Question

NNEED HELP!! GEETSS BRAINLIEST**

Answer 1

Answer:

D. $y=3\cos \left(\dfrac{4}{3}x-2\pi\right)+2$

Step-by-step explanation:

The period of the functions $y=\cos x,\ y=\sin x$ is $2\pi.$

The period of the function $y=a\cos (kx+b),\ y=a\sin(kx+b)$ is ALWAYS $\dfrac{2\pi}{k}$

In your case, you have function $y=-3\sin \left(\dfrac{2}{3}x-2\pi \right)+2$ and this function has the period

$\dfrac{2\pi}{\dfrac{2}{3}}=3\pi.$

You need to find the function that will have the period that is half of $3\pi,$ so

$\dfrac{3\pi}{2}=\dfrac{2\pi}{k}\\ \\3k=4\\ \\k=\dfrac{4}{3}.$

So, correct choice is

$y=3\cos \left(\dfrac{4}{3}x-2\pi\right)+2$

Answer 2

Answer:

The answer is y = 3 cos(4/3 x - 2π) + 2 ⇒ last answer

Step-by-step explanation:

* Lets revise the sine function

- If we have a sine function of the form f(x) = Asin(Bx + C) + D, where

 A, B , C and D are constant, then

# Amplitude is A

- The Amplitude is the height from the center line to the peak .

 Or we can measure the height from highest to lowest points and

 divide that by 2

# Period is 2π/B

- The period goes from one peak to the next

# phase shift is C (positive is to the left)

- The Phase Shift is how far the function is shifted horizontally  

 from the usual position.

# vertical shift is D

- The Vertical Shift is how far the function is shifted vertically from

 the usual position.

* Now lets solve the problem

∵ y = -3sin(2/3 x - 2π) + 2

- the period is 2π ÷ 2/3 = 2π × 3/2 = 3π

∴ The period of the function is 3π

- We look for a function has one-half (3π), means 3π/2

* Lets look to the answer to find the right one

- All of them have the same value of B except the last one, lets

 check it

∵ y = 3cos(4/3 x - 2π) + 2

∵ B = 4/3

∴ The period = 2π ÷ 4/3 = 2π × 3/4 = 6π/4 = 3π/2

∵ 3π/2 is half 3π

∴ The last answer is right