10.2 Translate and Reflect Trigonometric Graphs

How do you translate trigonometric graphs?How do you reflect trigonometric graphs?

p. 619

Graph y = 2 sin 4x + 3.

SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.Amplitude: a = 2 Horizontal shift: h = 0

Period: 2bπ= 2

4π=

π2

Vertical shift: k = 3

STEP 2 Draw the midline of the graph, y = 3.

STEP 3 Find the five key points.

Graph a vertical translation

On y = k: (0, 0 + 3) = (0, 3); π4( , 0 + 3) = ( , 3);π

4 ( , 0 + 3)π2

= ( , 3)π2

Maximum: ( , 2 + 3)π8 = ( , 5)π

8

Minimum: ( , –2 + 3)3π8 = ( , 1)

3π8

STEP 4 Draw the graph through the key points.

Graph y = 5 cos 2(x – 3π ).

SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.Amplitude: a = 5 Horizontal shift: h = 3π

Period: 2bπ 2

2π= π= Vertical shift: k = 0

STEP 2 Draw the midline of the graph. Because k = 0, the midline is the x-axis.

STEP 3 Find the five key points.

Graph a horizontal translation

On y = k: ( + 3π , 0)π4 = ( , 0);13π

4

( + 3π, 0)3π4 = ( , 0)15π

4

Maximum: (0 + 3π , 5) = (3π, 5)

(π + 3π , 5) = (4π, 5)

Minimum: ( + 3π, –5)π2 = ( , –5)7π

2

STEP 4 Draw the graph through the key points.

Ferris Wheel

Suppose you are riding a Ferris wheel that turns for 180 seconds. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by the equation π

20h = 85 sin (t – 10) + 90.

a. Graph your height above the ground as a functionof time.

b. What are your maximum and minimum heights?

Graph a model for circular motion

SOLUTIONThe amplitude is 85 and the period is = 40. The wheel turns = 4.5 times in 180 seconds, so the graph below shows 4.5 cycles. The five key points are (10, 90), (20, 175), (30, 90), (40, 5), and (50, 90).

a. π20

2 π

40180

Your maximum height is 90 + 85 = 175 feet and your minimum height is 90 – 85 = 5 feet.

b.

Graph the function.

y = cos x + 4.1.SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.Amplitude: 1 Horizontal shift: h = 0

Period: 2bπ 2

1π

= 2π= Vertical shift: k = 4

STEP 2 Draw the midline of the graph. y = 4.

STEP 3 Find the five key points.

Maximum: (0 , 1 + 4) = (0,5)

(2π ,1 + 4) = (2π, 5)

34( 2π, 0 + 4) = ( , 4)3π

2

Minimum: ( 2π, –1 + 4)12 = (π , 3)

On y = k: ( 2π , 0 +4)14 = ( , 4); π

2

SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.

Period: 2bπ 2

1π= 2π= Vertical shift: k = 0

STEP 2

STEP 3 Find the five key points.

Graph the function.y = 3 sin (x – )2. π

2

πAmplitude: 3 Horizontal shift: h = 2

Draw the midline of the graph.y = 0 Because k = 0, the midline is the x-axis.

On y = k: (0 + , 0)π2 = ( , 0); π

2

= ( , 0) 3π2( 2π + , 0)

12

π2

= ( , 0) 5π2( 2π + , 0)π

2Maximum: ( 2π + , 3) = (π, 3)

14

π2

Minimum: ( 2π + , –3)34 = ( , –3)2π

π2 ( + , –3)

3π4

π2=

Graph the function.f(x) sin (x + π) – 13.

SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.

Period: 2bπ 2

1π= 2π= Vertical shift: k = –1

STEP 2

STEP 3 Find the five key points.

Amplitude: 1 Horizontal shift: h = – π

Draw the midline of the graph. y = – 1.

On y = k: (0 – π , 0 – 1 ) = (– π, – 1);

= (0, –1) 12( 2π – π , –1)

= (π , – 1)( 2π – π , 0 – 1)

Maximum: π2

= (– ,0)( 2π – π , 1 + 1) 14

Minimum: ( 2π – π, –1 –1)34 ( , – 2) π

2=

• How do you translate trigonometric graphs?The graphs of y = and , where a>0 and b>0 are horizontal translations h units and vertical translations k units of the graphs of and respectively, and have amplitude a, period , and midline • How do you reflect trigonometric graphs?In general, when , the graphs of and are the reflections of the graphs of respectively, in the midline

10.2 Assignment, day 1

Page 623, 3-21 odd

10.2 Translate and Reflect Trigonometric Graphs, day 3

• How do you translate trigonometric graphs?

• How do you reflect trigonometric graphs?

p. 621

Graph y = –2 sin (x – ).23

π2

SOLUTION

STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift.Amplitude: a = –2 = 2 Horizontal shift: π

2h =

period : b2π 2π

322= 3π= Vertical shift: k = 0

STEP 2 Draw the midline of the graph. Because k = 0, the midline is the x-axis.

Combine a translation and a reflection

STEP 3 Find the five key points of y = –2 sin (x – ).23

π2

On y = k: (0 + , 0)π2

= ( , 0);π2 ( + , 0)

3π2

π2 = (2π, 0)

π2(3π + , 0) 7π

2 = ( , 0)

Maximum: ( + , 2)3π4

π2

5π4

= ( , 2)

Minimum: ( + , –2)9π4

π2

11π4( , –2)=

STEP 4 Reflect the graph. Because a < 0, the graph is reflected in the midline y = 0.

So, ( , 2) becomes ( , –2 )5π4

5π4

and becomes .11π4( , –2) 11π

4( , 2)

STEP 5 Draw the graph through the key points.

Combine a translation and a reflectionGraph y = –3 tan x + 5.

SOLUTION

STEP 1 Identify the period, horizontal shift, and vertical shift.Period: π Horizontal shift:h = 0Vertical shift: k = 5

STEP 2 Draw the midline of the graph, y = 5.

STEP 3 Find the asymptotes and key points of y = –3 tan x + 5.

Asymptotes: xπ

2 1–= = ;π2– x

π2 1= π

2=

On y = k: (0, 0 + 5) = (0, 5)

Halfway points: (– , –3 + 5)π4

(– , 2);π4= ( , 3 + 5)π

4 ( , 8)π4=

STEP 4 Reflect the graph. Because a < 0, the graph is reflected in the midline y = 5.

So, (– , 2) π4 (– , 8)π

4becomes and ( , 8)π4 ( , 2) .π

4becomes

STEP 5 Draw the graph through the key points.

Model with a tangent functionGlass Elevator

You are standing 120 feet from the base of a 260 foot building. You watch your friend go down the side of the building in a glass elevator. Write and graph a model that gives your friend’s distance d (in feet) from the top of the building as a function of the angle of elevation q .

SOLUTION

Use a tangent function to write an equation relating d and q .

Definition of tangenttan q oppadj= = 260 – d

120

Multiply each side by 120.120 tan q 260 – d =

Subtract 260 from each side.120 tan q – 260 – d=Solve for d.–120 tan q + 260 d=

The graph of d = –120 tan q + 260 is shown at the right.

Graph the function.4. y = – cos ( x + )π

2

SOLUTION

Graph the function.5. y = – 3 sin x + 2 1

2

SOLUTION

Graph the function.6. f(x) = – tan 2 x – 1

SOLUTION

Graphing tangent functions using translations and reflections is similar to graphing sine and cosine functions. When a tangent function has a horizontal shift, the asymptotes also have a horizontal shift., , and

10.2 Assignment, day 3

Page 623, 23-31 odd, 37-45 odd