How do you graph # 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 = #pi#. The graph is inserted.

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

This represents a cosine wave about y - 1. The period is #(2pi)/2=pi#

and the amplitude about y = 1 is 4.

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

period with # 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]}