EXAMPLE 1 Graph a vertical translation
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
=24π
=π2
Vertical shift: k = 3
STEP 2 Draw the midline of the graph, y = 3.
STEP 3 Find the five key points.
EXAMPLE 1 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.
EXAMPLE 2 Graph a horizontal translation
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.
EXAMPLE 2 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.
EXAMPLE 3 Graph a model for circular motion
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?
EXAMPLE 3 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
EXAMPLE 3 Graph a model for circular motion
Your maximum height is 90 + 85 = 175 feet and your minimum height is 90 – 85 = 5 feet.
b.
EXAMPLE 4 Combine a translation and a reflection
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.
EXAMPLE 4 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.
EXAMPLE 4 Combine a translation and a reflection
So, ( , 2) becomes ( , –2 )5π4
5π4
and becomes .11π
4( , –2)
11π4
( , 2)
STEP 5 Draw the graph through the key points.
EXAMPLE 5 Combine a translation and a reflection
Graph 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.
EXAMPLE 5 Combine a translation and a reflection
Asymptotes: xπ
2 1–= = ;π
2– xπ
2 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
EXAMPLE 5 Combine a translation and a reflection
STEP 5 Draw the graph through the key points.
EXAMPLE 6 Model with a tangent function
Glass 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 .
EXAMPLE 6 Model with a tangent function
SOLUTION
Use a tangent function to write an equation relating d and .
Definition of tangenttan oppadj= =
260 – d 120
Multiply each side by 120.120 tan 260 – d =
Subtract 260 from each side.120 tan – 260 – d=
Solve for d.–120 tan + 260 d=
The graph of d = –120 tan + 260 is shown at the right.
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