How do you graph $y = 4 cos 2 (x - π) + 1$ $y=4cos2(x-pi)+1$ ?

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Answer 1 · 2016-11-29T07:13:15

The graph is a cosine wave about y = 1, with amplitude = 4, period = $π$ $pi$ . The graph is inserted.

$y - 1 = 4 cos x (x - π) = 4 cos (2 x - 2 π) = 4 cos (2 π - 2 x) = 4 cos 2 x$ $y-1=4cosx(x-pi)=4cos(2x-2pi) =4cos(2pi-2x)=4cos2x$ .

This represents a cosine wave about y - 1. The period is $\frac{2 π}{2} = π$ $(2pi)/2=pi$

and the amplitude about y = 1 is 4.

Graph is inserted. Here, $π = 3.14$ $pi=3.14$ , nearly. Look for graph for one

period with $\in [- \frac{π}{2}, \frac{π}{2}] = [- 1.57, 1.57]$ $in [-pi/2, pi/2]=[-1.57, 1.57]$ Correspondingly, y reaching

crest y = 5 twice and minimum y =-3 also twice at the ends.

graph{y-1-4cos(2x)=0 [-10, 10, -5, 5]}

How do you graph y=4cos2(x-pi)+1y=4cos2(x−π)+1 y=4cos2(x-pi)+1?