This tutorial provides a step-by-step approach to graphing transformations of \( y = a \sin(b x + c) + d \) and \( y = a \cos(b x + c) + d \). We focus on the Interval Method to resolve period and phase shift efficiently.
From the unit circle, we identify five key points for one period \( [0, 2\pi] \). Range: \( [-1, 1] \).
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Similar to sine, but starting at its maximum. Period: \( 2\pi \). Range: \( [-1, 1] \).
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Vertical stretch by a factor of 3. Range: \( [-3, 3] \).
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Vertical stretch by 2. The negative sign indicates a reflection across the x-axis. Range: \( [-2, 2] \).
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Find one cycle of any transformed trig function by solving the double inequality: \[ 0 \le \text{Argument} \le 2\pi \]
Interval: \( 0 \le 2x \le 2\pi \implies \mathbf{0 \le x \le \pi} \). Period is \( \pi \).
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Solve: \( 0 \le 2x - \frac{\pi}{4} \le 2\pi \)
Add \( \pi/4 \): \( \frac{\pi}{4} \le 2x \le \frac{9\pi}{4} \)
Divide by 2: \( \mathbf{\frac{\pi}{8} \le x \le \frac{9\pi}{8}} \).
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Solve: \( 0 \le 3x + \frac{\pi}{3} \le 2\pi \)
Subtract \( \pi/3 \): \( -\frac{\pi}{3} \le 3x \le \frac{5\pi}{3} \)
Divide by 3: \( \mathbf{-\frac{\pi}{9} \le x \le \frac{5\pi}{9}} \).
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1. Interval: Solve \( 0 \le 2x - \pi/5 \le 2\pi \implies \mathbf{\frac{\pi}{10} \le x \le \frac{11\pi}{10}} \).
2. Reflection and Amplitude: The amplitude is 2. The negative sign results in a reflection on the x-axis (or midline).
3. Range: Shifted up 1 unit. \( [-2+1, 2+1] = \mathbf{[-1, 3]} \).
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Determine the range and the x-interval for one period of: \[ y = \frac{1}{2} \cos(3x - \pi/6) + 1 \]
Interval: Solve \( 0 \le 3x - \pi/6 \le 2\pi \implies \mathbf{\frac{\pi}{18} \le x \le \frac{13\pi}{18}} \).
Range: \( [1 - 0.5, 1 + 0.5] = \mathbf{[0.5, 1.5]} \).
Return to the Grade 11 Hub or explore Trig Function Properties.