# How do you write an equation of a sine function with amplitude 0.5, period 4pi, phase shift pi/6 to the left, vertical displacement 1 unit up?

Jan 2, 2018

$y = \frac{1}{2} \sin \left(\frac{1}{2} x + \frac{\pi}{12}\right) + 1$

#### Explanation:

We can express trig function in the following way:

$y = a \sin \left(b x + c\right) + d$

Where:

Amplitude = a.

Period $= \frac{2 \pi}{b}$. ( $2 \pi$ is the normal period of sin(x) )

Phase shift $= - \frac{c}{b}$.

Vertical shift $= d$.

For our example:

$a = \frac{1}{2}$

Period needs to be $\left(4 \pi\right)$

$\therefore$

$\frac{2 \pi}{b} = 4 \pi \implies b = \frac{1}{2}$

Phase shift needs to be $\frac{\pi}{6}$

$: ,$

$- \frac{c}{\frac{1}{2}} = \frac{\pi}{6} \implies c = - \frac{\pi}{12}$ ( this is $\frac{\pi}{12}$ for shift to the left )

Vertical shift needs to be 1

$d = 1$

So our equation is:

$y = \frac{1}{2} \sin \left(\frac{1}{2} x + \frac{\pi}{12}\right) + 1$