Graph two complete cycles of your function.
Answers are not unique.
a. Period of 6 hours with a minimum value of -2 at t=0 hr and a maximum value of 2 at t=3 hr.
b. Period of 24 hours with a maximum value of 20 at t=6 hr and a minimum value of 10 at t=18 hr.
I'm confused because what does this have to do with hours? Aren't the units supposed to be in radians? Also, do I just follow the $f(x)=A(\sin Bx+C)+D$ form?
Any help would be appreciated. :)
It is actually very easy if you understand how the periodic functions operate:
A sine function is in range $[0 \; 2\pi]$ with period $2\pi$ but in the question we see the period is mentioned 6 hours. For a sine function with period of 6, it has to finish its period $2\pi$ in 6 hours. So the function is:
$\sin(2\pi t / 6)$
If you plot this function the domain (on the y axis) of sine function is $[-1 \; 1]$ but in the question we see -2 and 2 so we have to multiply the function by 2 to increase the domain to $[-2 \; 2]$:
$2\sin(2\pi t / 6)$
We can also simplify the sine function as follows:
$2\sin(\pi t / 3)$
now it is only the mater of finding the phase. We can see that in the question it has been mentioned that at time 0 the output is -2. In a standard $\sin(x)$ the value in $x=0$ should be 0 so we can determine that the phase is $\pi/2$. Therefore the final function would be
$2\sin(\pi t / 3 + \pi/2) $
I hope you can figure out the other question by yourself.