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 -1.

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 4 is the vertical shift of the graph. A vertical shift of 4 means that the entire graph of the function will be moved up four units (in the positive y-direction).

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 3.

The phase shift is given by the value being added or subtracted inside the cosine 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. A vertical shift of -3 means that the entire graph of the function will be moved down three units (in the negative y-direction).

A normal  graph has its y-intercept at . This graph has its y-intercept at . Therefore, the graph was shifted down three units. Therefore the function of this graph is .

The correct answer is . There are no sign changes with vertical shifts; in other words, when the function includes , it directly translates to moving up three units. If you thought the answer was , you may have spotted the y-intercept at  and jumped to this answer. However, recall that the y-intercept of a regular  function is at the point . Beginning at  and ending at  corresponds to a vertical shift of 3 units. 

The general form for the secant transformation equation is .   represents the phase shift of the function.  When considering  we see that , so our vertical shift is  and we would shift this function  units up from the original secant function’s graph.

The graph of  with a vertical shift of  is shown below. This can also be expressed as .

Here is a graph that shows both  and , so that you can see the "before" and "after." The original function is in blue and the translated function is in purple.

The graphs of the incorrect answer choices are  (no vertical shift applied),  (shifted upwards instead of downwards),  (amplitude modified, and shifted upwards instead of downwards), and  (shifted downwards 3 units, but this is not the correct original graph of simply  since the amplitude was modified.)

We are looking for an answer choice that has one of the six trigonometric functions, as well as that function shifted up 3 units. The only answer choice that displays that is this graph of  (purple) and  (blue).

The incorrect answers depict  and ,  and , and  and .