), (2, -4), and (3, -1). Then, connect the points with a smooth curve that resembles a sine wave. Continue the pattern to the left and right to extend the graph beyond one cycle.The graph will oscillate between y = 2 and y = -4, with a midline at y = -1. It will complete one full cycle every 4 units along the x-axis, and the first maximum occurs at x = 0.