6.1 Graphs of the sine and cosine functions  (Page 4/13)

Let’s begin by comparing the equation to the general form $y=A\sin (Bx-C)+D.$

$A=3,$ so the amplitude is $\left|A\right|=3.$

Next, $B=2,$ so the period is $P=\frac{2\pi }{\left|B\right|}=\frac{2\pi }{2}=\pi .$

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

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

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

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, $\left(0,0\right).$ 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 $y=\mathrm{0.5.}$ This value, which is the midline, is $D$ in the equation, so $D=\mathrm{0.5.}$

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 $\left|A\right|=\mathrm{0.5.}$ 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 $\left|A\right|=\frac{1}{2}=\mathrm{0.5.}$ Also, the graph is reflected about the x -axis so that $A=-\mathrm{0.5.}$

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

Putting this all together,

With the highest value at 1 and the lowest value at $\mathrm{-5},$ the midline will be halfway between at $\mathrm{-2.}$ So $D=\mathrm{-2.}$

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

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, $P=\frac{2\pi }{\left|B\right|}=6.$ Using the positive value for $B,$ we find that

So far, our equation is either $y=3\sin \left(\frac{\pi }{3}x-C\right)-2$ or $y=3\cos \left(\frac{\pi }{3}x-C\right)-2.$ 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

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