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

Jul 12, 2015

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 $y = \frac{4}{3} \sin \left(\frac{2}{3}\right) x = 0.824 x$ ot's a linnear equation and it has no period nor amplitude, it's a linnear function.

If your equation is $y = \frac{4}{3} \sin \left(\frac{2}{3} x\right)$ 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 $3 \pi$, for x=0 the equation's value is 4/3, for $y = \frac{4}{3} \sin \left(2 \pi\right)$ 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 $2 \pi = \frac{2}{3} x$ you get $x = 3 \pi$ and $3 \pi - 0 = 3 \pi$ then $T = 3 \pi$