6.06 Graphing tangent functions

Complete the table with values in exact form:

Since the function f \left( \theta \right)= \tan \theta is represented by the y-coordinate divided by the x-coordinate, \dfrac{y}{x} on the unit circle, we can use the ordered pairs at each angle on the unit circle to complete the table of values.

Sketch a graph for f \left( \theta \right) = \tan \theta on the domain [-2\pi, 2\pi].

Plot the points for [0, 2 \pi] using the table.

Then, we can work backwards along the unit circle to each negative value of \theta which is coterminal to a positive angle between 0 and 2 \pi and match the values of tangent or \dfrac{y}{x} from the unit circle.

For the function from -2\pi to 0, \theta= \dfrac{-\pi}{4} is coterminal to \alpha= \dfrac{7 \pi}{4}, and since \tan \left( \dfrac{7\pi}{4} \right)=-1, \tan \left( \dfrac{-\pi}{4} \right) = -1 also. The following coterminal angles give the exact values:

\tan \left( \dfrac{-3\pi}{4} \right) = \sin \left( \dfrac{5\pi}{4} \right) = 1

\tan \left( - \pi \right) = \tan \left( \pi \right) = 0

\tan \left( \dfrac{-5 \pi}{4} \right) = \tan \left( \dfrac{3\pi}{4} \right) = -1

\tan \left( \dfrac{-7 \pi}{4} \right) = \tan \left( \dfrac{\pi}{4} \right) = 1

\tan \left( -2\pi \right) = \tan \left( 2 \pi \right) = 0

Describe the behavior of the graph as \theta \to \dfrac{\pi}{2} from the left. Explain how this relates to the unit circle.

We know that the graph of the tangent function has no amplitude and we can use this to describe the behavior of the function.

As \theta \to \dfrac{\pi}{2} from the left, the graph increases at an increasing rate and goes to \infty. There is a vertical asymptote at \theta=\dfrac{\pi}{2}, the function will approach but never intersect the line \theta=\dfrac{\pi}{2}, and will continue to increase forever.

This relates to the unit circle because the values of sine get closer to 1 and the values of cosine get closer to 0, as \theta approaches \dfrac{\pi}{2} from 0. Since \tan \theta = \dfrac{\sin \theta}{\cos \theta}, dividing 1 by smaller and smaller numbers results in the ratio becoming very large.

Eventually when \cos \theta = 0, tangent will be undefined, which appears as a vertical asymptote on the graph of the function.

Describe the behavior of the graph as \theta \to \pi from the left. Explain how this relates to the unit circle.

Use the x-intercepts of the tangent function occurring at multiples of \pi to describe the function's behavior.

As \theta \to \pi from the left, the graph increases at a decreasing rate until it reaches 0.

This relates to the unit circle because the y-values (values of sine) get closer to 0 from the positive direction as the x-values (values of cosine) get closer to -1 from the positive direction. This means that tangent will go from being very large and negative to zero.

At \pi on the unit circle, the ordered pair is \left( -1, 0 \right), and we know that \tan \theta = \dfrac{y}{x}. At this point, tangent is equal to zero because \dfrac{y}{x}=\dfrac{0}{-1}.

Identify the positive and negative intervals on the graph of f \left( \theta \right) = \tan \theta over the domain [0, 2 \pi]. Explain how this relates to the unit circle.

After describing the key features of f \left( \theta \right)= \tan \theta, use the signs of ordered pairs in each quadrant on the coordinate plane and the points on the unit circle to relate the function to the unit circle.

The positive intervals of f \left( \theta \right) = \tan \theta occur on \left(0, \dfrac{\pi}{2} \right) and \left( \pi, \dfrac{3 \pi}{2} \right).

The negative intervals of f \left( \theta \right) = \tan \theta occur on \left( \dfrac{\pi}{2}, \pi \right) and \left(\dfrac{3 \pi}{2}, 2 \pi \right).

Since \tan \theta= \dfrac{\sin \theta}{\cos \theta}, and the ordered pairs on the unit circle are \left( \cos \theta, \sin \theta \right), we can see that the sign of tangent is positive in the first and third quadrants where sine and cosine have the same sign, and these quadrants have the same intervals for the tangent function on the graph.

We can also see that the sign of tangent is negative in the second and fourth quadrants where sine and cosine have opposite signs, and these quadrants have the same intervals for the tangent function on the graph.

