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4.3 Vertical and Horizontal Translations. OBJ: Graph sine and cosine with vertical and horizontal translations. DEF: Vertical Translation. A function of the form y =c + a sin b x or of the form y = c + a cos b x is shifted vertically when compared with y = a sin b x or y =a cos b x. - PowerPoint PPT Presentation
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4.3 Vertical and Horizontal Translations
OBJ: Graph sine and cosine with vertical and horizontal
translations
DEF: Vertical Translation
A function of the form y =c + a sin b x or of the form y = c + a cos b x is shifted vertically when compared with y = a sin b x or y =a cos b x.
5 EX: Graph y = 2 – 2 sin x
0 π π 3π 2π
2 2
6 EX: Graph y = – 3 + 2 sin x
0 π π 3π 2π
2 2
DEF: Phase Shift
The function y=sin (x+d) has the shape of
the basic sine graph y = sin x, but with a
translation d units: to the right if d < 0
and to the left if d > 0. The number d is
the phase shift of the graph. The cosine
graph has the same function traits.
7 EX: Graph y = sin (x – π/3)
8 EX: Graph y = 3cos (x + π/4)
9 EX: Graph y = 4 – sin (x – π/3)
10 EX: Graph y =-3 + 3cos(x+π/4)
DEF: Period of Sine and Cosine
The graph of y = sin b x will look like that
of sin x, but with a period of 2 . b Also the graph of y = cos b x looks like
that of y = cos x, but with a period of 2
b
y = c + a(trig b (x + d)
a (amplitude) multiply a times (0 |1 0 -1 0 1)
b (period) 2π
b
c (vertical shift)
d (starting point)
11 EX: • Graph y = sin 2x
12 EX: • Graph y = -2cos 3x
13 EX: • Graph y = 3 – 2cos 3x
14 EX: Graph y = –2cos(3x+π)
15 EX: • Graph y = cos(2x/3)
16 EX: Graph y = –2 sin 3x
17 EX: Graph y = 3 cos ½ x
GRAPHING SINE AND COSINE FUNCTIONS
In previous chapters you learned that the graph ofy = a • f (x – h) + k is related to the graph of y = | a | • f (x) by horizontal and vertical translationsand by a reflection when a is negative. This also applies to sine, cosine, and tangent functions.
GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
To obtain the graph of
Transform the graph of y = | a | sin bx ory = | a | cos bx as follows.
y = a sin b (x – h) + k or y = a cos b (x – h) + k
GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
VERTICAL SHIFT
Shift the graph k units vertically.
y = a • sin bx
y = a • sin bx + k
k
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
GRAPHING SINE AND COSINE FUNCTIONS
HORIZONTAL SHIFT
Shift the graph h units Vertically.
y = a • sin b(x – h)
h
y = a • sin bx
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
GRAPHING SINE AND COSINE FUNCTIONS
REFLECTION
If a < 0, reflect the graph in the liney = k after anyvertical and horizontalshifts have been performed.
y = a • sin bx + k
y = – a • sin bx + k
38
8
4
2
Graphing a Vertical Translation
Graph y = – 2 + 3 sin 4 x.
SOLUTION
By comparing the given equationto the general equation y = a sin b(x – h) + k, you can see that h = 0, k = – 2, and a > 0.
Therefore translate the graph of y = 3 sin 4x down two units.
Because the graph is a transformation of the graph of
y = 3 sin 4x, the amplitude is 3 and the period is = . 24
2
38
8
4
2
Graphing a Vertical Translation
The graph oscillates 3 units up and down from its centerline y = – 2.
SOLUTION
38
8
4
2
Therefore, the maximum value of the function is – 2 + 3 = 1 and the minimum value of the function is – 2 – 3 = –5
y = – 2
Graph y = – 2 + 3 sin 4 x.
38
8
4
2
Graphing a Vertical Translation
The five key points are:
On y = k : (0, 2); , – 2 ; , – 2 4
2
Maximum: , 1 8
Minimum: , – 5 38
SOLUTION
Graph y = – 2 + 3 sin 4 x.
