Question

How do you find the domain, range, and asymptote for `y = 1 - tan ( x/2 - pi/8 )`?

Answer 1

Ask yourself where the function is defined.

Explanation:

The function `y=1-tan(x pi/2-pi/8)` is not defined where `tan(x pi/2-pi/8)` is not defined `=> x pi/2-pi/8 =k pi/2`, `k in ZZ`
`x/2-1/8=k/2`
`x=k+1/4`
The domain is `RR-{(k+1/4) pi/2}; k in ZZ`, the range is `RR`

An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.(from wikipedia)

In this case the asymptotes are vertical lines `x=(k+1/4) pi/2; k in ZZ`
Read more about asymptotes .

Some help:
1.
Remember about `tan(x)`:
`tan(x)` The domain is `RR-{k pi/2}; k in ZZ`, the range is `RR`

Look at the unit circle, the distance between the x-ax and the intersection of the green and blue line is `tan(x)`, where `x` is the angle. If `x rarr pi/2` there is no intersection of the green and blue line, there `tan(x)` is not defined.

graph{tan(x) [-5, 5, -2.5, 2.5]}

2.
`a*tan(b*x-c)+d`
`a`- change the slope of the graph
`b`-change the frequency (the "lines" are more close or away to each other)
`c`-translation parallel to y-ax, move the graph right or left
`d`-translation parallel to x-ax, move the graph up or down
You can use this graphing calculator write in `a*tan(b*x-c)+d` and play with sliders.