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:

The period of the function is indicated by the coefficient in front of ; here the period is unchanged.

The amplitude of the function is given by the coefficient in front of the ; here the amplitude is 2.

The phase shift is given by the value being added or subtracted inside the function; here the shift is  units to the right.

The only unexamined attribute of the graph is the vertical shift, so 3 is the vertical shift of the graph.

The amplitude of a sinusoidal function is  unless amplified by a constant in front of the equation. In this case, the amplitude is , so the front constant is .

The graph moves through the origin, so it is either a sine or a shifted cosine graph.

It repeats once in every , as opposed to the usual , so the period is doubled, the constant next to the variable is .

The only answer in which both the correct amplitude and period is found is:

Both sine and cosine functions go on infinitely to the left and right when viewed on a graph. For this reason, each of these functions has domains of "all real numbers."

Alternatively, each of these functions ranges between -1 and 1 in the y direction. The incorrect answers all include , which is the range of both the sine and the cosine functions.

The graph of  is shown in red below, and the graph of  is shown in blue below. Because the function is periodic, there are infinitely many transformations that could allow  to translate into , but there is only one answer choice below that is correct, and that is "shift  to the left  units." Per the graph, shifting  to the right  units would also be correct, but that is not an available answer choice.

The graph of  is:

This graph goes through three transformations. First, take the graph of , in blue below, and flip it over the x-axis. We do this because of the negative sign in front of the cosine function. You can see the resulting graph in green below. Next, we want to stretch the graph by a factor of 2, since our amplitude is 2 (we get this from the coefficient in front of the cosine function). You can see the resulting graph in purple, below. 

Finally, we need to shift the graph up 1 unit. This is represented by the black graph, below. 

The incorrect answers display the graphs of the functions , , and .

Looking at our graph, we can tell that the period is .  Using the formula 

 where  is the coefficient of  and  is the period, we can calculate 

This eliminates one answer choice.  We then retrun to our graph and see that the amplitude is 3.  Remembering that the amplitude is the number in front of the function, we can eliminate two more choices.

We then examine our graph and realize it contains the point .  Plugging 0 into our two remaining choices, we can determine which one gives us 4 for a result.

 is shown in red, and  is shown in blue.

 In order to graph , recall that .  First consider the graph .

Now anywhere this graph crosses the x-axis a vertical asymptote will form for the  graph because the denominator of  will be equal to zero and the function will be undefined.  At each maximum and minimum of , the graph of  will invert at that point.

And then we are left with the graph of  .