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4.3 Vertical and Horizontal Translations OBJ: Graph sine and cosine with vertical and horizontal translations.

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Presentation on theme: "4.3 Vertical and Horizontal Translations OBJ: Graph sine and cosine with vertical and horizontal translations."— Presentation transcript:

1 4.3 Vertical and Horizontal Translations OBJ: Graph sine and cosine with vertical and horizontal translations

2 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.

3 5 EX:  Graph y = 2 – 2 sin x 0ππ3π 2π 2

4 6 EX:  Graph y = – 3 + 2 sin x 0ππ3π 2π 2

5 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.

6 7 EX:  Graph y = sin (x – π/3)

7 8 EX:  Graph y = 3cos (x + π/4)

8 9 EX:  Graph y = 4 – sin (x – π/3)

9 10 EX: Graph y =-3 + 3cos(x+π/4)

10 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 

11 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)

12 11 EX: Graph y = sin 2x

13 12 EX: Graph y = -2cos 3x

14 13 EX: Graph y = 3 – 2cos 3x

15 14 EX: Graph y = –2cos(3x+π)

16 15 EX: Graph y = cos(2x/3)

17 16 EX: Graph y = –2 sin 3x

18 17 EX: Graph y = 3 cos ½ x

19 G RAPHING S INE AND C OSINE F UNCTIONS In previous chapters you learned that the graph of y = a f (x – h) + k is related to the graph of y = | a | f (x) by horizontal and vertical translations and by a reflection when a is negative. This also applies to sine, cosine, and tangent functions.

20 G RAPHING S INE AND C OSINE F UNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS To obtain the graph of Transform the graph of y = | a | sin bx or y = | a | cos bx as follows. y = a sin b (x – h) + k or y = a cos b (x – h) + k

21 G RAPHING S INE AND C OSINE F UNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS VERTICAL SHIFT Shift the graph k units vertically. y = a sin bx y = a sin bx + k k

22 TRANSFORMATIONS OF SINE AND COSINE GRAPHS G RAPHING S INE AND C OSINE F UNCTIONS HORIZONTAL SHIFT Shift the graph h units Vertically. y = a sin b ( x – h) h y = a sin bx

23 TRANSFORMATIONS OF SINE AND COSINE GRAPHS G RAPHING S INE AND C OSINE F UNCTIONS REFLECTION If a < 0, reflect the graph in the line y = k after any vertical and horizontal shifts have been performed. y = a sin bx + k y = – a sin bx + k

24 33 8  8  4  2 Graphing a Vertical Translation Graph y = – 2 + 3 sin 4 x. S OLUTION By comparing the given equation to 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 =. 22 4  2

25 33 8  8  4  2 Graphing a Vertical Translation The graph oscillates 3 units up and down from its center line y = – 2. S OLUTION 33 8  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.

26 33 8  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 33 8 S OLUTION Graph y = – 2 + 3 sin 4 x.

27 33 8  8  4  2 Graphing a Vertical Translation C HECK You can check your graph with a graphing calculator. Use the Maximum, Minimum and Intersect features to check the key points. Graph y = – 2 + 3 sin 4 x.

28 Graphing a Vertical Translation Graph y = 2 cos x –.  4 2 3 S OLUTION 3 Because the graph is a transformation of the graph of y = 2 cos x, the amplitude is 2 and the period is = 3 . 2 3 22 2

29 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. 2 3  4 Graph y = 2 cos x –. π 4 2 3 S OLUTION Notice that the maximum occurs unit to the right of the y-axis.  4

30 Graphing a Horizontal Translation The five key points are: Graph y = 2 cos x –.  4 2 3 S OLUTION  4 1 4 On y = k : 3  +, 0 = ( , 0); 3  +, 0 =, 0 55 2 3 4  4 Minimum: 3  +, – 2 =, – 2 77 4  4 1 2 Maximum: 0 +, 2 =, 2 ; 13  4 3  +, 2 =, 2 ;  4  4  4

31 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 the intercepts are the same as they are for the graph of y = 3 sin x. S OLUTION

32 Graphing a Reflection However, when the graph is reflected in the x-axis, the maximum becomes a minimum and the minimum becomes a maximum. Graph y = – 3 sin x. S OLUTION

33 Graphing a Reflection On y = k : (0, 0); (2 , 0); 1 2 2 , 0 = ( , 0) Minimum: 2 , – 3 =, – 3 1 4  2 Maximum: 2 , 3 =, 3 3 4 33 2 Graph y = – 3 sin x. S OLUTION The five key points are:

34 Modeling Circular Motion F ERRIS W HEEL 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 + 30  15 The Ferris wheel turns for 135 seconds before it stops to 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?

35 S OLUTION Modeling Circular Motion The amplitude is 25 and the period is = 30. 22  15 h = 25 sin t – 7.5 + 30  15 The wheel turns = 4.5 times in 135 seconds, so the graph shows 4.5 cycles. 130 30

36 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 + 30  15 S OLUTION

37 Modeling Circular Motion Since the amplitude is 25 and the graph is shifted up 30 units, the maximum height is 30 + 25 = 55 feet. The minimum height is 30 – 25 = 5 feet. h = 25 sin t – 7.5 + 30  15 S OLUTION

38 G RAPHING T ANGENT F UNCTIONS 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. ||

39 Combining a Translation and a Reflection Graph y = – 2 tan x +.  4 S OLUTION The graph is a transformation of the graph of y = 2 tan x, so the period is .

40 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 S OLUTION By comparing the given equation to y = a tan b (x – h) + k, you can see that h = –, k = 0, and a < 0.  4

41 Combining a Translation and a Reflection Asymptotes: On y = k: Halfway points: x = – – = – ; x = – =  2 12 1  4 33 4  4  4  2 12 1 (h, k) = –, 0  4 – –, 2 = –, 2 ; –, – 2 = (0, – 2)  4 14 1  4  24 14 1  4  Graph y = – 2 tan x +.  4


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