How do you find the domain, range, and asymptote for #y = 2 - sec ( 2x - pi/2 )#?

1 Answer
A. S. Adikesavan
Aug 9, 2018

See explanation and graph.

Explanation:

#y = 2 - sec ( 2x - pi/2 )= 2 - sec ( pi/2 - 2x )#

#= 2- csc 2x, 2x ne # asymptotic kpi, k =0, 1, 2, 3, ...#

#rArr x ne k(pi/2)#

Also, as csc values #notin ( - 1, 1 )#,

#y notin [-1 +2, 1 + 2 ) = ( 1, 3 )#.

The period is period of #sin 2x = (2pi)/2 = pi#.

Vertical shift = 2, giving midline #y = 2#.

See graph, depicting all these aspects.
graph{((2-y)sin (2x )-1)(x+0.0001y)(x-pi/2 + 0.0001y)(x+pi/2 + 0.0001y)(y-1+0x)(y-2+0x)(y-3+0x)=0[-10 10 -3 7]}

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