WEBVTT
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Let's take a look at the tangent function. Recall that the quotient identity for tangent x is equal to
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sine x over cosine x which tells us that the tangent function will be undefined when the cosine of x is equal to 0.
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Let's see when that happens.
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The cosine is equal to 0 when x is equal to 2n plus 1 over 2 times pi, which is another way of saying odd multiples
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of pi over 2. So as you can see in my table I know that this function is going to be undefined at pi over 2.
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Some other values, remember it's odd multiples of pi over 2, so that would be 3 pi over 2, 5 pi over 2,
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negative pi over 2, and so on. Next, the tangent function will be equal to zero, or the x-intercepts, when the sine
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function is equal to 0. Let's see when that happens. Well the sine is equal to 0 when x is equal to n times pi,
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and again in both of these cases n is any integer. So that means that the zeros or x-intercepts will occur at 0,
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pi, 2pi, negative pi, and so on. So you can see in my table I've got the value of (0, 0).
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So now I've listed some values for tangent function. The first four you should know. These are special values.
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I've plotted those points on my graph. What we want to see is that when we choose x-values closer and closer
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to pi over 2 the values of the tangent function increase without bound which means that there's a vertical asymptote
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at pi over 2. So let's put that on our graph. And so we know that this graph from 0 to pi over 2 looks like this.
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Now, the tangent function happens to be odd. Let's see why. If I look at the tangent of negative x, using the quotient
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identity again, that's going to be the sine of negative x over the cosine of negative x. You need to recall
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that the sine function is odd which means that the sine of negative x is negative sine x. Cosine function
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remember is even which means that the cosine of negative x is the same as the cosine of x. And I can see that
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that's going to equal negative tangent x. So we know the tangent function is odd. What does that mean?
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It means that the graph is symmetric about the origin, so we can get another piece of the graph.
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So remember it's undefined at negative pi over 2, and if I reflect this part about the origin the graph looks like this.
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So here's one cycle or period for the tangent graph between negative pi over 2 and pi over 2. We call this cycle
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the principal cycle, and there are a couple of key special points that you need to know for this graph.
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Let's draw those points right now. So at pi over 2 and negative pi over 2 the tangent function is undefined.
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It's zero at the origin, one of the special points. The point (pi over 4, 1) is an important point as well as
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(negative pi over 4, -1). So this graph you should know very well. It shows where the asymptotes are. It shows
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these three points. We call this the center point. It's obviously in the center of the cycle. These two points,
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we call them halfway points, and of course this is another halfway point. And this of course is the center point.
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So you need to know (negative pi over 4, -1), (0,0), (pi over 4, 1) for the tangent function and the asymptotes
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at negative pi over 2 and pi over 2. So now that we know what one cycle looks like, let's look at some more cycles
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and come up with some characteristics for the tangent function.
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Here I've drawn three cycles for the tangent function and we want to look at some of the characteristics of
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the graph. The first thing we want to observe is what the domain is. The domain is all real numbers, x, such that
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x cannot equal 2n plus 1 over 2 times pi, where n is any integer. Again, that's saying at all odd multiples
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of pi over 2 it's undefined. So that's the domain. So you can see pi over 2, 3 pi over 2, those are values that are
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not in the domain. The range of the function, look at the y-values, is negative infinity to infinity. The period,
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here we have three periods, here's another one. The period is pi, and the principal cycle we will call from
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negative pi over 2 to pi over 2. So these are very important concepts that you need to know about the
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tangent function. Let's look at some more. The tangent function remember is undefined and where it's
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undefined there are vertical asymptotes. They occur where x is equal to 2n plus 1 over 2 times pi, again odd
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multiples of pi over 2. Look at the graph. I have a vertical asymptote at pi over 2, 3 pi over 2, negative pi over 2,
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negative 3 pi over 2. The next one would be at ... 5 pi over 2. The y-intercept we can see is at the origin.
