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Graph y = –2 sin ( x – ). 2. 3. 2. π. π. 3. 2. Amplitude:. Horizontal shift:. a = –2 = 2. h =. 2. 2 π. 2 π. =. period :. =. 3 π. Vertical shift:. k = 0. 2. b. EXAMPLE 4. Combine a translation and a reflection. SOLUTION. STEP 1. - PowerPoint PPT Presentation
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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.
GUIDED PRACTICE for Examples 4, 5, and 6
Graph the function.
4. y = – cos ( x + )π2
SOLUTION
GUIDED PRACTICE for Examples 4, 5, and 6
Graph the function.
5. y = –3 sin x + 2 12
SOLUTION
GUIDED PRACTICE for Examples 4, 5, and 6
Graph the function.
6. f(x) = – tan 2 x – 1
SOLUTION
GUIDED PRACTICE for Examples 4, 5, and 6
7. What if ? In example 6, how does the model change if you are standing 150 feet from a building that is 400 feet tall ?
ANSWER
The graph is shifted 400 units up instead of 260. The new equation would be d = –150 tan + 400.