Find the Phase Shift of a Sine or Cosine Function - Pre-Calculus

This is the graph of sine, but shifted to the right units. To reflect this shift, should be subtracted from x.

Thus resulting in

.

The graph has an amplitude of 2 but has been shifted down 1:

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts \[is at 0\] where .

The phase shift is to the right, or .

Since is being added inside the parentheses, there will be a horizontal shift. The goal is to maintain zero within the parentheses so you will shift left radians.

To write this equation, it is helpful to sketch a graph:

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

The distance from the maximum to the minimum point is half the wavelength. In this case, the wavelength is . That means the full wavelength is , and the frequency is 1.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract from the maximum x-coordinate, :

.

Our equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

.

In order to write this equation, it is helpful to sketch a graph:

The dotted line is at , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right .

The distance from the maximum to the minimum is half the entire wavelength. Here it is .

Since half the wavelength is , that means the full wavelength is so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

.

The equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

To write the equation, it is helpful to sketch a graph:

From plotting the maximum and minimum, we can see that the graph is centered on with an amplitude of 3.

The distance from the maximum to the minimum is half the wavelength. For this graph, this distance is .

This means that the total wavelength is and the frequency is 1.

The graph starts behind the maximum point. To determine this x value, subtract from the x-coordinate of the maximum:

Our equation is:

.

In the formula,

.

represents the phase shift.

Plugging in what we know gives us:

.

Simplified, the phase is then .

This is the graph of sine, but shifted to the right units. To reflect this shift, should be subtracted from x.

Thus resulting in

.

The graph has an amplitude of 2 but has been shifted down 1:

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts \[is at 0\] where .

The phase shift is to the right, or .

Since is being added inside the parentheses, there will be a horizontal shift. The goal is to maintain zero within the parentheses so you will shift left radians.

To write this equation, it is helpful to sketch a graph:

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

The distance from the maximum to the minimum point is half the wavelength. In this case, the wavelength is . That means the full wavelength is , and the frequency is 1.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract from the maximum x-coordinate, :

.

Our equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

.

In order to write this equation, it is helpful to sketch a graph:

The dotted line is at , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right .

The distance from the maximum to the minimum is half the entire wavelength. Here it is .

Since half the wavelength is , that means the full wavelength is so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

.

The equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

To write the equation, it is helpful to sketch a graph:

From plotting the maximum and minimum, we can see that the graph is centered on with an amplitude of 3.

The distance from the maximum to the minimum is half the wavelength. For this graph, this distance is .

This means that the total wavelength is and the frequency is 1.

The graph starts behind the maximum point. To determine this x value, subtract from the x-coordinate of the maximum:

Our equation is:

.

In the formula,

.

represents the phase shift.

Plugging in what we know gives us:

.

Simplified, the phase is then .

This is the graph of sine, but shifted to the right units. To reflect this shift, should be subtracted from x.

Thus resulting in

.

The graph has an amplitude of 2 but has been shifted down 1:

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts \[is at 0\] where .

The phase shift is to the right, or .

Since is being added inside the parentheses, there will be a horizontal shift. The goal is to maintain zero within the parentheses so you will shift left radians.

To write this equation, it is helpful to sketch a graph:

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

The distance from the maximum to the minimum point is half the wavelength. In this case, the wavelength is . That means the full wavelength is , and the frequency is 1.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract from the maximum x-coordinate, :

.

Our equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

.

In order to write this equation, it is helpful to sketch a graph:

The dotted line is at , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right .

The distance from the maximum to the minimum is half the entire wavelength. Here it is .

Since half the wavelength is , that means the full wavelength is so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

.