Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the second quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, .

This is . Rounding, this is . However, since  is in the second quadrant, your value must be negative. (The tangent function is negative in that quadrant.) Therefore, the answer is .

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the fourth quadrant of the Cartesian plane. It would look like:

So, the tangent of an angle is:

  or, for your data,  or . However, since  is in the fourth quadrant, your value must be negative. (The tangent function is negative in that quadrant.) This makes the correct answer .

The standard period of a tangent function is  radians. In other words, it completes its entire cycle of values in that many radians. To alter the period of the function, you need to alter the value of the parameter of the trigonometric function. You multiply the parameter by the number of periods that would complete in  radians. With a period of , you are quadrupling your method. Therefore, you will have a function of the form:

Since  and  do not alter the period, these can be anything.

Therefore, among your options,  is correct.

The standard period of a tangent function is  radians. In other words, it completes its entire cycle of values in that many radians. To alter the period of the function, you need to alter the value of the parameter of the trigonometric function. You multiply the parameter by the number of periods that would complete in  radians. With a period of , you are multiplying your parameter by . Now, half of this would be a period of . Thus, you will have a function of the form:

Since  and  do not alter the period, these can be anything.

Therefore, among your options,  is correct.

The period of the tangent function defined in its standard form  has a period of . When you multiply the argument of the trigonometric function by a constant, you shorten its period of repetition. (Think of it like this: You pass through more iterations for each value  that you use.) If you have , this has one fifth of the period of the standard tangent function.  In the equation given, none of the other details matter regarding the period. They alter other aspects of the equation (its "width," its location, etc.). The period is altered only by the parameter. Thus, the period of this function is  of , or .

For tangent and cotangent the period is given by the formula:

 where  comes from .

Thus we see from our equation  and so
.

To find the period of a tangent funciton use the following formula:

 where  comes from .

thus we get that  so 

To find the period of a tangent or cotangent function use the following formula:

from the general tirogonometric formula: 

Since we have,

we have

.

Thus we get that 

Rewrite the tangent function in terms of cosine and sine.

Since the denominator cannot be zero, evaluate all values of theta where  on the interval .

These values of theta are asymptotes and will not exist on the tangent curve. They will not be included in the domain and parentheses will be used in the interval notation.

The correct solution is .

The domain for the parent function of tangent does not exist for:

The amplitude and the vertical shift will not affect the domain or the period of the graph.

The tells us that the graph will shift right  units.

Therefore, the asymptotes will be located at:

The locations of the asymptotes are: