Take the function  for example.  The graph for is

If we were to change the function to , our phase shift is .  This means we need to shift our entire graph  units to the left.

Our new graph  is the following

The general form for the secant transformation equation is .   represents the phase shift of the function.  When considering  we see that .  So our phase shift is  and we would shift this function  units to the left of the original secant function’s graph.

This is true because the graph  has a period of , meaning it repeats itself every  units.  So if  has a phase shift of any multiple of , then it will just overlay the original graph.  This is shown below.  In orange is the graph of and in purple is the graph of  .

Start this problem by graphing the function of tangent.

Now we need to shift this graph  to the right.

This gives us our answer

The form of the general cosecant function is .  So if we have  then , which represents the phase shift, is equal to .  This gives us a phase shift of .

The general form of the cotangent function is .  So first we need to get   into the form .

From this we see that  giving us our answer.

The amplitude is always a positive number and is given by the number in front of the trigonometric function.  In this case, the amplitude is 4.  The period is given by , where b is the number in front of x.  In this case, the period is .

The amplitude of the sine function is increased by 3, so this is the coefficient for . The +2 shows that the origin of the function is now at  instead of

The y-intercept is the value of y when .

Recall that cosine is the  value of the unit circle. Thus, , so it works.

Secant is the reciprocal of cosine, so it also works.

Also recall that . Thus, the only answer which is not equivalent is . 

The simplest way to solve a problem like this is to determine where a particular point on the graph would lie and then compare that to our answer choices. We should first find the y-value when the x-value is equal to zero. We will start by substituting zero in for the x-variable in our equation. 

Now that we have calculated the y-value we know that the correct graph must have the following point:

Unfortunately, two of our graph choices include this point; thus, we need to pick a second point.

Let's find the y-value when the x-variable equals the following:

We will begin by substituting this into our original equation.

Now we need to investigate the two remaining choices for the following point:

Unfortunately, both of our remaining graphs have this point as well; therefore, we need to pick another x-value. Suppose the x-variable equals the following:

Now, we must substitute this value into our given equation.

Now, we can look for the graph with the following point: 

We have narrowed in on our final answer; thus, the following graph is correct: