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

1 Answer
Apr 10, 2018

Answer:

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.
my_pic
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.