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 

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

y = | a | • f (x) by horizontal and vertical translations GRAPHING SINE AND COSINE FUNCTIONS 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.

GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS To obtain the graph of y = a sin b (x – h) + k or y = a cos b (x – h) + k Transform the graph of y = | a | sin bx or y = | a | cos bx as follows.

GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS VERTICAL SHIFT Shift the graph k units vertically. k y = a • sin bx + k y = a • sin bx

GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS HORIZONTAL SHIFT Shift the graph h units Vertically. h y = a • sin b(x – h) y = a • sin bx

GRAPHING SINE AND COSINE FUNCTIONS TRANSFORMATIONS OF SINE AND COSINE GRAPHS y = a • sin bx + k 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

 Graph y = – 2 + 3 sin 4 x. SOLUTION 2 4 2 Graphing a Vertical Translation Graph y = – 2 + 3 sin 4 x. Because the graph is a transformation of the graph of y = 3 sin 4x, the amplitude is 3 and the period is = . 2 4  2 SOLUTION 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. 3 8  4 2 Therefore translate the graph of y = 3 sin 4x down two units.

Graph y = – 2 + 3 sin 4 x. SOLUTION Graphing a Vertical Translation Graph y = – 2 + 3 sin 4 x. The graph oscillates 3 units up and down from its center line y = – 2. SOLUTION Therefore, the maximum value of the function is – 2 + 3 = 1 and the minimum value of the function is – 2 – 3 = –5 3 8  4 2 3 8  4 2 y = – 2

  Graph y = – 2 + 3 sin 4 x. SOLUTION 4 2 8 3 8 Graphing a Vertical Translation Graph y = – 2 + 3 sin 4 x. SOLUTION The five key points are: On y = k : (0, 2); , – 2 ; , – 2  4 2 3 8  4 2 Maximum: , 1  8 Minimum: , – 5 3 8

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

Graphing a Vertical Translation Graph y = 2 cos x – .  4 2 3 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 2 SOLUTION

Graphing a Vertical Translation Graph y = 2 cos x – . π 4 2 3 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 SOLUTION Therefore, translate the graph of y = 2 cos x right unit. 2 3  4 Notice that the maximum occurs unit to the right of the y-axis.  4

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

When you plot the five points on the graph, note that the 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. SOLUTION 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.

However, when the graph is reflected in the x-axis, the Graphing a Reflection Graph y = – 3 sin x. SOLUTION 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. SOLUTION 1 • 2, 0 = (, 0) 2 1 2 4 3 3 2 Graphing a Reflection Graph y = – 3 sin x. SOLUTION The five key points are: 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 2

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 + 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?

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

Modeling Circular Motion h = 25 sin t – 7.5 + 30  15 SOLUTION The key five points are (7.5, 30), (15, 55), (22.5, 30), (30, 5) and (37.5, 30).

Modeling Circular Motion h = 25 sin t – 7.5 + 30  15 SOLUTION 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.

• Shift the graph k units vertically and h units horizontally. GRAPHING TANGENT FUNCTIONS TRANSFORMATIONS OF TANGENT GRAPHS To obtain the graph of y = a tan b (x – h) + k transform the graph of y = a tan bx as follows. | • Shift the graph k units vertically and h units horizontally. • Then, if a < 0, reflect the graph in the line y = k.

 Graph y = – 2 tan x + . 4 SOLUTION 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 .

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