How do you graph #y=4csc2x#?

Given:

#color(red)(y = f(x) = 4 csc(2x)#

How to draw a graph for this trigonometric function?

Note that #color(green)(y = f(x) = csc(x)# is the base function.

Observe that #color(blue)(csc(x) = 1/sin(x)#

Analyze the graph below:

Note that the function #y=f(x)=sin(x)# has #zeros# at #x=kpi#, where #k# is an integer.

The function #y=f(x)=csc(x)# has #color(blue)"No "##color(blue)(zeros#.

For both the functions #sin(x) and csc(x)#, Period #= 2pi#.

Graph of the function #csc(x)# does not have a maximum or a minimum value, there is #color(blue)"No "##color(blue)(amplitude#.

If values of #sin(x)# is available, one can figure out point by point what the values of #csc(x)# are.

The function goes to infinity periodically and is symmetric with the origin.

At values of #x# for which #sin(x) = 0#, the function #csc(x)# is undefined.

The x-intercept of #y=sin(x)# and the asymptotes of #y=csc(x)# are the same.

Next, consider the given trigonometric function:

#color(blue)(y = f(x) = 4 csc(2x)#

Use the form:

A Csc(BX - C) + D.

The variables used gives us the Amplitude and Period.

#A=4; B=2; C=0 and D=0# (using the given trigonometric function).

Amplitude = None

Period #= (2pi)/B=(2pi)/|2| = pi#

Vertical Shift = D = 0

Frequency #=1/(Period) = 1/pi#

To draw the graph, we can select a few points as shown below:

#csc(x)# has only Vertical Asymptotes.

Vertical Asymptote = #x=(pi n)/2#, where n is an integer.

Graph of #color(blue)(y = f(x) = 4 csc(2x)#

x-intercepts and y-intercepts = None