Honors Trigonometry - Quiz Review: Graphing Trig Functions

Typical Formulas

  • Y=Asin(Bx+C)+DY = Asin(Bx + C) + D

  • Y=Acos(Bx+C)+DY = Acos(Bx + C) + D

Key Concepts

  • Amplitude (Amp): The height from the center line to the peak (or trough) of the function. It is the absolute value of A: ∣A∣|A|.

  • Period: The length of one complete cycle of the function. The formula for the period is 2πB\frac{2\pi}{B}. A larger B compresses the graph horizontally, while a smaller B stretches it.

  • Phase Shift: A horizontal shift of the function, determined by −CB-\frac{C}{B}. A positive value indicates a shift to the left, while a negative value indicates a shift to the right.

  • Vertical Translation: A vertical shift of the function, determined by DD. A positive value shifts the graph upward, and a negative value shifts the graph downward.

  • Maximum Value: The highest y-value of the function, calculated as D+∣A∣D + |A|.

  • Minimum Value: The lowest y-value of the function, calculated as D−∣A∣D - |A|.

  • Endpoints: Used to define one period of the function.

    • Left Endpoint: −CB-\frac{C}{B}

    • Right Endpoint: −CB+2πB-\frac{C}{B} + \frac{2\pi}{B}

  • Five Key Points: These points divide the period into four equal parts and are crucial for accurately graphing trigonometric functions.

    • The x-coordinate spacing between key points is: Δx=Period4=2π4B\Delta x = \frac{Period}{4} = \frac{2\pi}{4B}

Quiz Review - Graphing Trig Functions

General Instructions

  • Show all work.

  • Use a calculator.

  • Keep answers in terms of π\pi, using exact radians; no decimals.

Problem 1: Given Characteristics, Write the Equation of a Cosine Function

  • Amplitude = 3

  • Period = 6π\pi

  • Reflected about the x-axis

  • Equation: y=−3cos(13x)y = -3cos(\frac{1}{3}x)

Problem 2: Given y=−27−2cos(4x+π)y = -27 - 2cos(4x + \pi), List the Following

  • Maximum Value:

    • D+∣A∣=−27+∣−2∣=−27+2=−25D + |A| = -27 + |-2| = -27 + 2 = -25

  • Minimum Value:

    • D−∣A∣=−27−∣−2∣=−27−2=−29D - |A| = -27 - |-2| = -27 - 2 = -29

  • Vertical Shift:

    • D=−27D = -27

  • Amplitude:

    • ∣A∣=∣−2∣=2|A| = |-2| = 2

  • Period:

    • 2πB=2π4=π2\frac{2\pi}{B} = \frac{2\pi}{4} = \frac{\pi}{2}

  • Phase Shift:

    • −CB=−π4-\frac{C}{B} = -\frac{\pi}{4}, which is to the left.

  • Endpoints in Exact Radians:

    • Left Endpoint: −π4-\frac{\pi}{4}

    • Right Endpoint: −π4+π2=π4-\frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4}

  • Five Key Points:

    • Calculate Δx=Period4=π/24=π8\Delta x = \frac{Period}{4} = \frac{\pi/2}{4} = \frac{\pi}{8}

    • (-π4\frac{\pi}{4}, -29), (-π8\frac{\pi}{8}, -27), (0, -25), (π8\frac{\pi}{8}, -27), (π4\frac{\pi}{4}, -29)

Problem 3: Given y=1−2sin(12x+π)y = 1 - 2sin(\frac{1}{2}x + \pi), List the Following

  • Amplitude:

    • ∣A∣=∣−2∣=2|A| = |-2| = 2

  • Period:

    • 2πB=2π1/2=4π\frac{2\pi}{B} = \frac{2\pi}{1/2} = 4\pi

  • Phase Shift:

    • −CB=−π1/2=−2π-\frac{C}{B} = -\frac{\pi}{1/2} = -2\pi, shifted to the left.

  • Endpoints in Exact Radians:

    • Left Endpoint: −2π-2\pi

    • Right Endpoint: −2π+4π=2π-2\pi + 4\pi = 2\pi

  • Five Key Points:

    • (−2π,1),(−π,−1),(0,1),(π,3),(2π,1)(-2\pi, 1), (-\pi, -1), (0, 1), (\pi, 3), (2\pi, 1)

Problem 4: Period Compression or Stretch

  • If the period of a sine function is 6π6\pi, the graph is being horizontally stretched because 6\pi > 2\pi.

  • If the amplitude of a cosine function is 5, the graph is being vertically stretched because 5 > 1.

Problem 5: Given y=−3cos(2x−π)y = -3cos(2x - \pi), List the Following

  • Amplitude:

    • ∣A∣=∣−3∣=3|A| = |-3| = 3

  • Period:

    • 2πB=2π2=π\frac{2\pi}{B} = \frac{2\pi}{2} = \pi

  • Phase Shift:

    • −CB=−−π2=π2-\frac{C}{B} = -\frac{-\pi}{2} = \frac{\pi}{2}, shifted to the right.

  • Find the third key point of one period (both x and y coordinate).

    • To find the third key point, we need to understand that it represents the value at half the period. since the period is π\pi, half the period is π2\frac{\pi}{2}. Adding the phase shift to it leads to π2+π2=π\frac{\pi}{2} + \frac{\pi}{2} = \pi. Hence the coordinates are (π\pi, 3).

Problem 6: Given Characteristics, Write the Equation of a Sine Function

  • Reflected about the x-axis

  • Amplitude = 56\frac{5}{6}

  • Period = π\pi

  • Phase Shift = π2\frac{\pi}{2} to the left

  • Equation: y=−56sin(2x+π)y = -\frac{5}{6}sin(2x + \pi)

Problem 7: Given Characteristics, Write the Equation of a Cosine Function

  • Amplitude = 2

  • Period = 4π4\pi

  • Vertical Shift down by 1

  • Phase Shift = π3\frac{\pi}{3} to the right

  • Equation: y=2cos(12x−π6)−1y = 2cos(\frac{1}{2}x - \frac{\pi}{6}) - 1

Problem 8-10: Given the Graph, Write the Equation of the Function (No Horizontal Shift, Compression, or Stretch)

  • These problems refer to graphs, and based on those hypothetical graphs, the equations are:

    • y=3cos(4x)y = 3cos(4x)

  • Based on another hipotethical graph:

    • y=3cos(12x)y = 3cos(\frac{1}{2}x)

  • Given another hypothetical graph, find the amplitude of the function.

    • Amplitude is 2