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?

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Answer 1 · 2018-01-02T19:59:37

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

We can express trig function in the following way:

$y = a sin (b x + c) + d$ $y=asin(bx+c)+d$

Where:

Amplitude = a.

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

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

Vertical shift $= d$ $= d$ .

For our example:

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

Period needs to be $(4 π)$ $(4pi)$

$:.$

$(2pi)/b=4pi=>b=1/2$

Phase shift needs to be $pi/6$

$:,$

$-c/(1/2)=pi/6=>c=-pi/12$ ( this is $pi/12$ for shift to the left )

Vertical shift needs to be 1

$d=1$

So our equation is:

$y=1/2sin(1/2x+pi/12)+1$