Question

determine the amplitude, period, and phase shift for ( y=\frac{1}{3}\tan(\theta + 30^{circ}) ) and use them to plot the graph of the function. amplitude = does not exist; period ( =180^{circ} ); phase shift ( =-30^{circ} ) amplitude = does not exist; period ( =180^{circ} ); phase shift ( =-30^{circ} )

Explanation:

Step1: Recall the general form of the tangent function

The general form of the tangent function is \(y = A\tan(B(\theta - C))+D\). For the function \(y=\frac{1}{3}\tan(\theta + 30^{\circ})=\frac{1}{3}\tan(1\times(\theta-(- 30^{\circ})))+0\), where \(A=\frac{1}{3}\), \(B = 1\), \(C=-30^{\circ}\), \(D = 0\).

Step2: Determine the amplitude

The amplitude of the tangent function is not defined. Because the range of \(y = \tan x\) is \((-\infty,\infty)\), and multiplying by a non - zero constant \(A\) (in \(y=A\tan(B(\theta - C))+D\)) only vertically stretches or compresses the graph but does not bound it above and below in the same way as for \(y = A\sin(B(\theta - C))+D\) or \(y=A\cos(B(\theta - C))+D\).

Step3: Calculate the period

The period of the tangent function \(y = A\tan(B(\theta - C))+D\) is given by the formula \(T=\frac{180^{\circ}}{|B|}\). Since \(B = 1\), then \(T=\frac{180^{\circ}}{|1|}=180^{\circ}\).

Step4: Calculate the phase shift

The phase shift of the function \(y = A\tan(B(\theta - C))+D\) is given by the formula \(h = C\). Here, \(C=-30^{\circ}\), so the phase shift is \(-30^{\circ}\) (a shift to the left by \(30^{\circ}\)).

Answer:

amplitude = does not exist; period = \(180^{\circ}\); phase shift=\(-30^{\circ}\)