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

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Answer 1 · 2018-03-20T05:47:36

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

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]}

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