We are asked to write a sine function that has an amplitude of $4,$ period of $3 π,$ phase shift of $π,$ and vertical shift of $- 5 .$ Let's first review the properties of sine functions!

Properties of Sine Functions
Suppose $y = a sin b x,$ with $a \neq = 0,$ $b > 0,$ and $x$ in radians. $∙ ∙ ∙ ∣ a ∣ is the amplitude of the function b is the number of cycles in the interval from 0 to 2 π \frac{2 π}{b} is the period of the function$

We will start by writing the sine function prior to any translation. Since we want the amplitude of our function to be

4,

we can choose

a

to be either

4

or

- 4 .

We will arbitrarily choose it to be

4 .

y = 4 sin b x

Next, we will find

b .

We are given that the period is

3 π,

so we can set it equal to

\frac{2 π}{b}

and solve for

b .

Period = \frac{2 π}{b}

Substitute

$Period = 3 π$

3 π = \frac{2 π}{b}

▼

Solve for

b

MultEqn

$LHS \cdot b = RHS \cdot b$

3 π \cdot b = 2 π

DivEqn

$LHS / 3 π = RHS / 3 π$

b = \frac{2 π}{3 π}

CrossCommonFac

Cross out common factors

b = \frac{2 π}{3 π}

SimpQuot

Simplify quotient

b = \frac{2}{3}

We can now substitute this value into our function.

y = 4 sin \frac{2}{3} x

Finally, we will translate the function. Let's review the possible translations for sine and cosine functions.

Translations of Sine and Cosine Functions
$Parent Function y = sin x y = cos x Transformed Function y = a sin b (x - h) + k y = a cos b (x - h) + k$ $∙ ∙ h = phase shift, or horizontal shift k = vertical shift$

We are given that the phase shift is

π

and the vertical shift is

- 5 .

We can write the transformed function using these values!

Parent Function y = sin x Transformed Function y = 4 sin \frac{2}{3} (x - π) - 5

Finally, we will take a look at the translation of this function.

Please note that we could have also chosen

a

to be

- 4 .

In that case, the graph would have been reflected over the

x -

axis.

Exercise