Question

What is the rule of a function of the form f(t)= a sin (bt+c) +d whose graph appears to be identical to the given graph?

Answer 1

The function which represent the graph that is given to us is:

a.
`-4\sin ((1)/(3)t+(\pi)/(6))+3`

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)

`-4\sin ((1)/(3)t-(\pi)/(6))-3`

when t=0 we have:

`-4\sin (0-(\pi)/(6))-3\\\\\\=-4\sin (-(\pi)/(6))-3\\\\\\=4\sin ((\pi)/(6))-3\\\\\\=4* (1)/(2)-3\\\\\\=2-3\\\\\\-1\\eq 1`

Hence, option: b is incorrect.

c)

`4\sin ((1)/(3)t+(\pi)/(6))-3`

when t=0 we have:

`4\sin (0+(\pi)/(6))-3\\\\\\=4\sin ((\pi)/(6))-3\\\\\\=4* (1)/(2)-3\\\\\\=2-3\\\\\\-1\\eq 1`

Hence, option: c is incorrect.

d)

`-4\sin (3t+(\pi)/(6))+3`

The period of the function is:

`(2\pi)/(3)\\eq 6\pi`

since the general function of the type:

`f(t)=a\sin (bt+c)+d`

The period of the function is given by:

`(2\pi)/(b)`

Hence, option: d is incorrect.

Hence, we are left with option: a

a)

`-4\sin ((1)/(3)t+(\pi)/(6))+3`

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.

Answer 2

Option A.

f(x) = -4*sin((1/3)*t + (π/6)) + 3

Explanation:

We can easily solve this problem by using a graphing calculator or plotting tool.

The function is

f(t) = a*sin (b*t +c) + d

Please, see attached picture below.

By looking at the picture with all the possible cases, we can tell that the correct option is A.

The function has a period of T = 6π

Max . Amplitude = 7

Min . Amplitude = -1