How do you graph, identify the domain, range, and asymptotes for $y = (\frac{1}{2}) sec 2 (x - \frac{π}{2}) + 1$ $y=(1/2)sec2(x-pi/2)+1$ ?

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Answer 1 · 2018-07-21T17:27:11

See graph and explanation.

$y = \frac{1}{2} sec (2 (x - \frac{π}{2}) + 1 = \frac{1}{2} sec (2 x - π) + 1$ $y = 1/2 sec( 2 ( x - pi/2 ) + 1 = 1/2 sec( 2 x - pi ) + 1$

$= \frac{1}{2} sec (π - 2 x) + 1$ $= 1/2 sec( pi - 2 x ) + 1$

$= \frac{1}{2} sec (2 x) + 1$ $= 1/2 sec ( 2 x ) + 1$ .

$= \frac{1}{2 cos (2 x)} + 1$ $= 1/(2 cos ( 2 x )) + 1$ .

The period = period of cos( 2 x ) = $2 π \frac{)}{2} = π$ $2pi )/2 = pi$ .

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

$\Rightarrow 2 x = (2 k + 1) \frac{π}{2} \Rightarrow x = (2 k + 1) \frac{π}{4}$ $rArr 2 x = ( 2 k + 1 )pi/2 rArr x = ( 2 k + 1 )pi/4$ ,

$k == 0, +-1, +-2, +-3,...$ 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]}

How do you graph, identify the domain, range, and asymptotes for y=(1/2)sec2(x-pi/2)+1y=(12)sec2(x−π2)+1y=(1/2)sec2(x-pi/2)+1?