# How do you find the amplitude, period, and shift for y=sin (3x-pi/4)?

Oct 20, 2016

$A m p l i t u \mathrm{de} = 1$
$P e r i o d = \frac{2 \pi}{3}$
phase (shift) $= \frac{\pi}{4}$ to the right

#### Explanation:

Let $y = a \sin \left(b x + c\right)$
The amplitude is the coefficient $a$ of the sine, here it is 1
The period depends on the coefficient b of x, So here we have $3$
So the peroid $= \frac{2 \pi}{3}$, if the coefficient was 1, the period is $2 \pi$
The phase shift depends on the value and sign of c: Here it is $- \frac{\pi}{4}$ So it is $\frac{\pi}{4}$ to the right