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The y-intercept is 0. For each cycle there are three important points. One of them is called the center point,
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so on these periods I can see I have a center point at (0, 0), center point at (pi, 0), center point at (negative pi, 0).
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The x-coordinates of those center points, the x-intercepts or the zeros, are of the form n pi, again where
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n is any integer, so again 0, pi, negative pi, 2 pi. And the y-coordinate of all those center points is 0. Next we have
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halfway points. For each cycle there are two halfway points. The halfway point to the left of the center point
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is located midway between the x-coordinate of the left vertical asymptote and the x-coordinate of the center point,
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and the y-coordinate of that halfway point to the left is negative 1. The halfway point to the right of the center point
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is located midway between the x-coordinate of the center point and the x-coordinate of the right vertical asymptote,
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and the y-coordinate of those halfway points is 1. Next, we see that the tangent function is odd. That means the
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graph is symmetric about the origin. And lastly, the graph of each cycle we can see is one-to-one. Remember your
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horizontal line test. If I look at one cycle, it passes the horizontal line test, so each cycle is one-to-one.
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You need to be very familiar with these characteristics and be able to graph one period for the tangent function.
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Let's take a look at sketching functions of the form y equals A tan (Bx minus C). Recall that the principal cycle
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for y equals tangent x was negative pi over 2 to pi over 2. When you're sketching functions in this form, the
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principal cycle will occur when (Bx minus C) is between negative pi over 2 and pi over 2. So in this example
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we have y equals 2 tan (x minus pi over 4), and we want to sketch this graph. So the first thing we're going to
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do is notice that B is positive. In this case B is 1. C is pi over 4. If B is negative you would have to rewrite it.
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Remember that the tangent function is odd to make the B positive, and then you would begin. So in this case
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our B is positive, so we take our (Bx minus C) which in this case is (x minus pi over 4), and we put it between
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negative pi over 2 to pi over 2 to determine the principal cycle for this function. So we want to solve for x.
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We want to add pi over 4 to each side, so that would give me negative pi over 4 less than x. Pi over 2 plus
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pi over 4 would be 3 pi over 4. So here's what I know. I know the principal cycle for this function is going to be
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negative pi over 4 to 3 pi over 4. Notice that the period, the length of that interval, is still pi. You can always
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find the period by taking pi and dividing by B. Remember the normal period for tangent is pi, and of course
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in this case, B is 1. So we can see the period is pi. So now we know the principal cycle. We can figure out where
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the left and the right vertical asymptotes occur for this principal cycle. The left vertical asymptote is going to be
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at x equals negative pi over 4, and the right vertical asymptote will occur when x is equal to 3 pi over 4.
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So let's go ahead and put that on our graph. So let's see, we have negative pi over 4 and 3 pi over 4, and
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I know those are where my vertical asymptotes are. Next we need to find those three key points. Let's start
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with the center point. Recall that the center point is located, the x-coordinate is going to be located midway
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between the x-coordinate of the left vertical asymptote and the x-coordinate of the right vertical asymptote.
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So that means our x-coordinate will be, you can add up these coordinates and divide by 2, which gives me
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2 pi over 4 divided by 2, which is pi over 4. So the x-coordinate of the center point is pi over 4.
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The y-coordinate of the center point is 0, so the point that I have now, my center point, let's write it up here, is
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(pi over 4, 0). So let's go ahead and plot that point (pi over 4, 0) is my center point. Next we need to find the
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two halfway points. Let's find the halfway point to the left of this center point. To get the x-coordinate of the
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halfway point to the left, it's going to be located halfway between this left vertical asymptote and the x-coordinate
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of the center point. So the x-coordinate will be negative pi over 4 plus pi over 4, and again you divide that by 2.
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Well, the numerator is 0. 0 divided by 2 is 0. The y-coordinate, well, remember the y-coordinate for the normal
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tangent was going to be down at negative 1. So what you do to get the y-coordinate of this halfway point,
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you're going to multiply that y-coordinate of negative 1 for the basic tangent, and you're going to multiply it by A.
