Unit 3

Transformations

Determining the Equation of a Sinusoidal Function from its Graph

If the graph of a sinusoidal function is provided, a sinusoidal equation for that function can be determined from the graph.

Notice the sine function is a translation of the cosine function, and vice versa.

When looking at the graph of a sinusoidal function, there is a possible sine function and a possible cosine function.

Example 5

Determine two possible equations for the function represented in the graph below.

Write the equations of the function in the following ways.

y = a \cos b (x - c) + d

y = a \sin b (x - c) + d

Solution

If the equation of the function is of the form

y = a \cos b (x - c) + d

, refer to

y = \cos x

to determine the transformations that occurred.

Determine the amplitude of the transformed function.

\begin{array}{rcl} amplitude & = & \frac{maximum y - minimum y}{2} \\ = & \frac{1 - (- 3)}{2} \\ = & \frac{4}{2} \\ = & 2 \\ |a| & = & 2 \end{array}

The amplitude is

2

. There was a vertical stretch by a factor of

2

applied to the cosine function.

y = 2 \cos b (x - c) + d

Determine the period, and then the horizontal stretch factor.

On the given graph, one cosine cycle is complete in

\frac{2 π}{3}

, so

\begin{array}{rcl} Period & = & \frac{2 π}{|b|} \\ \frac{2 π}{3} & = & \frac{2 π}{|b|} \\ 2 π |b| & = & 2 π (3) \\ |b| & = & 3 \end{array}

So, the equation is now

y = 2 \cos 3 (x - c) + d

Determine the phase shift.

The graph repeats itself regularly, so there are an infinite number of possible phase shifts. Two possibilities are to the left by

\frac{π}{6}

, or to the right by

\frac{π}{2}

For this Example,

c = - \frac{π}{6}

will be used.

y = 2 \cos 3 (x + \frac{π}{6}) + d

Last, determine the vertical displacement.

The location of the midline is the best way to determine the vertical displacement.

\begin{array}{rcl} Midline & = & d \\ d & = & \frac{maximum y + minimum y}{2} \\ = & \frac{1 + (- 3)}{2} = \frac{- 2}{2} = - 1 \end{array}

The vertical displacement is

- 1

.

The equation of the function is

y = 2 \cos 3 (x + \frac{π}{6}) - 1

.

Graph with technology to confirm this is the correct function.

If the equation of the function is of the form

y = a \sin b (x - c) + d

, refer to

y = \sin x

to determine the transformations that occurred.

The amplitude, period, and vertical displacement are the same for the sine function as they are for the cosine function. The only difference is the phase shift.

On the graph of the base sine function, the graph starts at the midline and goes up to the maximum

y

-value.

On the graph of the transformed function, the first positive

x

-value for which this occurs is at the point

(\frac{π}{3}, - 1)

.

The base sine function was translated

\frac{π}{3}

units to the right, so

c = \frac{π}{3}

.

A possible sine function is

y = 2 \sin 3 (x - \frac{π}{3}) - 1

.

Graphing the sine function with technology verifies the equation.

Summary

\begin{array}{rcl} |a| & = & amplitude \\ = & \frac{\max - \min}{2} \end{array}

\frac{1}{|b|} =

horizontal stretch factor

period = \frac{2 π}{|b|} or |b| = \frac{2 π}{period}

c =

phase shift or horizontal translation

\begin{array}{rcl} d & = & vertical displacement \\ = & \frac{\max + \min}{2} \end{array}