How do you graph, identify the domain, range, and asymptotes for #y=(1/2)sec2(x-pi/2)+1#?

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Answer 1 · 2018-07-21T17:27:11

See graph and explanation.

#y = 1/2 sec( 2 ( x - pi/2 ) + 1 = 1/2 sec( 2 x - pi ) + 1#

#= 1/2 sec( pi - 2 x ) + 1#

#= 1/2 sec ( 2 x ) + 1#.

# = 1/(2 cos ( 2 x )) + 1#.

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

The asymptotes are given by #cos ( 2 x ) = 0#

# rArr 2 x = ( 2 k + 1 )pi/2 rArr x = ( 2 k + 1 )pi/4#,

# k == 0, +-1, +-2, +-3,...# So, the domain is

#x in ...U (-5/4pi, -3/4pi ) U (-3/4pi, -pi/4 ) U (-pi/4, pi/4 ) #

#U ( pi/4, 3pi/4 ) U (3pi/4, 5pi/4 ) U ...#

Range is given by

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

See illustrative graph, indicating range and two asymptotes,

near O..
graph{(2(y-1) cos (2x)-1)(y-1/2)(y-3/2)(x^2-(pi)^2/16 )=0[-8 8 -3 5]}