Trigonometric Graphs - Trigonometry
Card 0 of 20
Let 
be a function defined as follows:
The 4 in the function above affects what attribute of the graph of 
?
Let
The 4 in the function above affects what attribute of the graph of
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
The amplitude of the function is given by the coefficient in front of the
The phase shift is given by the value being added or subtracted inside the
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).
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Let be a function defined as follows:
What is the vertical shift in this function?
Let
What is the vertical shift in this function?
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 period of the function is indicated by the coefficient in front of
The amplitude of the function is given by the coefficient in front of the
The phase shift is given by the value being added or subtracted inside the cosine function; here the shift is
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).
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This graph shows a translated cosine function. Which of the following could be the equation of this graph?
This graph shows a translated cosine function. Which of the following could be the equation of this graph?
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 correct answer is 
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The graph below shows a translated sine function. Which of the following functions could be shown by this graph?
The graph below shows a translated sine function. Which of the following functions could be shown by this graph?
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 
.
A normal 
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Consider the function 
. What is the vertical shift of this function?
Consider the function 
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 general form for the secant transformation equation is 
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Which of the following is the graph of 
with a vertical shift of 
?
Which of the following is the graph of 
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.)
The graph of 
Here is a graph that shows both 
The graphs of the incorrect answer choices are 
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Which of the following graphs shows one of the original six trigonometric functions with a vertical shift of 
applied?
Which of the following graphs shows one of the original six trigonometric functions with a vertical shift of 
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 
.
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 
The incorrect answers depict 
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This is the graph of what function of x?
This is the graph of what function of x?
is shown in red, and 
is shown in blue.
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Give the equation of the following graph.
Give the equation of the following graph.
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.
Looking at our graph, we can tell that the period is
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
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Which of the following is the graph of 
?
Which of the following is the graph of 
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 
.
In order to graph 
Now anywhere this graph crosses the x-axis a vertical asymptote will form for the 
And then we are left with the graph of 
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Considering the general form of the cosecant transformation function 
, what does each letter, (A, B, C, and D) correspond to?
Considering the general form of the cosecant transformation function 
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.
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.
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Which of the following is the graph of 
?
Which of the following is the graph of 
Knowing that the graph of 
is
we can use the general form of the cosecant transformation equation, 
, and apply these transformations. We can ignore 
because in this case 
. In this case 
and so our period is:
Period = 
Period = 
Period = 
is the normal period for cosecant graphs and so we do not have to worry about lengthening or shortening the period. 
and so we need to apply a phase shift of 
. This will cause our graph to shift left a total of 
units.
Lastly, we must apply the transformation for 
, so we will have an upward vertical shift of 1 unit.
The application of these transformations leaves us with our graph of 
.
Knowing that the graph of 
we can use the general form of the cosecant transformation equation, 
Period =
Period =
Period =
Lastly, we must apply the transformation for 
The application of these transformations leaves us with our graph of 
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Which of the following is the graph for 
?
Which of the following is the graph for 
In order to understand the graph of secant, recall that 
. First consider the graph of 
.
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 
.
In order to understand the graph of secant, recall that 
Anywhere this graph crosses the x-axis a vertical asymptote will form for the 
And then we are left with the graph 
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True or False: Amplitude must be considered when graphing the transformation of a secant graph.
True or False: Amplitude must be considered when graphing the transformation of a secant graph.
When looking at the graph of 
, it extends infinitely upwards and downwards from each local maximum and minimum. This will be true for all transformed secant graphs as well. Due to this, there is no amplitude for secant graphs. However, secant is the reciprocal of cosine graphs which do rely on amplitude for transformations. For this reason amplitude must be considered as a vertical shift.
When looking at the graph of 
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Which of the following is the graph of 
?
Which of the following is the graph of 
Knowing that the general form of the graph 
is:
We can use the general form of the cosecant transformation equation, 
, and apply these transformations. 
because secant graphs extend infinitely upwards and downwards and does not have an amplitude, we must think of the secant graph being a reciprocal of the cosine graph. So we will consider 
for cosine.
We will shift our secant graph to invert at the maximums and minimums of the cosine graph.
Next, we will factor 
in order to get our equation into the form 
.
And so 
. We can now solve for our period,
Period = 
Period = 
Period = 
This shortens our original period of 
to 
.
Now we must consider 
. This will give us a phase shift of 
units to the left. Since our period has also been shortened this does not change the graph visually. 
in this case so we do not need to consider a vertical shift.
And we are left with the graph of 
.
Knowing that the general form of the graph 
We can use the general form of the cosecant transformation equation, 
We will shift our secant graph to invert at the maximums and minimums of the cosine graph.
Next, we will factor 
And so 
Period =
Period =
Period =
This shortens our original period of 
Now we must consider 
And we are left with the graph of 
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This is the graph of what function?
This is the graph of what function?
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 of the sine function is increased by 3, so this is the coefficient for
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The function shown below has an amplitude of ___________ and a period of _________.
The function shown below has an amplitude of ___________ and a period 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 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
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Which of the following graphs does not have a 
-intercept at 
?
Which of the following graphs does not have a
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 y-intercept is the value of y when
Recall that cosine is the
Secant is the reciprocal of cosine, so it also works.
Also recall that
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What is an equation for the above function, enlarged below?
What is an equation for the above function, enlarged below?
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 amplitude of a sinusoidal function is
The graph moves through the origin, so it is either a sine or a shifted cosine graph.
It repeats once in every
The only answer in which both the correct amplitude and period is found is:
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Which graph correctly illustrates the given equation?
Which graph correctly illustrates the given equation?
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 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:
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