Question

How do you graph, identify the domain, range, and asymptotes for `y=sec(1/2)x`?

Answer 1

Domain; `x in ( - oo, oo )`. Range of.
`y = sec (x/2 ) in ( - oo, 1] U [ 1, oo ) rArr y notin ( - 1, 1 )`.
Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...#

Explanation:

The period of `y = sec (x/2 )` is `(2pi)/(1/2) = 4pi`.

Cosine range is `[ - 1, 1 ]` Its reciprocal

secant range is `( - oo, 1] U [ 1, oo )`. So,

`y = sec (x/2 ) in ( - oo, 1] U [ 1, oo ) rArr y notin ( - 1, 1 )`.

Domain; `x in ( - oo, oo )`

Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...#,

by solving the denominator `cos ( x/2 ) = 0`.

See illustrative graph, with the markings os asymptotes near O and

the out-of- bounds range #( - 1, 1 ):.
graph{(y cos ( x/2 ) - 1)(y^2-1)(x^2-(pi)^2) = 0[-10 10 -5 5]}

Slide the graph `rarr uarr larr darr`, to view the extended graph.