Question

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

Answer 1

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]}