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

Jul 20, 2018

Domain; $x \in \left(- \infty , \infty\right)$. Range of.
$y = \sec \left(\frac{x}{2}\right) \in \left(- \infty , 1\right] U \left[1 , \infty\right) \Rightarrow y \notin \left(- 1 , 1\right)$.
Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...

#### Explanation:

The period of $y = \sec \left(\frac{x}{2}\right)$ is $\frac{2 \pi}{\frac{1}{2}} = 4 \pi$.

Cosine range is $\left[- 1 , 1\right]$ Its reciprocal

secant range is $\left(- \infty , 1\right] U \left[1 , \infty\right)$. So,

$y = \sec \left(\frac{x}{2}\right) \in \left(- \infty , 1\right] U \left[1 , \infty\right) \Rightarrow y \notin \left(- 1 , 1\right)$.

Domain; $x \in \left(- \infty , \infty\right)$

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

by solving the denominator $\cos \left(\frac{x}{2}\right) = 0$.

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 $\rightarrow \uparrow \leftarrow \downarrow$, to view the extended graph.