Question

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

Answer 1

`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]}