1 / 15

EXAMPLE 1

π. π. 2. 2. b. 4. h = 0. Amplitude:. a = 2. Horizontal shift:. π. 2. Period:. Vertical shift:. k = 3. =. =. EXAMPLE 1. Graph a vertical translation. Graph y = 2 sin 4 x + 3. SOLUTION. Identify the amplitude, period, horizontal shift, and vertical shift. STEP 1.

nsill
Download Presentation

EXAMPLE 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. π π 2 2 b 4 h = 0 Amplitude: a = 2 Horizontal shift: π 2 Period: Vertical shift: k = 3 = = EXAMPLE 1 Graph a vertical translation Graphy = 2 sin 4x + 3. SOLUTION Identify the amplitude, period, horizontal shift, and vertical shift. STEP 1 STEP 2 Draw the midline of the graph, y = 3. Find the five key points. STEP 3

  2. On y = k: (0, 0 + 3) = (0, 3); ( , 0 + 3) π = ( , 3); = ( , 3) ( , 0 + 3) 2 = ( , 5) Maximum: ( , 2 + 3) 3π ( , –2 + 3) 3π 8 Minimum: = ( , 1) π π π π π 8 4 8 2 8 4 EXAMPLE 1 Graph a vertical translation Draw the graph through the key points. STEP 4

  3. π π 2 2 b 2 h = 3π Amplitude: a = 5 Horizontal shift: Period: Vertical shift: π k = 0 = = EXAMPLE 2 Graph a horizontal translation Graphy = 5 cos 2(x – 3π ). SOLUTION Identify the amplitude, period, horizontal shift, and vertical shift. STEP 1 STEP 2 Draw the midline of the graph. Because k = 0, the midline is the x-axis. Find the five key points. STEP 3

  4. 15π 13π On y = k: = ( , 0); ( + 3π,0) 4 4 3π ( + 3π,0) 4 = ( , 0) = (3π, 5) Maximum: (0 + 3π, 5) (π+ 3π, 5) = (4π, 5) π π 7π 4 ( + 3π, –5) = ( , –5) Minimum: 2 2 EXAMPLE 2 Graph a horizontal translation Draw the graph through the key points. STEP 4

  5. Ferris Wheel Suppose you are riding a Ferris wheel that turns for 180seconds. Your height h(in feet) above the ground at any time t (in seconds) can be modeled by the equation h = 85 sin (t – 10) + 90. π a. Graph your height above the ground as a function 20 of time. EXAMPLE 3 Graph a model for circular motion b. What are your maximum and minimum heights?

  6. 2π a. 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). π 20 40 180 EXAMPLE 3 Graph a model for circular motion SOLUTION

  7. b. Your maximum height is 90 + 85 = 175feet and your minimum height is 90 – 85 = 5feet. EXAMPLE 3 Graph a model for circular motion

  8. Graphy = –2 sin (x – ). 2 3 2 π π 3 2 Amplitude: Horizontal shift: a = –2 = 2 h = 2 2π 2π = period : = 3π Vertical shift: k = 0 2 b EXAMPLE 4 Combine a translation and a reflection SOLUTION STEP 1 Identify the amplitude, period, horizontal shift, and vertical shift. Draw the midline of the graph. Because k = 0, the midline is the x-axis. STEP 2

  9. STEP 3 Find the five key points of y = –2 sin (x – ). 3π 9π 7π 5π 3π π π π π π π ( + ,2) ( + ,0) ( + ,–2) On y = k: = ( , 0); (0 + ,0) = (2π, 0) 4 4 2 4 2 2 2 2 2 2 2 = ( , 0) (3π+ ,0) 2 π = ( , 2) 2 3 Maximum: 11π ( , –2) Minimum: = 4 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

  10. 5π So, ( , 2) becomes ( , –2 ) 4 5π 11π 4 and becomes . ( , 2) 4 11π ( , –2) 4 EXAMPLE 4 Combine a translation and a reflection STEP 5 Draw the graph through the key points.

  11. π Period: h = 0 Horizontal shift: k = 5 Vertical shift: Find the asymptotes and key points of y = –3 tanx + 5. EXAMPLE 5 Combine a translation and a reflection Graph y= –3tanx + 5. SOLUTION STEP 1 Identify the period, horizontal shift, and vertical shift. STEP 2 Draw the midline of the graph, y = 5. STEP 3

  12. π Asymptotes: = x = x = ; – = 2 1 On y = k: = (0, 5) (0, 0 + 5) (– , –3 + 5) Halfway points: = π π (– , 2); 2 2 π π π π π π = π – 4 4 4 4 4 4 2 1 ( , 3 + 5) ( , 8) So, (– , 2) becomes π (– , 8) 4 ( , 8) and π becomes ( , 2) . 4 EXAMPLE 5 Combine a translation and a reflection STEP 4 Reflect the graph. Because a < 0, the graph is reflected in the midline y = 5.

  13. EXAMPLE 5 Combine a translation and a reflection STEP 5 Draw the graph through the key points.

  14. 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 q . EXAMPLE 6 Model with a tangent function

  15. opp 260 – d adj tanq = = 120 120 tanq 260 – d = 120 tan q– 260 – d = d –120 tan q+ 260 = EXAMPLE 6 Model with a tangent function SOLUTION Use a tangent function to write an equation relating dand q . Definition of tangent Multiply each side by 120. Subtract 260 from each side. Solve for d. The graph of d = –120 tanq + 260 is shown at the right.

More Related