How do you find the amplitude, period, vertical and phase shift and graph #y=1/4cos(2theta-150)+1#?

Question

Quinn Anderson · Accepted Answer

Amplitude # = 1/4#

Period # = pi#

Phase shift #= 75# to the right

Vertical shift # = 1# to the top.

Use the form #y = a cos(bx - c) + d# to find the variables used to find the amplitude, period, phase shift, and vertical shift Given : (1/4) cos (2theta - 150) + 1# graph{(1/4) sin(2x - 150) + 1 [-10, 10, -5, 5]} Amplitude #a = 1/4# Period # = - 2pi / |b| = 2pi / 2 = pi# Phase shift # = c / b = 150 / 2 = 75# to the right Vertical shift # = d = 1# to the top.

Quinn Anderson · Answer

undefined To find the amplitude, period, vertical shift, phase shift, and graph $ y = \frac{1}{4} \cos(2\theta - 150^\circ) + 1 $, follow these steps:

1. **Amplitude**: The amplitude of a cosine function is the absolute value of the coefficient of the cosine term. In this case, the amplitude is $ \frac{1}{4} $.

2. **Period**: The period of a cosine function is $ \frac{2\pi}{b} $, where $ b $ is the coefficient of $ \theta $. Here, $ b = 2 $, so the period is $ \frac{2\pi}{2} = \pi $.

3. **Vertical Shift**: The vertical shift is the constant term added or subtracted to the function. Here, the vertical shift is $ +1 $, indicating that the graph is shifted upward by 1 unit.

4. **Phase Shift**: To find the phase shift, set the argument of the cosine function equal to zero and solve for $ \theta $. Here, we have $ 2\theta - 150^\circ = 0 $. Solving for $ \theta $, we get $ \theta = 75^\circ $. Since the phase shift is opposite to what is usually thought (rightward instead of leftward), the phase shift here is $ -75^\circ $.

5. **Graph**: Plot the points and graph the function using the amplitude, period, vertical shift, and phase shift.

- Amplitude: $ \frac{1}{4} $
   - Period: $ \pi $
   - Vertical Shift: $ +1 $
   - Phase Shift: $ -75^\circ $

You can start by plotting the key points:
- At $ \theta = -75^\circ $ (the phase shift), the cosine function reaches its maximum or minimum (depending on the sign of the amplitude). Since the amplitude is positive, it reaches a maximum. So, one point is $ (-75^\circ, 1 + \frac{1}{4}) = (-75^\circ, \frac{5}{4}) $.
- At $ \theta = -75^\circ + \frac{\pi}{2} $, the cosine function crosses the midline. So, another point is $ (-75^\circ + \frac{\pi}{2}, 1) $.
- Continue plotting points using the period $ \pi $ and the amplitude $ \frac{1}{4} $.

Then, sketch the graph, noting the periodic nature of the cosine function and the shifts in amplitude, vertical position, and phase.