How do you draw the graph of #y=-tanx# for #0<=x<2pi#?

1 Answer
A. S. Adikesavan
Jul 16, 2018

See the required graph, in-between #x = 0 and x = 2pi#.

Explanation:

For x in #( 0, 2pi )#, y= 0, at #x = 0, pi, 2pi#
,
There are two asymptotes # x = pi/2 and x = (3/2)pi#, within the

interval.

The graph is not on uniform scale, for better visual effect.

graph{ (y+tan x)(x+0y)(x-2pi+0y) = 0[-0.1 6.3 -16 16]}

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