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Notice in this case, my A is 2, so I'm going to do negative 1 times 2, which gives me negative 2. So this
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halfway point would be (0, -2). Now we need to find the halfway point to the right of the center point. Well again,
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the x-coordinate will be located midway between this x-coordinate of the center point and the x-coordinate
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of the right vertical asymptote. So that's going to be pi over 4 plus 3 pi over 4, and again divide by 2 so that
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gives you pi divided by 2 is going to be the x-coordinate of this halfway point. The y-coordinate is going to be,
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recall the y-coordinate to the right of the normal tangent, y equals tan x, is positive 1. So you're going to multiply
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that positive 1 times A, which in this case is 2, so the y-coordinate of this halfway point is going to be 2.
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So this point will be (pi over 2, 2). So we'll plot that point. So I can see that this sketch, the graph will look like this.
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This is what this one cycle will look like for this particular tangent function. It's very important that you remember
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how to find these points and how you find the principal cycle and the left and the right vertical asymptotes.
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Let's look at this example. We have y equals negative tan (1/2 x plus pi), and we want to sketch this graph.
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The first thing we want to determine is the principal cycle. The principal cycle will occur when (Bx minus C),
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which in this case is (1/2 x plus pi), is between negative pi over 2 and pi over 2, which you recall negative pi over 2
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to pi over 2 is the principal cycle for tangent. Now we want to solve this for x. So first I want to subtract pi.
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So let's see. Now I need to multiply by 2. So now I know that the principal cycle for this function is negative 3 pi
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to negative pi. Now that I know the principal cycle I know that the left vertical asymptote is x equals negative 3 pi
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and the right vertical asymptote will be x equals negative pi.
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So again to determine the principal cycle, you take the argument (Bx minus C), you put it between negative
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pi over 2 and pi over 2, and you solve. That will give you the principal cycle. The left of the principal cycle tells
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you the left vertical asymptote, and the right gives you the right vertical asymptote. Now let's find the three
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special points. Let's find the center point first. Well, recall the center point is located midway between the
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left and the right, the x-coordinate of the center point is located midway between the left and the right
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vertical asymptotes. So if I take negative 3 pi plus a negative pi divided by 2, that gives me negative 2 pi
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for the x-coordinate of the center point. The y-coordinate of the center point is 0. So my center point let's write
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as (negative 2 pi, 0). So let's go over to our graph, and let's show what we have so far. Let's put in the left
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vertical asymptote. So let's see, that's negative pi. That's negative 3 pi, so I know I have a vertical asymptote
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here and here. And I know the center point is (negative 2 pi, 0). Now we need to find the halfway points.
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Well, the halfway point to the left of the center point, the x-coordinate is located midway between this
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vertical asymptote and the x-coordinate of the center point. So that will be, the x-coordinate will be negative 3 pi
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plus a negative 2 pi, divided by 2. Well that's negative 5 pi over 2 is the x-coordinate of this halfway point.
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The y-coordinate, we have to remember what the principal cycle is for tan x. Remember the halfway point to
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the left in y equals tan x, the y-coordinate to the left of the center point is negative 1, so we take that
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negative 1 value and multiply it by our A. Notice in this case our A is negative 1, so we have a negative 1
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times a negative 1 which of course is 1, so this halfway point is (negative 5 pi over 2, 1). So let's show that.
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Ok, now we have one more halfway point to find, the one of course to the right of the center point. Its
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x-coordinate is going to be located midway between the x-coordinate of the center point and this vertical
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asymptote, so that's going to be negative 2 pi plus a negative pi, over 2, that's negative 3 pi over 2 is the
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x-coordinate of this halfway point. And the y-coordinate you find by taking the y-coordinate for the basic tan x,
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which is positive 1, times A and again A is negative 1, so the y-coordinate of this halfway point is negative 1.
