Problem 1 Answer the following questions and graph the function \( y=2 \cos (x-\pi) \) - Amplitude: \( \qquad \) - Period: \( \qquad \) and Xscale: \( \qquad \) - Vertical Shift: \( \qquad \) Horizontal Shift \( \qquad \) - Starting point \( \qquad \) and Ending Point \( \qquad \)
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For the function \( y = 2 \cos(x - \pi) \): - Amplitude: The amplitude is the coefficient of the cosine function, which is \( 2 \). This means the function will oscillate between \( 2 \) and \( -2 \). - Period: The standard period of the cosine function is \( 2\pi \). Since there’s no horizontal scaling factor, the period remains \( 2\pi \). Xscale can be considered \( \frac{2\pi}{4} = \frac{\pi}{2} \) if you're charting on a typical 4x unit system. - Vertical Shift: There’s no vertical shift in this function, so it is \( 0 \). Horizontal Shift: The expression \( (x - \pi) \) indicates a horizontal shift of \( \pi \) units to the right. - Starting point: At \( x = \pi \), the cosine function starts at its maximum value, so the starting point is \( (\pi, 2) \). - Ending Point: To find the ending point, we can determine where the function has completed one full cycle. The function returns to its original position at \( x = \pi + 2\pi = 3\pi \), which gives us the ending point \( (3\pi, 2) \). So, when you graph \( y = 2 \cos(x - \pi) \), plot the specified points and use the identified features to create a wave that oscillates between \( 2 \) and \( -2 \) over intervals defined by the period!