# How do you find the amplitude, period, and phase shift of y=sin 4x + 5?

Mar 8, 2018

Amplitude = 1, Period = $\frac{\pi}{2}$,

Phase shift $= - \left(\frac{5}{4}\right)$, $\textcolor{w h i t e}{a a a} \left(\frac{5}{4}\right)$ to the left

#### Explanation:

$y = \sin \left(4 x + 5\right)$

$A m p l i t u \mathrm{de} = a = 1$

$P e r i o d = \frac{2 \pi}{|} b | = \frac{2 \pi}{4} = \frac{\pi}{2}$

Phase shift $= - \frac{c}{b} = - \frac{5}{4}$, $\textcolor{w h i t e}{a a a} \left(\frac{5}{4}\right)$ to the left

graph{sin (4x + 5) [-10, 10, -5, 5]}