# Graph y=3csc(2x+π/4) Include vertical asymptotes and intercepts and label with points on the unit circle.

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### 1 Answer

Graph `y=3csc(2x+pi/4) ` :

Rewrite as `y=3csc[2(x+pi/8)] `

This is in the form of `y=Acsc[B(x-h)]+k `

With the base function `y=csc(x) ` we have the following transformations:

A: A performs a vertical dilation. Since A=3 we have a dilation of factor 3.

B: B affects the period. the period p is given by `p=(2pi)/B=(2pi)/2=pi ` . (This is equivalent to a horizontal compression of factor 2 or horizontal stretch factor 1/2.)

h: h performs a horizontal translation. The base graph is shifted ` pi/8 ` units left.

k: k performs a vertical translation. Here k=0 so there is no vertical shift.

So the base graph of y=csc(x) will be transformed with a period of `pi ` , shifted `pi/8 ` units left and stretched vertically by a factor of 3.

Thus the asymptotes will be at` `` ` `x=-pi/8+(npi)/2,n in ZZ ` (n an integer; that is the asymptotes are at `x=...,(-5pi)/8,-pi/8,(3pi)/8,(7pi)/8,... `

The y-intercept is at `y=3csc(pi/4)=3*sqrt(2)~~4.24 `

There are no x-intercepts.

The minimums, y=-3, occur at ` ``x=pi/8+npi ` and the maximums, y=3, at `x=(5pi)/8+npi `

Please see the attached image.