Let’s begin by comparing the equation to the general form [latex]y=A\sin(Bx)[/latex].

In the given equation, [latex]B =\frac{π}{6}[/latex], so the period will be

[latex]\begin{align}P&=\frac{\frac{2}{\pi}}{|B|} \\ &=\frac{2\pi}{\frac{x}{6}} \\ &=2\pi\times \frac{6}{\pi} \\ &=12 \end{align}[/latex]

Let’s begin by comparing the function to the simplified form [latex]y=A\sin(Bx)[/latex].

In the given function, A = −4, so the amplitude is |A|=|−4| = 4. The function is stretched.

Analysis of the Solution

The negative value of A results in a reflection across the x-axis of the sine function, as shown in Figure 10.

Figure 10

Let’s begin by comparing the equation to the general form [latex]y=A\sin(Bx−C)+D[/latex].

In the given equation, notice that B = 1 and [latex]C=−\frac{π}{6}[/latex]. So the phase shift is

[latex]\begin{align}\frac{C}{B}&=−\frac{\frac{x}{6}}{1} \\ &=−\frac{\pi}{6} \end{align}[/latex]

or [latex]\frac{\pi}{6}[/latex] units to the left.

Analysis of the Solution

We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before C. Therefore [latex]f(x)=\sin(x+\frac{π}{6})−2[/latex] can be rewritten as [latex]f(x)=\sin(x−(−\frac{π}{6}))−2[/latex]. If the value of C is negative, the shift is to the left.

Let’s begin by comparing the equation to the general form [latex]y=A\cos(Bx−C)+D[/latex]. In the given equation, [latex]D=-3[/latex], so the shift is 3 units downward.

Let’s begin by comparing the equation to the general form [latex]y=A\sin(Bx−C)+D[/latex]. A = 3, so the amplitude is |A| = 3.

Next, B = 2, so the period is [latex]P=\frac{2π}{|B|}=\frac{2π}{2}=π[/latex].

There is no added constant inside the parentheses, so C = 0 and the phase shift is [latex]\frac{C}{B}=\frac{0}{2}=0[/latex].

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

Analysis of the Solution

Inspecting the graph, we can determine that the period is π, the midline is y = 1,and the amplitude is 3. See Figure 14.

Figure 14

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

[latex]y=A\sin\left(Bx-C\right)+D[/latex]

[latex]y=A\cos\left(Bx-C\right)+D[/latex]

The graph could represent either a sine or a cosine function that is shifted and/or reflected. When [latex]x=0[/latex], the graph has an extreme point, [latex](0,0)[/latex]. Since the cosine function has an extreme point for [latex]x=0[/latex], 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 [latex]y=0.5[/latex]. 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|=[latex]\frac{1}{2}[/latex]. Also, the graph is reflected about the x-axis so that A=0.5.

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

Putting this all together,

[latex]g(x)=0.5\cos\left(x\right)+0.5[/latex]

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, [latex]\text{P}=\frac{2\pi}{|B|}=6[/latex]. Using the positive value for B, we find that

[latex]B=\frac{2π}{P}=\frac{2π}{6}=\frac{π}{3}[/latex]

So far, our equation is either [latex]y=3\sin(\frac{\pi}{3}x−C)−2[/latex] or [latex]y=3\cos(\frac{\pi}{3}x−C)−2[/latex]. 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

[latex]y=3\cos(\frac{π}{3}x−\frac{π}{3})−2[/latex] or [latex]y=−3\cos(\frac{π}{3}x+\frac{2π}{3})−2[/latex]

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

Let’s begin by comparing the equation to the form [latex]y=A\sin(Bx)[/latex].

Step 1. We can see from the equation that A=−2,so the amplitude is 2.

|A| = 2

Step 2. The equation shows that [latex]B=\frac{π}{2}[/latex], so the period is

[latex] \begin{align}P&=\frac{2\pi}{\frac{\pi}{2}}\\&=2\pi\times\frac{2}{\pi}\\&=4 \end{align}[/latex]

Step 3. Because A is negative, the graph descends as we move to the right of the origin.

