Im stuck Question a bout graphing sin and plotting points

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Homework Help Overview

The discussion revolves around graphing the function y = -4 sin x, specifically focusing on how to divide the interval [0, 2π] into four equal sub-intervals for plotting points. Participants are exploring the concepts of amplitude, period, and the relationship between the sine function and its graphical representation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of dividing the interval into four equal parts and express confusion about how to determine the specific points for plotting. Some refer to the unit circle for guidance on sine values at key points.

Discussion Status

There is ongoing exploration of how to accurately divide intervals and plot points on the graph. Some participants provide guidance on marking midpoints and evaluating sine values, while others express uncertainty about their understanding of the process.

Contextual Notes

Participants are grappling with the concept of interval division and the implications for graphing, particularly when extending the discussion to a different interval, [π/2, 13π/2]. There is mention of needing a coordinate system for accurate graphing.

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Homework Statement



Im doing a question where I have to graphc y = -4 sin x

Ok well I got the period and amplitude, so I have the interval of [0, 2pi]

Now, in order to plot points, I have to divide the interval into 4 sub intervals.

I don't understand how I do that! This is where I get stuck, and don't know how to plot the points now.

Any help would be great!

Homework Equations





The Attempt at a Solution



In my book it says " We divide the interval [0, 2pi] into 4 sub intervals, each of length 2pi/4 = pi/2

I don't get that.
 
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Do just what it says, divide the interval [0,2\pi ] into 4 intervals. They even tell you what points to use. Now just evaluate sin at each of those points, this will tell how to draw the graph.
 
Integral said:
Do just what it says, divide the interval [0,2\pi ] into 4 intervals. They even tell you what points to use. Now just evaluate sin at each of those points, this will tell how to draw the graph.

I am not confident on how to divide the intervals.

I know for this one I can just refer to the unit circle, and get,

pi/2, pi, 3pi/2, 2pi

And those are my x cords right? and my y cords would just be -4, 0, 4 0

But given the fact that I am not 100% confident in dividing the intervals, how would I divide the following interval into 4 sub intervals?

[pi/2, 13pi/2]
 
When they say divide into four intervals, they mean four equal intervals. Mark 0 and 2\pi on our x-axis, then mark the midpoint- which divide the interval from 0 to 2\pi into two intervals, then mark the midpoint of each of those to get four intervals.

Look at a graph of sin(x) itself. It starts at (0, 0), goes up to (\pi/2, 1), down to (\pi, 0), continues down to (3\pi/2, -1), then goes back up to (2\pi, 0)- and then repeats.

So when you have decided where you want 0 to 2\pi line, mark first the midpoint. That will be, of course, at (2\pi)/2= \pi. That divides the line into two intervals, from 0 to \pi and from \pi to 2\pi. Now mark the midpoint of each those intervals, at \pi/2 and 3\pi/2. At those last two points, mark 4 units below and 4 units above the axis.

Your graph will start at (0, 0), go down to (\pi/2, -4) at that first "quarter point" you marked,up to the axis at the middle point, (\pi, 0), up to (3\pi/2, 4) and finally back down to (2\pi, 0)

For \pi/2 to 13\pi/2, the midpoint will be the average of the two numbers: (\pi/2+ 13\pi/2)/2= 14\pi/4= 7\pi/2 so mark that point. Now average \pi/2 and 7\pi/2, (\pi/2+ 7\pi/2)/2= 8\pi/4= 2\pi, and average 7\pi/2 and 13\pi/2, (7\pi/2+ 13\pi/2)/2= 20\pi/4= 5\pi.

BUT when you are drawing the graphs, what you really want to do is "eyeball" those points. It should be pretty easy to mark the midpoint of an interval without calculating anything.
 
Last edited by a moderator:
HallsofIVY - Great, thanks! But I have a question.

in regards to: For π/2 to 13π/2, the midpoint will be the average of the two numbers: (π/2+13π/2)/2=14π/4=7π/2 so mark that point. Now average π/2 and 7π/2, (π/2+7π/2)/2=8π/4=2π, and average 7π/2 and 13π/2, (7π/2+13π/2)/2=20π/4=5π.

As you stated, the midpoint wuold be 7pi/2 - How and where do I know to put that point??
 
How do you know where to put "0" and "13\pi/2"? In order to graph anything, you have to have a coordinate system which means you must have and x-axis, marked with x-values.

Are you simply saying that you don't know what number 7\pi/2?

\pi= 3.1715926... so 7\pi= 21.991148575128552669238503682957 and 7\pi/2=10.995574287564276334619251841478

(Using the Windows calculator.)
 
You can still use a calculator can't you?
 
HallsofIvy said:
How do you know where to put "0" and "13\pi/2"? In order to graph anything, you have to have a coordinate system which means you must have and x-axis, marked with x-values.

Are you simply saying that you don't know what number 7\pi/2?

\pi= 3.1715926... so 7\pi= 21.991148575128552669238503682957 and 7\pi/2=10.995574287564276334619251841478

(Using the Windows calculator.)


I would just put 7pi/2 half way between the 2 intervals. And to get the other 2 intervals I just average the starting point and midpoint, and the midpoint and the end interval?
 

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