The given graph is the result of what transformations on f \left( \theta \right) = \tan \theta?

Use the graph of f \left( \theta \right) = \tan \theta to identify changes to the function visually.

The graph together with the y-intercept of f \left( \theta \right) = \tan \theta has been translated up by 4 units.

The period of the function has decreased from \pi to \dfrac{\pi}{2}, so the function is horizontally compressed by a factor of \dfrac{1}{2}.

State the equation of the graphed function.

Use the transformations stated in part (a) and apply them to the general form of the tangent function, f \left( \theta \right) = a \tan \left[b \left(\theta - c \right)\right] + d.

The translation of 4 units up will change the value of d to 4. The horizontal compression from a period of \pi to \dfrac{\pi}{2} will change the value of b to 2. The equation of the function is f \left( \theta \right) = \tan 2 \theta + 4.

We can choose a few strategic values of \theta such as \dfrac{\pi}{4} and \dfrac{\pi}{2} to confirm that they match the graph.

The function should be undefined when \theta = \dfrac{\pi}{4}. We can test the value of the graph algebraically:

The function should be 4 when \theta = \dfrac{\pi}{2}. We can test the value of the graph algebraically:

Algebraically, these values match what we see on the graph of the function, so we can confirm that the equation of the function is correct.

State the domain and range.

Start by considering the domain and range of the tangent function, then determine how the transformed function changes the domain and range:

• Domain: x \neq \dfrac{\pi}{2} + n \pi, n is any integer
• Range: \left( -\infty, \infty \right)

The domain of the function excludes any values of x where the function would equal \pm\dfrac{\pi}{4},\,\pm\dfrac{3\pi}{4},\,\pm\dfrac{5\pi}{4}, etc. So, x \neq \dfrac{\pi}{4} + \dfrac{n \pi}{2}, where n is any integer. The range of the function remains the same, \left( -\infty, \infty \right).

Graph the function over the domain \left[-2 \pi, 2 \pi\right].

The value of a=-1, so the graph of f \left( \theta \right) = \tan \theta has been reflected across the x-axis. The value of c=\dfrac{\pi}{2}, so the graph has been shifted by \dfrac{c}{b}=\dfrac{\frac{\pi}{2}}{1}=\dfrac{\pi}{2} units to the left.

First, apply the reflection across the x-axis:

Then, we can apply the phase shift to the left by \dfrac{\pi}{2} units:

Identify the key features of f \left( \theta \right) = -\tan \left( \theta + \dfrac{\pi}{2} \right).

• Domain: x \neq n \pi, where n is any integer
• Range: \left( -\infty, \infty \right)
• x-intercept: x=\dfrac{\left(2n+1\right)\pi}{2}, where n is any integer
• y-intercept: None
• Period: \pi
• Amplitude: None
• Midline: y=0

Since the graph is not shifted up or down, the midline will remain the same as the parent function.

g \left( \theta \right) = \tan \dfrac{1}{2} \theta

The value of b for g \left( \theta \right) is \dfrac{1}{2}, so the period of the tangent function changes.

The new period for the function is \dfrac{\pi}{\frac{1}{2}} = 2 \pi, meaning that the period increases and we see a horizontal stretch.

The domain of g \left( \theta \right) is x \neq \left(2n+1\right) \pi, n is any integer, and the range is the same as the parent function, \left( -\infty, \infty \right).

The x-intercepts of g \left( \theta \right) occur at x=2 n \pi, where n is any integer, while the y-intercept and midline remain the same as the parent function, at y=0.

h \left( \theta \right) = 6 \tan \theta

The value of a for h \left( \theta \right) is 6, so the graph should experience a vertical stretch.

The period for the function is the same as the parent function, since the value of b does not change.

The domain of h \left( \theta \right) is the same as the parent function, x \neq \dfrac{\pi}{2} + n \pi, where n is any integer. The range is the same as the parent function, \left( -\infty, \infty \right).

The x-intercepts of h \left( \theta \right) occur at the same points as the parent function, at x=n \pi, wheren is any integer. The y-intercept and midline remain the same as the parent function, at y=0.

When only changing the value of a, the location of the key features, such as the midline, asymptotes, and axis-intercepts, remain the same as f \left( \theta \right) = \tan \theta.