How do you graph two cycles of #y=-0.5tan(2theta)#?

1 Answer
Mar 20, 2018

Answer:

#A = -.5; " " B = 2; " period " = pi/2#

Explanation:

Given: #y = -0.5 tan(2 theta)#

To graph the function you need to figure out the amplitude, period, phase shift and vertical shift.

The standard form is #y = A tan (Bx - C) + D#

where #A = #amplitude; #" "#period = #pi/|B|#
phase Shift = #C/B#; #" vertical shift" = D#

From the given function: #A = -0.5; " period" = pi/2#
There is no vertical or phase shift.

From the graph: #y = tan x# note the direction.
graph{tan x [-10, 10, -5, 5]}

Since the given function has a negative value, the function is flipped about the #y#-axis.

The function is center at (0,0).

Vertical asymptotes are at half a period on each side of (0,0).

1/2 period about (0,0) = #+-1/2 * pi/2 = +-pi/4#

The next vertical asymptote is one period away: #pi/4 + pi/2 = (3 pi)/4#

The x-intercepts are at #(0,0), (0, pi/2), ...#

Graph: #y = -0.5 tan (2 theta):#
graph{-0.5 tan(2x) [-3.897, 3.9, -1.95, 1.947]}