Answer:
The function which represent the graph that is given to us is:
a. [tex]-4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})+3[/tex]
Step-by-step explanation:
By looking at the function we see that the period of the function is 6π i.e. it repeats itself after every 6π value.
Also, when t=0 the function attains the value 1.
So, we will check in each of the options whose period is 3 and which attains the value 1 when t=0
b)
[tex]-4\sin (\dfrac{1}{3}t-\dfrac{\pi}{6})-3[/tex]
when t=0 we have:
[tex]-4\sin (0-\dfrac{\pi}{6})-3\\\\\\=-4\sin (-\dfrac{\pi}{6})-3\\\\\\=4\sin (\dfrac{\pi}{6})-3\\\\\\=4\times \dfrac{1}{2}-3\\\\\\=2-3\\\\\\-1\neq 1[/tex]
Hence, option: b is incorrect.
c)
[tex]4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})-3[/tex]
when t=0 we have:
[tex]4\sin (0+\dfrac{\pi}{6})-3\\\\\\=4\sin (\dfrac{\pi}{6})-3\\\\\\=4\times \dfrac{1}{2}-3\\\\\\=2-3\\\\\\-1\neq 1[/tex]
Hence, option: c is incorrect.
d)
[tex]-4\sin (3t+\dfrac{\pi}{6})+3[/tex]
The period of the function is:
[tex]\dfrac{2\pi}{3}\neq 6\pi[/tex]
since the general function of the type:
[tex]f(t)=a\sin (bt+c)+d[/tex]
The period of the function is given by:
[tex]\dfrac{2\pi}{b}[/tex]
Hence, option: d is incorrect.
Hence, we are left with option: a
a)
[tex]-4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})+3[/tex]
The period of this function is: 6π
and at x=0 the value of function is 1.
Also, the graph of this function matches the given graph.
