00:01
In this problem, we're asked to take the sign function and do some shifts on it and change the amplitude and period, such that we have an amplitude of 16 units, a period of 5 units, a vertical displacement of 3 units, and a phase shift of 2 .5 units left.
00:24
So let's first look at what does the basic sign function look like.
00:31
So the basic sign function is periodic as such.
00:47
And of course it just keeps going, right? where without any shifts or amplitude changes or period changes or vertical displacement or anything, it would look like this, right? this would be x and this would be y.
01:14
Okay, so the amplitude, let's talk about what that is first.
01:20
So the amplitude is from here to here.
01:22
This is the amplitude.
01:28
And the period is from here to here.
01:37
Okay.
01:47
And then a vertical shift, had i drawn this a little bit better, i would have done this to make it look a little more symmetrical, right? so it's the same on both sides.
02:13
So then the vertical shift is this axis shifting up or down.
02:19
Or it's the center.
02:28
Okay.
02:29
And then a phase shift will be here, where we'll shift that peak, one way or another.
02:48
Now, if i were to write this function, i would write y equals a sign of b, x, plus c, plus d, where the amplitude here is a, the period is 2 pi over b.
03:28
The phase shift is c, and the vertical displacement would be d.
03:34
So if i was going to write this as an equation, i would write y equals a is 16 and sine of b.
03:54
So b, i want a period of five units.
04:00
Well, if five units is 2 pi over b, then b is 2 pi over 5.
04:11
So this is 2 pi over 5 times x.
04:16
And then i want a vertical displacement of three units.
04:26
I forgot to write this is up.
04:28
So this is plus 3 because i'm going positive.
04:31
But then i'm going left.
04:42
That's supposed to be the phase shift...