Questions: Amplitude, Period, and Phase Shift

Question 1

The function y = 3 sin(2x − π) is rewritten in standard form. What is the phase shift?

• A. π units to the left — the minus sign indicates a leftward shift
• B. π units to the right — the constant being subtracted from the argument is π
• C. π/2 units to the right — factor out B = 2 to get y = 3 sin(2(x − π/2))
• D. π/2 units to the left — dividing by B reverses the direction of the shift

Answer: (C)

To find the phase shift, the expression inside sin must be written as B(x − C). Factor out B = 2: 2x − π = 2(x − π/2). Now C = π/2, giving a rightward shift of π/2. Option B is the most common error: reading the raw constant π as the phase shift without factoring out B first. Always extract B before reading C. The phase shift is C in the factored form B(x − C), not the raw constant in the unfactored expression.

Question 2

For the function y = −4 cos(x) + 3, what is the amplitude?

• A. −4, because A is the coefficient of cosine and it equals −4
• B. 4, because amplitude is |A| and is always a positive quantity
• C. 3, because the midline is at y = 3 and amplitude is measured from the midline to zero
• D. 7, because the function reaches a maximum value of 3 + 4 = 7

Answer: (B)

Amplitude is defined as |A|, the absolute value of the vertical stretch coefficient. A = −4 means the graph is reflected over the midline (flipped upside down) but the waves still reach 4 units above and below the midline. The negative sign changes orientation, not height. Option C confuses the midline value (3) with the amplitude. Option D gives the maximum y-value, not the amplitude — the maximum is D + |A| = 3 + 4 = 7, but amplitude is just |A| = 4.

Question 3

In the function y = sin(Bx), increasing the value of B stretches the graph horizontally, producing a longer period.

• T. True
• F. False

Answer: False

Increasing B compresses the graph horizontally, producing a shorter period. The period is 2π/|B| — a larger B makes the denominator larger, so the period shrinks. Think of B as the wave's speed: B = 2 means the wave completes its cycle in half the usual horizontal distance (period = π instead of 2π). The common confusion is that multiplying the input 'feels like' stretching, but multiplying the input compresses the horizontal scale, just as multiplying the output stretches the vertical scale.

Question 4

For y = A sin(B(x − C)) + D, the graph oscillates between the values D − |A| and D + |A|.

• T. True
• F. False

Answer: True

The vertical shift D moves the midline from y = 0 to y = D. The amplitude |A| is the maximum distance from the midline in either direction. So the wave reaches a maximum of D + |A| and a minimum of D − |A|. This formula holds regardless of the sign of A — a negative A flips the wave but doesn't change where the extremes are. For example, y = −3 sin(x) + 5 oscillates between 5 − 3 = 2 and 5 + 3 = 8.

Question 5

In y = sin(x − C), why does a positive value of C shift the graph to the right, even though subtracting seems like it should move the graph left?

Model answer: The phase shift moves the graph so that the pattern which began at x = 0 now begins at x = C. In y = sin(x), the wave starts its cycle where the argument equals zero — at x = 0. In y = sin(x − C), the argument equals zero when x = C, so the starting point has moved to x = C: a rightward shift. To get the same output that occurred at x = 0, you now need x = C — you must go further right to reach the same input value.

The confusion arises from thinking of the transformation as acting on the output (like a vertical shift), when it actually acts on the input. With input transformations, the effect is always reversed from intuition: subtracting C from the input means you need a larger x to reach the same argument value, so the graph shifts right. This is opposite to output transformations, where adding D shifts the graph up. The rule: input changes shift horizontally opposite to their sign; output changes shift vertically matching their sign.