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So this point would be (negative 3 pi over 2, -1). Let's plot that point. Ok, so now we've got
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the center point, our halfway points, our vertical asymptotes, so here is one cycle for this tangent function.
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Let's take a look at the cotangent function, y equals cotangent x. First thing I want to use is, remember that
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the cotangent function using the quotient identity is the same thing as cosine x over sine x. So this means
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that the cotangent function will be undefined when sine x is equal to 0. And when does that happen?
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That occurs when x is equal to n times pi, where n is an integer. So it's undefined for example at 0, pi, 2 pi,
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negative pi. I've listed those in my table. The next thing is it's equal to 0, or the x-intercepts occur, when
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cosine x is equal to 0, and that occurs when x is equal to 2n plus 1 over 2 times pi, or odd multiples of pi over 2.
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So at pi over 2 it's 0. The next one would be 3 pi over 2, 5 pi over 2, and so on. So I've got that point.
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Pi over 4 and 3 pi over 4, those are some special values you should know. The cotangent of pi over 4 is 1.
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The cotangent of 3 pi over 4 is negative 1. So I've plotted those three points. Where the cotangent function is
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undefined, that's where your vertical asymptotes will be. So there's one here on the y-axis, and there's one
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here at x equals pi. So one cycle for this cotangent will look like this. We can see the length of that interval is pi,
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so the period is pi. And we call this cycle from 0 to pi, we call this the principal cycle for cotangent.
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You need to know where those asymptotes are, and then let's look at these points. There are three special points
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for the cotangent. We have the center point located at (pi over 2, 0). We have a halfway point to the left of that,
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(pi over 4, 1), and our other halfway point to the right of the center point is (3 pi over 4, -1). It's very important
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that you learn this graph, that you can draw one cycle including the asymptotes and the three important points.
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Here I've drawn three cycles for the cotangent function, and we want to look at the characteristics of the
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cotangent function. The first is the domain, that's the values of x. x can be any real number except n pi,
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where n is any integer. So that means for example it's undefined at 0, pi, 2 pi, negative pi, and all multiples of pi.
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So that's the domain. Next let's look at the range, the y-values. Well we can look at a cycle, and we can see that
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the y-values go from negative infinity to infinity. The period for the cotangent function we can see is pi. This is
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what we call the principal cycle. We can see another period over here, so you need to know that the period
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for cotangent is pi, and you need to know that the principal cycle for the cotangent function is 0 to pi.
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From here to here is the principal cycle. Now, let's look at some other characteristics. Because it's undefined,
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there are vertical asymptotes at all values x equals n pi, so on this graph you can see I have a vertical asymptote
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at x equals 0, x equals pi, x equals 2 pi, and x equals negative pi. All multiples of pi there will be vertical
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asymptotes, and you need to know where they occur. We can see from the graph that there's no y-intercept
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since x cannot equal 0. Now for each cycle there are three special points, and you need to learn these
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points and be able to plot them. Let's talk about the center point first. The center point is located halfway
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between the left vertical asymptote and the right vertical asymptote. So the x-coordinate is right in the middle
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between the left and the right vertical asymptote. And the y-coordinate of the center point is 0. So all of these,
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you can see three center points. They always occur, the x-coordinate is midway between the left vertical
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asymptote and the right vertical asymptote and the y-coordinate is always 0. Now let's look at the halfway points
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for each cycle. Let's look at the halfway point to the left of the center point. Let's look at this one.
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The x-coordinate of this halfway point is located midway between the left vertical asymptote and the x-coordinate
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of the center point, and the y-coordinate of the halfway point is 1. Next, let's look at the halfway point to the right
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of the center point. Let's look at this one. Its x-coordinate is located midway between the x-coordinate of the
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center point and the right vertical asymptote, and the y-coordinate of this halfway point is negative 1.
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So you need to know how to plot and find those halfway points and center points.
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Next we can see that the graph of the cotangent is odd. I can see that the graph is symmetric about the origin.
