# How do you write an equation of a sine function with amplitude 4, period pi, phase shift pi/2 to the right, and vertical displacement 6 units down?

Jan 23, 2017

$y = 4 \sin \left[2 \left(x - \frac{\pi}{2}\right)\right] - 6$.

#### Explanation:

The standard form of a sine function is

$y = a \sin \left[b \left(x - h\right)\right] + k$

where

• $a$ is the amplitude,
• $\frac{2 \pi}{b}$ is the period,
• $h$ is the phase shift, and
• $k$ is the vertical displacement.

We start with classic $y = \sin x$:
graph{(y-sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
(The circle at (0,0) is for a point of reference.)

The amplitude of this function is $a = 1$. To make the amplitude 4, we need $a$ to be 4 times as large, so we set $a = 4$.

Our function is now $y = 4 \sin x$, and looks like:
graph{(y-4sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
The period of this function—the distance between repetitions—right now is $2 \pi$, with $b = 1$. To make the period $\pi$, we need to make the repetitions twice as frequent, so we need $b = \left[\text{normal period"]/["desired period}\right] = \frac{2 \pi}{\pi} = 2$.

Our function is now $y = 4 \sin \left(2 x\right)$, and looks like:
graph{(y-4sin(2x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}

This function currently has no phase shift, since $h = 0$. To induce a phace shift, we need to offset $x$ by the desired amount, which in this case is $\frac{\pi}{2}$ to the right. A phase shift right means a positive $h$, so we set $h = \frac{\pi}{2}$.

Our function is now $y = 4 \sin \left[2 \left(x - \frac{\pi}{2}\right)\right]$, and looks like:
graph{(y-4sin(2(x-pi/2)))((x-pi/2)^2+y^2-0.075)=0 [-15, 15, -11, 5]}
Finally, the function currently has no vertical displacement, since $k = 0$. To displace the graph 6 units down, we set $k = - 6$.

Our function is now $y = 4 \sin \left[2 \left(x - \frac{\pi}{2}\right)\right] - 6$, and looks like:
graph{(y-4sin(2(x-pi/2))+6)((x-pi/2)^2+(y+6)^2-0.075)=0 [-15, 15, -11, 5]}