MHB How do I find A,B,C, and D in a sinusoidal function?

[Randi]

I really need someone to break it down for me. I think I understand A and D, but I am confused on B and C. I have some example problems. But first, the equation my pre-calculus teacher has given us is y=Asin(2π/B(θ-C))+D. But I am still having a lot of trouble.

Find amplitude, period, a phase shift, and the mean of the following sinusodial functions.
a.) y=sin(2x-π)+1
b.) y=6sin(πx)-1

The answers to a.) are 1. π. π/2. 1. and the answers to b.) are 6. 2. 0. -1.

I just don't understand how these answers were found. PLEASE HELP.

 

[MarkFL]

If we use the form:

$$y=A\sin\left(B(x-C)\right)+D$$

The amplitude is defined as $|A|$.

The period is:

$$T=\frac{2\pi}{B}$$

The phase shift is $C$. This comes from the horizontal shift of a function.

Finally, let's look at the mean. We may begin with:

$$-1\le\sin(\theta)\le1$$

Multiply through by $0\le A$ (if $A$ is negative, we would just reverse the inequality):

$$-A\le A\sin(\theta)\le A$$

Add through by $D$:

$$D-A\le A\sin(\theta)+D\le D+A$$

And so the mean will be the average of the boundaries:

$$\frac{(D-A)+(D+A)}{2}=\frac{2D}{2}=D$$

Now, for the first problem, we may write it as:

$$y=1\cdot\sin\left(2\left(x-\frac{\pi}{2}\right)\right)+1$$

So, we identify:

$$A=1,\,B=2,\,C=\frac{\pi}{2},\,D=1$$

And so the values are:

Amplitude: $$|1|=1$$

Period: $$T=\frac{2\pi}{2}=\pi$$

Phase shift: $$C=\frac{\pi}{2}$$

Mean: $$D=1$$

Now, see if you can do the second one. :D