Question

How do you find the amplitude and period of `y= 4/3 sin (2/3)x`?

Answer 1

This question has no sense to me, the x should be INSIDE the sinusfor being a periodic function. If the x is really out the parentesis then is a tricky question. It has no period nor amplitude..

Explanation:

If you really wrote it correctly then `y=4/3sin(2/3)x=0.824x` ot's a linnear equation and it has no period nor amplitude, it's a linnear function.

If your equation is `y=4/3sin(2/3x)` then you need to know that the sin oscillates between -1 and 1 so if it's multipliyed by 4/3 your equaion is oscillating between 4/3 and -4/3 so you have you amplitude, it's 4/3.

for the period it is `3pi`, for x=0 the equation's value is 4/3, for `y=4/3sin(2pi)` you again have y=4/3 betwwen this two point there's no other point with x=4/3 so between these two has passed a period.

The difference between the x's will be the period, then isolating x form `2pi=2/3x` you get `x=3pi` and `3pi-0=3pi` then `T=3pi`