10.2 translate and reflect trigonometric graphs how do you translate trigonometric graphs? how do...
TRANSCRIPT
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