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

Answer:

Option A.

f(x) = -4*sin((1/3)*t + (π/6))  + 3

Step-by-step 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

Answer 2

Answer:

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

           a.   $-4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})+3$

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)

$-4\sin (\dfrac{1}{3}t-\dfrac{\pi}{6})-3$

when t=0 we have:

$-4\sin (0-\dfrac{\pi}{6})-3\\\\\\=-4\sin (-\dfrac{\pi}{6})-3\\\\\\=4\sin (\dfrac{\pi}{6})-3\\\\\\=4\times \dfrac{1}{2}-3\\\\\\=2-3\\\\\\-1\neq 1$

Hence, option: b is incorrect.

c)

$4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})-3$

when t=0 we have:

$4\sin (0+\dfrac{\pi}{6})-3\\\\\\=4\sin (\dfrac{\pi}{6})-3\\\\\\=4\times \dfrac{1}{2}-3\\\\\\=2-3\\\\\\-1\neq 1$

Hence, option: c is incorrect.

d)

$-4\sin (3t+\dfrac{\pi}{6})+3$

The period of the function is:

$\dfrac{2\pi}{3}\neq 6\pi$

since the general function of the type:

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

The period of the function is given by:

$\dfrac{2\pi}{b}$

Hence, option: d is incorrect.

Hence, we are left with option: a

a)

 $-4\sin (\dfrac{1}{3}t+\dfrac{\pi}{6})+3$

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.