Problem 60 - Precalculus Essentials

Question

Use a vertical shift to graph one period of the function. $$y=-3 \sin 2 \pi x+2$$

Step-by-Step Solution

Answer

The graph of the function \(y = -3 sin(2 \pi x) + 2\) starts at the point (0,-1), rises to the point (0.25,2), continues to rise to the peak at (0.5,5), falls to the point (0.75,2), then returns to the initial point at (1,-1).

Step 1: Identify transformations

The amplitude of the sine function is 3, which means the maximum and minimum values will be \(-3\) and \(3\). The sine function has been reflected due to the negative sign, so instead of starting from the maximum, it will start from the minimum. The function has also been vertically shifted upwards by \(2\) units.

Step 2: Sketch the graph

Begin sketching the graph by first drawing a horizontal line at \(y=2\), this is our midline due to the vertical shift. Mark five equally spaced points on the x-axis starting at \(x=0\) and ending at \(x=1\); these represent one period of the function which is \(1\). For a normal sine curve, we will start at the maximum, drop to the midline halfway through, reach the minimum at the end, then return to the midline. However, because our sine function is reflected about the x-axis, we would start at the minimum, rise to the midline halfway through, reach the maximum at the end, then return to the midline.

Step 3: Apply transformations

The minimum and maximum values are given by the amplitude, \(3\), and therefore will be found at \(y=-1\) and \(y=5\), respectively. To sketch the full period of the function, start at \(y=-1\) for \(x=0\), rise to \(y=2\) at \(x=0.25\), continue to rise to \(y=5\) at \(x=0.5\), drop to \(y=2\) at \(x=0.75\) and finally drop back to \(y=-1\) at \(x=1\). Connect the points to sketch one period of the curve.