How do you find the important points to graph #y=3 tan (4x-pi/3)#?

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Answer 1 · 2018-07-11T03:41:51

See explanation and graphs.

The period of #tan ( k x + c ) = pi/k#.

So, the period of #y = pi/4 = 0.7854#, nearly, and #pi/3 = 1.0472#,

nearly..

As # 4 x - pi/3 to ( 2 k + 1 )pi/2, y to +- oo, k = 0, +-1, +-2, +-3, ...#.

Upon setting #k = - 1 and 0#, one period about origin is

#x in ( - pi/24, 5pi/24 ) = ( - 0.1309, 0.6545 )#, nearly.

Direct graph for 10 periods, #x in ( 0, 7.854 0 3.927)#

(Slide the graph (#larr#) to see further periods)
graph{y - 3 tan ( 4 x - 1.0472 ) = 0[ 0 7.854 0 3.927] ]}

The inverse is # x = 1/4(arctan( y/3 ) +pi/3 )#

One-period graph using the inverse:
graph{x - 1/4(arctan( y/3 ) + 1.0472) = 0}