10.2 Translate and Reflect Trigonometric Graphs
How do you translate trigonometric graphs?How do you reflect trigonometric graphs?
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);
π4 ( , 0 + 3)
π2
= ( , 3)π2
Maximum: ( , 2 + 3)π8 = ( , 5)
π8
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);
13π4
( + 3π, 0)3π4 = ( , 0)
15π4
Maximum: (0 + 3π , 5) = (3π, 5)
(π + 3π , 5) = (4π, 5)
Minimum: ( + 3π, –5)π2 = ( , –5)
7π2
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
SOLUTION
The 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)
3π2
Minimum: ( 2π, –1 + 4)12 = (π , 3)
On y = k: ( 2π , 0 +4)14 = ( , 4);
π2
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. π2
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);
π2
= ( , 0) 3π2( 2π + , 0)
12
π2
= ( , 0) 5π2( 2π + , 0)π
2
Maximum: ( 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 Translate and Reflect Trigonometric Graphs, day 3
• How do you translate trigonometric graphs?
• How do you reflect trigonometric graphs?
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: π2h =
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 .11π
4( , –2) 11π4( , 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 = 0
Vertical 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) .π4becomes
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 qoppadj= = 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.
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