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U8P2D6 Have out: Bellwork:

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Presentation on theme: "U8P2D6 Have out: Bellwork:"— Presentation transcript:

1 U8P2D6 Have out: Bellwork:
Pencil, red pen, highlighter, assignment, calculator Bellwork: Graph one positive period of each function. Using radians, and be sure label the axes. y x +2 y1 5 +1 |a| = ____ 5 y1 b = ____ 4 +1 +2 y2 +1 y2 p = ____ +1 h = ____ To the left +2 labeled x–axis +1 scale –5 total: +1 labeled y–axis

2 Identify and/or describe the major parts of the function
y = a sin (b (x – h)) + k Amplitude |a| = ________________ b = ________________ Horizontal Shift Frequency Vertical Shift b = ________________ p = ________________ Period p = ________________

3 Is there a “sine” name for ? ________________
Sketch one period for each pair of functions on the same axes. Use radian measure. y 2 2 . . x Is there a “sine” name for ? ________________

4 Is there a “cosine” name for ? ________________
y 3 3 . . x Is there a “cosine” name for ? ________________

5 Is there a simple name for ? ________________
y 4 4 . . x Is there a simple name for ? ________________

6 Is there a “cosine” name for ? ________________
y . . x Is there a “cosine” name for ? ________________

7 Work on the practice.

8 Old Slides

9 Solving Trigonometric Equations
Trigonometric equations can be solved by using methods from algebra. Example #1: Solve for x. Solve for all values of θ between 0 ≤ θ < 2π. y x 1 4 4 4 4 θ θ =

10 Example #2: Solve for x. Solve for all values of θ between 0 ≤ θ < 2π. y x (0, 1) x = 0 x + 1 = 0 (–1, 0) θ x = –1 (0, –1) θ =

11 Example #3: Solve for x. Solve for all values of θ between 0 ≤ θ < 2π. y x 3x3 = x 3x3 – x = 0 (1, 0) (–1, 0) x(3x2 – 1) = 0 x = 0 3x2 – 1 = 0 θ = 0, 3x2 = 1 y x 2 1 θ θ =

12 Example #4: Solve for x and y. Solve for all values of θ between 0 ≤ θ < 2π. y x 2xy – x = 0 x(2y – 1) = 0 (1, 0) (–1, 0) x = 0 2y – 1 = 0 2y = 1 θ = 0, y x 1 θ θ =

13 Example #5: Solve for x. Solve for all values of θ between 0 ≤ θ < 2π. y x 2x2 + 3x + 1 = 0 2 3 2 1 (–1, 0) x + 1 θ = 2x 2x2 2x 1x 1 y x + 1 1 (x + 1)(2x + 1) = 0 θ x + 1 = 0 2x + 1 = 0 x = –1 2x = –1 θ =


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