# How do you graph y=4csc2x?

Mar 10, 2018

#### Explanation:

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 $y = f \left(x\right) = \sin \left(x\right)$ has $z e r o s$ at $x = k \pi$, where $k$ is an integer.

The function $y = f \left(x\right) = \csc \left(x\right)$ has $\textcolor{b l u e}{\text{No }}$color(blue)(zeros.

For both the functions $\sin \left(x\right) \mathmr{and} \csc \left(x\right)$, Period $= 2 \pi$.

Graph of the function $\csc \left(x\right)$ does not have a maximum or a minimum value, there is $\textcolor{b l u e}{\text{No }}$color(blue)(amplitude.

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

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

At values of $x$ for which $\sin \left(x\right) = 0$, the function $\csc \left(x\right)$ is undefined.

The x-intercept of $y = \sin \left(x\right)$ and the asymptotes of $y = \csc \left(x\right)$ 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 $= \frac{2 \pi}{B} = \frac{2 \pi}{|} 2 | = \pi$

Vertical Shift = D = 0

Frequency $= \frac{1}{P e r i o d} = \frac{1}{\pi}$

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

$\csc \left(x\right)$ has only Vertical Asymptotes.

Vertical Asymptote = $x = \frac{\pi n}{2}$, where n is an integer.

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

x-intercepts and y-intercepts = None