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Remember, because it's odd that means that the cotangent of negative x is the opposite of the cotangent of x.
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And lastly, we can see that for each cycle for the cotangent function that it passes the horizontal line test
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which means that for each cycle the cotangent function is one-to-one on each cycle.
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You need to learn all these characteristics of the cotangent function and be able to graph at least one period
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showing the asymptotes and showing the center point and the halfway points.
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Let's take a look at sketching functions of the form y equals A cotangent of (Bx minus C). In this example,
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we want to sketch the graph of 3 cotangent (x plus pi over 3). When you have functions in this form the
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first thing you want to note is if B is positive. If B is not positive, then you would rewrite this function using the
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fact that the cotangent is odd. In this case, I can see my B is 1, so we don't need to rewrite it. Recall that
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the cotangent function the principal cycle for y equals cotangent x was between 0 and pi. So in order to find
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the principal cycle for functions in this form, we're going to let (Bx minus C), in this case (x plus pi over 3),
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we're going to put it between 0 and pi. This will give us the principal cycle for this function. Solving for x,
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subtracting the pi over 3, we can see that the principal cycle for this function is negative pi over 3 to 2 pi over 3.
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Now the length of that interval would be the period, and the length of that interval is pi. We can always find the
00:28:08.058 --> 00:28:17.075
period by taking pi, the normal period for cotangent x, and dividing by B. In this case B of course is 1, so I can see
00:28:17.077 --> 00:28:31.339
that that's pi. Now that we have the principal cycle we can find the left vertical asymptote and the right.
00:28:31.341 --> 00:28:46.356
The left vertical asymptote will occur when x is equal to negative pi over 3, and the right vertical asymptote
00:28:46.358 --> 00:29:00.700
would be when x is equal to 2 pi over 3. Now we want to find the center point. Well, the center point is
00:29:00.702 --> 00:29:06.622
located midway between the left and the right vertical asymptotes. Let's go ahead and put our asymptotes
00:29:06.624 --> 00:29:21.133
on the graph now. So we have negative pi over 3. Let's see, if this is 2 pi over 3, then about right here. So I have
00:29:21.135 --> 00:29:31.387
an asymptote here, and I have an asymptote here. So the center point is located midway between the left
00:29:31.389 --> 00:29:37.407
and the right vertical asymptote. Let's find the x-coordinate. Well, I don't need a right coordinate. Let's just write
00:29:37.409 --> 00:29:52.413
x equals. So the x-coordinate will be negative pi over 3 plus 2 pi over 3, divided by 2. Pi over 3 divided by
00:29:52.415 --> 00:30:07.663
2 is pi over 6, and the y-coordinate of this center point is 0. So the center point is (pi over 6, 0). So remember
00:30:07.665 --> 00:30:19.370
it's midway between so let's try to eyeball. That looks about right, so (pi over 6, 0) is my center point.
00:30:19.372 --> 00:30:28.213
Now we need to find the halfway points. Well, let's find the halfway point to the left of the center point.
00:30:28.215 --> 00:30:35.963
The x-coordinate to the left of the center point is located midway between the left vertical asymptote and this
00:30:35.965 --> 00:30:43.215
x-coordinate of the center point. Well, let's see what we get. That's negative 2 pi over 6. So negative pi over 6
00:30:43.217 --> 00:30:57.230
divided by 2 is negative pi over 12. Now we need to find the y-coordinate. Well in order to find the y-coordinate
00:30:57.232 --> 00:31:03.991
we have to recall when you look at the graph of y equals cotangent of x the halfway point to the left has a
00:31:03.993 --> 00:31:12.993
y-coordinate of positive 1, so we want to take that y-coordinate of 1 and we want to multiply it by A. In this case
00:31:12.995 --> 00:31:25.015
my A is 3, so I want to multiply 1 times 3 which of course is 3, so this halfway point is (negative pi over 12, 3).
