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

Mar 20, 2018

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

#### Explanation:

Given: $y = - 0.5 \tan \left(2 \theta\right)$

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

The standard form is $y = A \tan \left(B x - C\right) + D$

where $A =$amplitude; $\text{ }$period = $\frac{\pi}{|} B |$
phase Shift = $\frac{C}{B}$; $\text{ vertical shift} = D$

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) = $\pm \frac{1}{2} \cdot \frac{\pi}{2} = \pm \frac{\pi}{4}$

The next vertical asymptote is one period away: $\frac{\pi}{4} + \frac{\pi}{2} = \frac{3 \pi}{4}$

The x-intercepts are at $\left(0 , 0\right) , \left(0 , \frac{\pi}{2}\right) , \ldots$

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