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Identifying the variations of a sinusoidal function from an equation

Determine the midline, amplitude, period, and phase shift of the function y = 3 sin ( 2 x ) + 1.

Let’s begin by comparing the equation to the general form y = A sin ( B x C ) + D .

A = 3 , so the amplitude is | A | = 3.

Next, B = 2 , so the period is P = 2 π | B | = 2 π 2 = π .

There is no added constant inside the parentheses, so C = 0 and the phase shift is C B = 0 2 = 0.

Finally, D = 1 , so the midline is y = 1.

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Determine the midline, amplitude, period, and phase shift of the function y = 1 2 cos ( x 3 π 3 ) .

midline: y = 0 ; amplitude: | A | = 1 2 ; period: P = 2 π | B | = 6 π ; phase shift: C B = π

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Identifying the equation for a sinusoidal function from a graph

Determine the formula for the cosine function in [link] .

A graph of -0.5cos(x)+0.5. The graph has an amplitude of 0.5. The graph has a period of 2pi. The graph has a range of [0, 1]. The graph is also reflected about the x-axis from the parent function cos(x).

To determine the equation, we need to identify each value in the general form of a sinusoidal function.

y = A sin ( B x C ) + D y = A cos ( B x C ) + D

The graph could represent either a sine or a cosine function    that is shifted and/or reflected. When x = 0 , the graph has an extreme point, ( 0 , 0 ) . Since the cosine function has an extreme point for x = 0 , let us write our equation in terms of a cosine function.

Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below y = 0.5. This value, which is the midline, is D in the equation, so D = 0.5.

The greatest distance above and below the midline is the amplitude. The maxima are 0.5 units above the midline and the minima are 0.5 units below the midline. So | A | = 0.5. Another way we could have determined the amplitude is by recognizing that the difference between the height of local maxima and minima is 1, so | A | = 1 2 = 0.5. Also, the graph is reflected about the x -axis so that A = 0.5.

The graph is not horizontally stretched or compressed, so B = 1; and the graph is not shifted horizontally, so C = 0.

Putting this all together,

g ( x ) = 0.5 cos ( x ) + 0.5
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Determine the formula for the sine function in [link] .

A graph of sin(x)+2. Period of 2pi, amplitude of 1, and range of [1, 3].

f ( x ) = sin ( x ) + 2

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Identifying the equation for a sinusoidal function from a graph

Determine the equation for the sinusoidal function in [link] .

A graph of 3cos(pi/3x-pi/3)-2. Graph has amplitude of 3, period of 6, range of [-5,1].

With the highest value at 1 and the lowest value at −5 , the midline will be halfway between at −2. So D = −2.

The distance from the midline to the highest or lowest value gives an amplitude of | A | = 3.

The period of the graph is 6, which can be measured from the peak at x = 1 to the next peak at x = 7 , or from the distance between the lowest points. Therefore, P = 2 π | B | = 6. Using the positive value for B , we find that

B = 2 π P = 2 π 6 = π 3

So far, our equation is either y = 3 sin ( π 3 x C ) 2 or y = 3 cos ( π 3 x C ) 2. For the shape and shift, we have more than one option. We could write this as any one of the following:

  • a cosine shifted to the right
  • a negative cosine shifted to the left
  • a sine shifted to the left
  • a negative sine shifted to the right

While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes

y = 3 cos ( π 3 x π 3 ) 2  or  y = 3 cos ( π 3 x + 2 π 3 ) 2

Again, these functions are equivalent, so both yield the same graph.

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Practice Key Terms 5

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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