Algebra 2 Graphs of Trigonometric Functions and Identities Test Solutions and Scoring (PDF)

Amplitude 4. Period \(\frac{2\pi}{3}\). Midline \(y = -2\).

4; 2π/3; y = -2

Scoring: 1 point for the amplitude and midline, 1 point for the period.

\(A = 3\). Period \(4\pi = \frac{2\pi}{B}\), so \(B = \frac{1}{2}\).

\(y = 3\cos\left(\frac{x}{2}\right)\)

Scoring: 1 point for the amplitude, 1 point for B, 1 point for the equation.

\(h(5) = 25 – 20\cos\frac{\pi}{3} = 25 – 10 = 15\). The maximum is \(25 + 20 = 45\).

15 feet; maximum 45 feet

Scoring: 1 point for the height at 5 seconds, 1 point for the maximum.

\(\cos^{2}\theta = 1 – \frac{25}{169} = \frac{144}{169}\). Negative in quadrant II.

\(\cos\theta = -\frac{12}{13}\)

Scoring: 1 point for the identity, 1 point for the sign and value.

\(\cos^{2}x + \cos^{2}x\cdot\frac{\sin^{2}x}{\cos^{2}x} = \cos^{2}x + \sin^{2}x = 1\).

1

Scoring: 1 point for substituting the tangent, 1 point for the result.

Amplitude \(\frac{10 – 2}{2} = 4\). Midline \(\frac{10 + 2}{2} = 6\). \(B = \frac{2\pi}{8} = \frac{\pi}{4}\).

\(y = 4\cos\left(\frac{\pi}{4}x\right) + 6\)

Scoring: 1 point for the amplitude, 1 point for the midline, 1 point for B and the equation.

Period \(\frac{2\pi}{\pi/6} = 12\) hours. Maximum \(5 + 3 = 8\).

12 hours; 8 feet

Scoring: 1 point for the period, 1 point for the maximum.

\(\sin^{2}\theta = 1 – \frac{9}{25} = \frac{16}{25}\); sine is negative in quadrant IV, so \(\sin\theta = -\frac{4}{5}\).

\(\tan\theta = -\frac{4}{3}\).

sin = \(-\frac{4}{5}\), tan = \(-\frac{4}{3}\)

Scoring: 1 point for the sine, 1 point for the tangent.

Midline \(y = 3\), amplitude 2: maximum \(5\), minimum \(1\). She is correct.

Yes: 3 ± 2 gives 1 and 5

Scoring: 1 point for the midline and amplitude, 1 point for the conclusion.