2. A changes height: amplitude, stretch, compression, reflection

Compare y = 4 sin x with y = sin x. Use these key outputs of y = sin x: at the inputs x = 0, π2$\frac{\pi }{2}$, π, 3π2$\frac{3\pi }{2}$, 2π, its outputs are 0, 1, 0, −1, 0. (a) Name A and find the amplitude. (b) Decide whether the height is stretched, compressed or unchanged, and whether the graph is reflected across the x-axis. (c) Find the outputs of y = 4 sin x at the same five inputs and list the key points. (d) State the midline, the maximum and minimum, the range and the period. (e) Confirm the amplitude from the maximum and minimum heights.

1. Find A, the constant multiplying sine on the outside. In y = 4 sin x, A = 4.A is the number written in front of sin x, and it multiplies the output. In y = sin x it would be the invisible 1, but here it is 4.
2. Compute the amplitude: |A| = |4| = 4.The amplitude is |A|, the size of A with any minus sign removed. It measures a distance from the midline, so it is never negative.
3. Compare |A| with 1. Because 4 > 1, the graph has a vertical stretch by a factor of 4. Because A = 4 is positive, there is no reflection across the x-axis.|A| > 1 stretches vertically, 0 < |A| < 1 compresses vertically, and |A| = 1 leaves the height unchanged. Only A < 0 adds a reflection.
4. Multiply each key output of sin x by A = 4: 4 · 0 = 0, 4 · 1 = 4, 4 · 0 = 0, 4 · (−1) = −4, 4 · 0 = 0. The key points are (0, 0), (π2$\frac{\pi }{2}$, 4), (π, 0), (3π2$\frac{3\pi }{2}$, −4), (2π, 0).A multiplies on the outside, so it changes only the outputs (heights). The inputs 0, π2$\frac{\pi }{2}$, π, 3π2$\frac{3\pi }{2}$, 2π stay fixed, and every height becomes 4 times as far from the x-axis.
5. Write the midline y = 0. The peak at x = π2$\frac{\pi }{2}$ gives the maximum 4, and the valley at x = 3π2$\frac{3\pi }{2}$ gives the minimum −4. So the range is [−4, 4].Multiplying by A keeps the wave centered on y = 0, and the outputs run from −|A| to |A|. The highest output of sin x, 1, becomes 4, and the lowest, −1, becomes −4.
6. Confirm the amplitude from the extreme heights: maximum − minimum2$\frac{maximum-minimum}{2}$ = 4 − (−4)2$\frac{4-(-4)}{2}$ = 82$\frac{8}{2}$ = 4.The wave is centered at zero, so the full top-to-bottom distance of 8 spans two amplitudes. The amplitude is half of it, the distance from the midline to a peak, so 8 is not the amplitude.
7. Check whether A is zero. Here A = 4 ≠ 0, so the graph is a true wave, not the flat line y = 0, and its period is T = 2π.An outside multiplication never changes the input, so the wave repeats every 2π, exactly like y = sin x. Only A = 0 would make every output 0 and leave no smallest positive period.

In an audio editor, a test tone is drawn as the curve y = sin x, with x in radians. The editor's Invert command multiplies every output by −1, so the inverted tone is y = −sin x. The original tone passes through these seven points of y = sin x: (−π, 0), (−π2$\frac{\pi }{2}$, −1), (−π6$\frac{\pi }{6}$, −12$\frac{1}{2}$), (0, 0), (π6$\frac{\pi }{6}$, 12$\frac{1}{2}$), (π2$\frac{\pi }{2}$, 1) and (π, 0). (a) Name A in y = −sin x and find the amplitude. (b) Decide whether Invert stretches, compresses or keeps the height of the wave, and whether it reflects the graph. (c) Find the point of y = −sin x at each of the same seven inputs. (d) State the midline, the range and the period T of y = −sin x. (e) For −π ≤ x ≤ π, find the maximum and the minimum of y = −sin x and the inputs where they occur. Then use them to check the amplitude.

