# Question: If ` `` `` ``f(theta)=sin4theta` where `0<=theta<3pi` , then the Number of vertical asymptotes in the graph of`(1)/(f(theta)) ` is A.8 B.9 C.12 ANSWER D.13 Answer: The graph...

Question: If ` `` `` ``f(theta)=sin4theta` where `0<=theta<3pi` , then the Number of vertical asymptotes in the graph of`(1)/(f(theta)) ` is

A.8

B.9

C.12 ANSWER

D.13

Answer: The graph is `f(theta)=sin4theta` is shown at the top. There are`12-theta` intercepts between `0<=theta<3pi` (Note that you don't include the one at `3pi` ) So, there are 12 asymptotes.

Somebody can explainme more how to read and understand the graph! Please

### 1 Answer | Add Yours

Given the graph of `f(theta)=sin4theta` on `[0,3pi)` you are asked for the number of vertical asymptotes for the graph of `1/(f(theta))` .

The graph we are interested in is the graph of `1/(f(theta))=1/(sin4theta)=csc4theta` . Consider `1/(sin4theta)` : `sin4theta` is continuous (roughly that means you can draw the graph without lifting your pencil), so its reciprocal will also be continuous except when the denominator is zero. Dividing by zero is undefined and the graph will have a vertical asymptote. (Note that some rational functions can have a denominator of zero that results in a "hole" in the graph instead of an asymptote -- regardless the function is undefined at the point where you divide by zero.)

So `1/(f(theta))` will have vertical asymptotes anywhere the parent function has value zero.

The graph of `y=1/(sin4theta)` :

The vertical bars between the "U's" are the vertical asymptotes. They occur precisely where the parent function (in red) is zero.

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If you are asking about the original function `f(theta)=sin4theta` then realize that this is the sine function with period `(2pi)/4=pi/2` ; the sine curve is repeated every `pi/2` units. Since the sine is zero twice in every period, and there are 6 periods from 0 to `3pi` there will be 12 zeros of the function.