00:31:25.017 --> 00:31:37.520
So, looks about right here, negative pi over 12, and we want to be up at 3. Now we need to find the halfway
00:31:37.522 --> 00:31:43.518
point to the right of the center point. Well the x-coordinate for this halfway point is going to be located midway
00:31:43.520 --> 00:31:53.266
between the x-coordinate of the center point and the right vertical asymptote, so let's see what that gives us.
00:31:53.268 --> 00:32:03.324
Pi over 6 plus 2 pi over 3 is the same thing as 4 pi over 6. I have pi over 6 plus 4 pi over 6 is 5 pi over 6.
00:32:03.326 --> 00:32:14.081
Well that gives me 5 pi over 12 is the x-coordinate. And then to find the y-coordinate we need to remember
00:32:14.083 --> 00:32:21.339
again the y-coordinate for the halfway point for the basic cotangent is negative 1, so we need to do
00:32:21.341 --> 00:32:33.588
negative 1 times our A, and again my A is 3, so this will be negative 3. So this halfway point is (5 pi over 12, -3),
00:32:33.590 --> 00:32:45.309
and it's located halfway between there. So this would be 5 pi over 12, and then I need negative 3. So this is the
00:32:45.311 --> 00:32:58.875
sketch of this particular cotangent function. Once again for cotangent it's very important that you remember that
00:32:58.877 --> 00:33:06.630
your (Bx minus C) has to be between 0 and pi because that's the principal cycle for the normal cotangent.
00:33:06.632 --> 00:33:12.131
Also make sure you know how to find the center point and the halfway points.
00:33:13.887 --> 00:33:20.633
Let's take a look at the cosecant function. Recall that the cosecant function is the reciprocal of the sine
00:33:20.635 --> 00:33:29.388
function, so it's equal to 1 over the sine of x. In order to get the graph of the cosecant, what we're going to do
00:33:29.390 --> 00:33:36.635
is first look at the sine function and then use the reciprocal identity. So let's do this first. Let's lightly sketch
00:33:36.637 --> 00:33:43.423
the sine function over here. You should remember the sine function very well. Its period is 2 pi.
00:33:43.425 --> 00:34:08.948
It's (0, 0), (pi over 2, 1), (pi, 0), (3 pi over 2, -1), and then (2 pi, back to 0). Of course we can draw another period.
00:34:08.950 --> 00:34:18.197
Ok, so we've sketched sine x, two periods for the sine x. If the cosecant, which it is, is the reciprocal of the sine
00:34:18.199 --> 00:34:31.960
function. First of all, notice that the cosecant is going to be undefined when the sine of x is 0,
00:34:31.962 --> 00:34:46.983
and the sine of x is equal to 0, we can see, when x is equal to n times pi, where n is an integer.
00:34:46.985 --> 00:34:54.732
Now when the sine of x is 0, that's going to give you your vertical asymptotes. So we know at all multiples of n pi
00:34:54.734 --> 00:35:00.951
there are going to be vertical asymptotes. So I know there's going to be an asymptote here on the y-axis, here
00:35:00.953 --> 00:35:19.014
at pi, here at 2 pi, negative pi, and then negative 2 pi. Now, notice that the maximum point for the sine function is
00:35:19.016 --> 00:35:28.776
(pi over 2, 1). The reciprocal of 1 is 1, so this point will also be a point on the cosecant function. Look at all
00:35:28.778 --> 00:35:36.032
these values for sine x between 0 and pi, the sine function, the values are between 0 and 1. If you take the
00:35:36.034 --> 00:35:41.785
reciprocal of those values you get numbers bigger than 1, and it has to approach those asymptotes. So this portion
00:35:41.787 --> 00:35:49.535
of the graph looks like this. Similarly, over here at 3 pi over 2, the sine function is negative 1. The reciprocal
00:35:49.537 --> 00:35:57.043
of negative 1 is also negative 1. So this is going to be a point on the cosecant function. And all these values of sine
00:35:57.045 --> 00:36:03.060
between negative 1 and 0, the reciprocal of those values are numbers less than negative 1, and it of course
00:36:03.062 --> 00:36:10.016
approaches the asymptote, so this portion of the cosecant graph looks like this. So here's one cycle.