Step 4–7. The x-intercepts are at the beginning of one period, x = 0, the horizontal midpoints are at x = 2 and at the end of one period at x = 4.

The quarter points include the minimum at x = 1 and the maximum at x = 3. A local minimum will occur 2 units below the midline, at x = 1, and a local maximum will occur at 2 units above the midline, at x = 3. Figure 19 shows the graph of the function.

Step 1. The function is already written in general form: [latex]f(x)=3\sin\left(\frac{π}{4}x−\frac{π}{4}\right)[/latex]. This graph will have the shape of a sine function, starting at the midline and increasing to the right.

Step 2. |A|=|3|=3. The amplitude is 3.

Step 3. Since [latex]|B|=|\frac{π}{4}|=\frac{π}{4}[/latex], we determine the period as follows.

[latex]P=\frac{2π}{|B|}=\frac{2π}{\frac{π}{4}}=2π\times\frac{4}{π}=8[/latex]

The period is 8.

Step 4. Since [latex]\text{C}=\frac{π}{4}[/latex], the phase shift is

[latex]\frac{C}{B}=\frac{\frac{\pi}{4}}{\frac{\pi}{4}}=1[/latex].

The phase shift is 1 unit.

Step 5. Figure 20 shows the graph of the function.

Begin by comparing the equation to the general form and use the steps outlined in Example 9.

[latex]y=A\cos(Bx−C)+D[/latex]

Step 1. The function is already written in general form.

Step 2. Since A = −2, the amplitude is|A| = 2.

Step 3. [latex]|B|=\frac{\pi}{2}[/latex], so the period is [latex]P=\frac{2π}{|B|}=\frac{2\pi}{\frac{\pi}{2}}\times2\pi=4[/latex]. The period is 4.

Step 4. [latex]C=−\pi[/latex], so we calculate the phase shift as [latex]\frac{C}{B}=\frac{−\pi}{\frac{\pi}{2}}=−\pi\times\frac{2}{\pi}=−2[/latex]. The phase shift is −2.

Step 5. D = 3, so the midline is y = 3, and the vertical shift is up 3.

Since A is negative, the graph of the cosine function has been reflected about the x-axis.

Figure 21 shows one cycle of the graph of the function.

Recall that, for a point on a circle of radius r, the y-coordinate of the point is [latex]y=r\sin(x)[/latex], so in this case, we get the equation [latex]y(x)=3\sin(x)[/latex]. The constant 3 causes a vertical stretch of the y-values of the function by a factor of 3, which we can see in the graph in Figure 22.

Analysis of the Solution

Notice that the period of the function is still 2π; as we travel around the circle, we return to the point (3,0) for [latex]x=2\pi,4\pi,6\pi,\dots[/latex] Because the outputs of the graph will now oscillate between –3 and 3, the amplitude of the sine wave is 3.

Sketching the height, we note that it will start 1 ft above the ground, then increase up to 7 ft above the ground, and continue to oscillate 3 ft above and below the center value of 4 ft, as shown in Figure 24.

Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.

Second, we see that the graph oscillates 3 above and below the center, while a basic cosine has an amplitude of 1, so this graph has been vertically stretched by 3, as in the last example.

Finally, to move the center of the circle up to a height of 4, the graph has been vertically shifted up by 4. Putting these transformations together, we find that

[latex]y=−3\cos(x)+4[/latex]

With a diameter of 135 m, the wheel has a radius of 67.5 m. The height will oscillate with amplitude 67.5 m above and below the center.

Passengers board 2 m above ground level, so the center of the wheel must be located 67.5 + 2 = 69.5 m above ground level. The midline of the oscillation will be at 69.5 m.

The wheel takes 30 minutes to complete 1 revolution, so the height will oscillate with a period of 30 minutes.

Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve.

• Amplitude: 67.5, so A = 67.5
• Midline: 69.5, so D = 69.5
• Period: 30, so [latex]B=\frac{2\pi}{30}=\frac{\pi}{15}[/latex]
• Shape: −cos(t)

An equation for the rider’s height would be

[latex]y=−67.5\cos\left(\frac{\pi}{15}t\right)+69.5[/latex]

where t is in minutes and y is measured in meters.