38
8
4
2
Graphing a Vertical Translation
CHECK
You can check your graph with a graphing calculator. Use theMaximum, Minimum and Intersect features to check the key points.
Graph y = – 2 + 3 sin 4 x.
Graphing a Vertical Translation
Graph y = 2 cos x – .4
23
SOLUTION
3
Because the graph is a transformation of the graph of
y = 2 cos x, the amplitude is 2 and the period
is = 3 .
23
22
Graphing a Vertical Translation
By comparing the given equation to the general equation
y = a cos b (x – h) + k, you can see that h = ,
k = 0, and a > 0.
4
Therefore, translate the
graph of y = 2 cos x
right unit.
23
4
Graph y = 2 cos x – .π4
23
SOLUTION
Notice that the maximum
occurs unit to the right of
the y-axis.
4
Graphing a Horizontal Translation
The five key points are:
Graph y = 2 cos x – .4
23
SOLUTION
4
14
On y = k : • 3 + , 0 = (, 0);
• 3 + , 0 = , 052
34
4
Minimum: • 3 + , – 2 = , – 2 74
4
12
Maximum: 0 + , 2 = , 2 ;
134
3 + , 2 = , 2 ;
4
4
4
Graphing a Reflection
Graph y = – 3 sin x.
Because the graph is a reflection of the graph of y = 3 sin x, the amplitude is 3 and the period is 2.
When you plot the five points on the graph, note that theintercepts are the same as they are for the graph of y = 3 sin x.
SOLUTION
Graphing a Reflection
However, when the graph is reflected in the x-axis, themaximum becomes a minimum and the minimum becomes a maximum.
Graph y = – 3 sin x.
SOLUTION
Graphing a Reflection
On y = k : (0, 0); (2, 0);
12
• 2, 0 = (, 0)
Minimum: • 2, – 3 = , – 3 14
2
Maximum: • 2, 3 = , 3 34
32
Graph y = – 3 sin x.
SOLUTION The five key points are:
Modeling Circular Motion
FERRIS WHEEL You are riding a Ferris wheel. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by the following equation:
h = 25 sin t – 7.5 + 3015
The Ferris wheel turns for 135 seconds before it stopsto let the first passengers off.
Graph your height above the ground as a function of time.
What are your minimum and maximum heights above the ground?
SOLUTION
Modeling Circular Motion
The amplitude is 25 and the period is = 30.2
15
h = 25 sin t – 7.5 + 3015
The wheel turns = 4.5 times in 135 seconds,
so the graph shows 4.5 cycles.
13030
Modeling Circular Motion
The key five points are (7.5, 30), (15, 55), (22.5, 30),(30, 5) and (37.5, 30).
h = 25 sin t – 7.5 + 3015
SOLUTION
Modeling Circular Motion
Since the amplitude is 25 and the graph is shifted up 30units, the maximum height is 30 + 25 = 55 feet.
The minimum height is 30 – 25 = 5 feet.
h = 25 sin t – 7.5 + 3015
SOLUTION
GRAPHING TANGENT FUNCTIONS
TRANSFORMATIONS OF TANGENT GRAPHS
• Shift the graph k units vertically and h units horizontally.
• Then, if a < 0, reflect the graph in the line y = k.
To obtain the graph of y = a tan b (x – h) + k transform the graph of y = a tan bx as follows. ||
Combining a Translation and a Reflection
Graph y = – 2 tan x + .4
SOLUTION
The graph is a transformation of the graph of y = 2 tan x, so the period is .
Combining a Translation and a Reflection
Therefore translate the graph of
y = 2 tan x left unit and then
reflect it in the x-axis.
4
Graph y = – 2 tan x + .4
SOLUTION
By comparing the given equation to
y = a tan b (x – h) + k, you can see
that h = – , k = 0, and a < 0. 4
Combining a Translation and a Reflection
Asymptotes:
On y = k:
Halfway points:
x = – – = – ; x = – = 2 •1
4
34
4
4
2 •1
(h, k) = – , 04
– – , 2 = – , 2 ; – , – 2 = (0, – 2) 4 •1
4
2 4 •1
4
Graph y = – 2 tan x + .4