Questions: Determine the amplitude and phase shift of y=2 sin (πx)-1. (You do not need to graph it, but if you do, it could be worth up to 5 points of extra credit.)

Solution Steps

The given function is \( y = 2 \sin (\pi x) - 1 \). We can express it in the general form of a sine function: \[ y = A \sin(Bx + C) + D \] where:

• \( A = 2 \)
• \( B = \pi \)
• \( C = 0 \)
• \( D = -1 \)

The amplitude of the sine function is given by the absolute value of \( A \): \[ \text{Amplitude} = |A| = |2| = 2 \]

The phase shift is calculated using the formula: \[ \text{Phase Shift} = -\frac{C}{B} \] Substituting the values of \( C \) and \( B \): \[ \text{Phase Shift} = -\frac{0}{\pi} = 0 \]

We apply the transformation to a set of points to verify the function's behavior. The original points and their transformed counterparts are:

• Point \((0, 0) \Rightarrow (0, -1)\)
• Point \((0.25, 0) \Rightarrow \left( \frac{0.25}{\pi}, -1 \right)\)
• Point \((0.5, 0) \Rightarrow \left( \frac{0.5}{\pi}, -1 \right)\)
• Point \((0.75, 0) \Rightarrow \left( \frac{0.75}{\pi}, -1 \right)\)
• Point \((1, 0) \Rightarrow \left( \frac{1}{\pi}, -1 \right)\)

From the analysis, we conclude:

• The amplitude of the function is \( 2 \).
• The phase shift of the function is \( 0 \).

Final Answer