1. Find A. Write y = −sin x as y = (−1) · sin x, so A = −1.A is the constant multiplying sine on the outside. A lone minus sign in front of sin x is a hidden factor of −1, just as plain sin x hides A = 1.
2. Find the amplitude: |A| = |−1| = 1.The amplitude keeps the size of A and removes its minus sign. It measures how far the wave reaches from the midline, which is a distance and is never negative. The amplitude is 1, not −1.
3. Compare |A| = 1 with 1. They are equal, so there is no vertical stretch and no vertical compression, and the height is unchanged. Since A = −1 < 0, the graph is also reflected across the x-axis.|A| > 1 would stretch the wave and 0 < |A| < 1 would compress it; |A| = 1 does neither. The size of A decides stretch or compression. The sign of A, separately, decides the reflection.
4. Keep each input and multiply each output by A = −1: (−π, 0) → (−π, 0), (−π2$\frac{\pi }{2}$, −1) → (−π2$\frac{\pi }{2}$, 1), (−π6$\frac{\pi }{6}$, −12$\frac{1}{2}$) → (−π6$\frac{\pi }{6}$, 12$\frac{1}{2}$), (0, 0) → (0, 0), (π6$\frac{\pi }{6}$, 12$\frac{1}{2}$) → (π6$\frac{\pi }{6}$, −12$\frac{1}{2}$), (π2$\frac{\pi }{2}$, 1) → (π2$\frac{\pi }{2}$, −1), (π, 0) → (π, 0).The −1 multiplies outside the sine, so only the outputs change and every input stays fixed. Each point moves straight across the x-axis to the same distance on the other side. Points with output 0 lie on the x-axis, so they do not move.
5. Write the midline y = 0 and the range [−|A|, |A|] = [−1, 1]. Because A = −1 ≠ 0, the period is T = 2π.Multiplying a zero-centered wave by A keeps it centered on y = 0, and its outputs then run from −|A| up to |A|. A only rescales heights, so the period of sine stays 2π. Only A = 0 would flatten the graph to y = 0 with no smallest positive period.
6. Read the extremes from the new points. The maximum is 1, at x = −π2$\frac{\pi }{2}$, and the minimum is −1, at x = π2$\frac{\pi }{2}$. Check the amplitude: maximum − minimum2$\frac{maximum-minimum}{2}$ = 1 − (−1)2$\frac{1-(-1)}{2}$ = 22$\frac{2}{2}$ = 1.The maximum and minimum are the highest and lowest outputs, and the range [−1, 1] shows that no output goes beyond them. The amplitude is half the top-to-bottom distance, not the whole distance of 2, and it matches |A| = 1 from step 2. The flip also swaps peak and valley: sin x peaks at x = π2$\frac{\pi }{2}$, so −sin x has its valley there.

This asks for the new size and a sample height in y = 34$\frac{3}{4}$ sin x. Find amplitude, range, and the output at π6$\frac{\pi }{6}$.

1. A = 34$\frac{3}{4}$, so amplitude = 34$\frac{3}{4}$ and range is [−34$\frac{3}{4}$, 34$\frac{3}{4}$].0 < 34$\frac{3}{4}$ < 1 makes a Vertical compression.
2. At π6$\frac{\pi }{6}$, y = 34$\frac{3}{4}$ × 12$\frac{1}{2}$ = 38$\frac{3}{8}$.Multiply the numerator by the numerator and denominator by the denominator to scale the known sine height.

This asks you to separate size from sign in y = −6 sin x. Find amplitude, range, and the heights at π2$\frac{\pi }{2}$ and 3π2$\frac{3\pi }{2}$.

1. A = −6; amplitude = |−6| = 6.The magnitude determines size while the minus sign determines reflection.
2. At π2$\frac{\pi }{2}$, y = −6 × 1 = −6; at 3π2$\frac{3\pi }{2}$, y = −6 × (−1) = 6.The old top becomes the new bottom, and the old bottom becomes the new top.
3. Write range [−6, 6] and midline y = 0.The scaled output extremes are equally far below and above zero.