Question

Little help?
state the amplitude period and transformation of the function
y=2-3sin(pi/3x-pi)

Answer 1

\[y=2-3\sin (\frac{\pi}{3}x-\pi)\]
Is that it?

Answer 2

yes it is

Answer 3

The general equation is:
\[y=A \sin (Bx+C)+D\]
Amplitude = |A|
\[Period\ T=\frac{2 \pi}{B}\]
\[Phase\ shift=\frac{C}{B}\]
\['y'\ shift=D\]
Using that, can you identify the amplitude?

Answer 4

ok is the amp -3?

Answer 5

It might help if I rearrange the given equation as follows:
\[y=-3\sin (\frac{\pi}{3}x-\pi)+2\]

Answer 6

The amplitude is |-3| = ?

Answer 7

3 got it thanks!!

Answer 8

so the period is 2pi/pi/3?

Answer 9

Good work! Can you simplify it?

Answer 10

so its 6 and the horizontal shift is pi?

Answer 11

The period is 6.
The phase shift is:
\[\frac{- \pi}{\frac{\pi}{3}}\]

Answer 12

ok thanks got it!

Answer 13

Good work!
You're welcome :)