00:01
All right, so let's take a look here at some of these problems in terms of looking at our amplitude period and frequencies, given some of these functions.
00:08
I want to just give a brief overview and a reminder of what we mean by some of these terms.
00:16
So our amplitude can really be thought of as our height when it comes to these problems.
00:23
Our period can be thought of as the distance between two equal points.
00:28
You can see here that we are looking at our maximums here, where i'm drawing these little dots.
00:35
And our period can be represented by 2 pi divided by b.
00:40
The frequency, in order for us to calculate frequency, we can think of that as one over our period.
00:47
And so one over our period will give us frequency, and then one over frequency will actually end up giving us our period.
00:59
So we can utilize this equation in order for us to kind of calculate some of these.
01:07
So looking at this first sketch here, what exactly, you know, does a two in front of a cosine and then a one -third in front of our x due to our graph? well, if we utilize this formula right here, we can see that whenever we make some changes, to some of these numbers from our base functions, that there will be indicative of a change in some of these.
01:38
So amplitude period and phase shifts.
01:41
So for this problem, 1a, we can see that a 2 correlates to our amplitude.
01:48
So our amplitude in this first problem is 2, which means that it's going to be twice as high as what we typically think of whenever we are looking at a regular graph of cosine.
01:58
So instead of cosine fluctuating between negative 1 and 1, it's actually going to fluctuate between negative 2 and 2 in terms of it fluctuating to x.
02:10
Or, excuse me, fluctuating in terms of y.
02:13
So our period in this instance is indicated by 2 pi over b.
02:22
B in this case is 1 3rd.
02:26
So we're going to take 2 pi over one third, which is going to give us 2 pi times 3 over 1, which is going to give us 6 pi.
02:42
And then our frequency is just 1 over our period, which is 6 pi.
02:51
So that is the answer for our first questions here.
02:55
In terms of sketching a graph, what we can do is we can actually pick out, some points that we can typically think of whenever we're looking at cosine in particular.
03:09
So for cosine, a lot of the points that we can think of will typically end up being, you know, so i'm thinking in the frame that they're giving us, just for the sake of our graphing, we can really think of negative three pi over two.
03:28
We can think of 0 you can also think 3 pi 2 and then we can think of potentially some others as well in particular we can actually just do negative 3 pi and 3 pi so what do we get whenever we plug these into our equation well negative 3 pi will end up giving us negative 2 negative 3 pi divided by 2 will end up giving us 0.
04:08
0 is going to end up giving us 2.
04:15
3 pi over 2 is going to give us 0, and then we will go back to negative 2 when we get to 3 .5.
04:25
So you can see here that if we were to actually kind of sketch this, that we would have a graph that looks something like this.
04:34
So if they give us our limits here, we can say that this will go to i believe they said four, correct? yes.
04:42
So this top value will go to four.
04:44
It's a negative value.
04:45
We'll go to negative four.
04:48
There we go of negative three pi.
04:53
Three pi.
04:54
And so negative three pi is again going to give us negative two.
04:59
So we can say that negative two is here.
05:03
I'm actually going to draw this in a different color so we can see it better.
05:08
Negative 3 pi over 2 is going to give us 0.
05:18
0 is going to end up giving us 2.
05:24
3 pi over 2 is going to give us 0.
05:29
And then our 3 pi is going to end up giving us negative 2.
05:36
So you can see here that then we can trace our graph through and that is kind of what our graph will look like for this first problem.
05:46
And we can do the exact same thing for the second problem as well, except in our second problem, we are looking at y equals negative three halves, sine 2x.
06:01
So in this instance, i'm going to take this.
06:07
I'm going to take this image that we have up here, move this down for us so that where we can use it for this particular problem.
06:19
Well.
06:21
So in this problem, we can see here that our amplitude in this instance is going to be one half.
06:35
Why do i say it's going to be one half? that's because we have this negative three halves right here and we're really looking and we're really interested in that three half.
06:43
So it's going to go 1 .5 is going to be what it ranges from.
06:49
So from negative 1 .5 to 1 .5 in terms of it fluctuating in between our ys.
06:55
Our period in this instance is going to be 2 pi over b.
07:03
B here is actually this two value.
07:07
So our period is going to be pi.
07:10
And then our frequency is going to be one over our period.
07:14
So one over pi.
07:16
So in this instance, whenever we're looking at this particular problem, and we're thinking about some of the shifts that we are, are doing, we are going to be flipping our equation is going to be mirrored over the x -axis because we have this negative value.
07:40
And we know that this two right here is going to indicate that we are going to have a difference in the standard period that we have.
07:50
So if we were to look at this value, again, we're going to just pick some values.
07:57
And what i like to do is i kind of like to start at, you know, some tried and true different points.
08:04
So like negative pi, for instance, we may also want to look at, you know, negative pi over two...