Trigonometric Graphs - Trigonometry

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).

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.

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 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 .

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 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 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 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:

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.

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 .

is shown in red, and is shown in blue.

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.

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 .

Since cosecant is a reciprocal of sine, it uses the same general formula of the sine function with the letters corresponding to the same transformations. Note that while A does correspond to amplitude, the cosecant function extends infinitely upwards and downwards so there is no amplitude for the graphs.