Question

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

Answer 1

See explanation and graphs.

Explanation:

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}