Precalculus Graphs of Trigonometric Functions Quiz: 10 Questions with Explanations

A 7B -7C 1D \(2\pi\)

Amplitude is the absolute value of a, so 7. The negative only reflects the graph.

A \(2\pi\)B \(\frac{\pi}{2}\)C \(8\pi\)D \(4\pi\)

Period = 2 pi / 4 = pi/2.

A \(y=5\)B \(y=2\)C \(y=7\)D \(y=0\)

The vertical shift k = 5 gives the midline y = 5.

A \(y=3\sin x\)B \(y=3\sin2x\)C \(y=2\sin3x\)D \(y=3\cos2x\)

Amplitude 3 and period pi mean b = 2. It starts at 0 and rises, so it is sine.

A \(\pi\)B \(2\pi\)C \(\frac{\pi}{2}\)D \(4\pi\)

For tangent the period is pi divided by |b|, and b = 1/2 gives 2 pi.

A \(x=\frac{\pi}{4}\)B \(x=\pi\)C \(x=0\)D \(x=\frac{\pi}{2}\)

Tangent is undefined where cosine is 0, which is pi/2 in this interval. At 0 and pi the graph passes through zero.

A \(\frac{\pi}{4}\) rightB \(\frac{\pi}{2}\) leftC \(\frac{\pi}{4}\) leftD \(\frac{\pi}{4}\) up

A plus inside shifts left: x + pi/4 = x – (-pi/4).

A \(d=5+2\cos\frac{\pi t}{6}\)B \(d=5+2\sin\frac{\pi t}{6}\)C \(d=2+5\cos\frac{\pi t}{6}\)D \(d=5+2\cos12t\)

Midline 5, amplitude 2, b = 2 pi/12 = pi/6, and it starts at a maximum, so cosine.

A \([-3,3]\)B \([-1,3]\)C \([-4,2]\)D \([-2,4]\)

Cosine ranges over [-1, 1], so 3 cos x covers [-3, 3]; subtract 1.

A 5B 8C 10D 4

Amplitude = (9 – 1)/2 = 4. The midline is 5.