00:36:10.018 --> 00:36:23.328
Let's draw another one. Now let's look at some characteristics of the cosecant function.
00:36:25.030 --> 00:36:32.290
Let's look at the characteristics of the cosecant function. First of all, remember that when the sine of x is 0
00:36:32.293 --> 00:36:41.549
it's undefined, so the domain is all real numbers except x equals n pi, where n is an integer. The range...
00:36:41.551 --> 00:36:51.807
The domain is all real numbers except n pi. The range, we need to look at the y-values. So let's look at the y-values.
00:36:51.809 --> 00:37:00.069
Goes down to negative infinity, goes up to negative 1. And then it picks up at 1 and goes to infinity.
00:37:00.071 --> 00:37:09.579
So the range is this. The period is the same as the period for sine which is 2 pi, and because it's undefined it has
00:37:09.581 --> 00:37:17.579
infinitely many vertical asymptotes at x equals n pi, again where n is an integer. You can see these on the graph.
00:37:17.581 --> 00:37:26.598
So asymptotes here, I've shown at negative 2 pi, negative pi, 0, pi, and 2 pi. Now there are a couple of other points.
00:37:26.600 --> 00:37:34.359
For each cycle there are two important points for the cosecant. Let's look at this one first. This point is a
00:37:34.361 --> 00:37:41.862
relative maximum. This value of negative 1 is a relative max, because in this little neighborhood that is the highest
00:37:41.864 --> 00:37:51.369
y-value. Where do they occur? Well it occurs at 3 pi over 2, plus of course 2 pi times n, where n is any integer.
00:37:51.371 --> 00:38:00.626
Adding the period gets me to the next one. The maximum value is of course the y-value, which is negative 1.
00:38:00.628 --> 00:38:09.137
Now let's look at this point. This point is where the relative min occurs. The x-coordinate where it occurs is at
00:38:09.139 --> 00:38:16.390
pi over 2, and of course to get to the next one you add 2 pi, the period, times n, again where n is an integer.
00:38:16.392 --> 00:38:27.657
The minimum value is 1. So those are two important points to know about. So let's see. Let's underline them,
00:38:27.659 --> 00:38:39.156
the relative max and the relative min, and it's important what the y-value is. And the function is odd,
00:38:39.158 --> 00:38:44.906
which means that the cosecant of negative x is negative cosecant of x, which means the graph is symmetric
00:38:44.908 --> 00:38:52.406
about the origin. You need to learn this graph and be able to graph it with showing the two points on each cycle
00:38:52.408 --> 00:38:54.406
and the asymptotes.
00:38:55.906 --> 00:39:03.942
Let's look at sketching functions of the form y equals A cosecant of (Bx minus C). Here my example is y equals
00:39:03.944 --> 00:39:11.198
negative 3 cosecant of (x minus pi). To graph functions in this form, we're going to graph its reciprocal function,
00:39:11.200 --> 00:39:19.951
A sine (Bx minus C), that you should be very good at graphing. So our first step is going to be to sketch the graph
00:39:19.953 --> 00:39:32.471
of negative 3 sine of (x minus pi). For this graph we know the period is 2 pi because B is 1. We know there's a
00:39:32.473 --> 00:39:46.724
phase shift of pi to the right, and we know the amplitude is the absolute value of negative 3 which is 3.
00:39:46.726 --> 00:39:55.728
So let's sketch that graph lightly first. So the shift is pi. So remember, let me make some tick marks here.
00:39:55.730 --> 00:40:15.974
So let's see, pi and we need 3. Ok, so since the phase shift is pi, that means (pi, 0) is going to be a point
00:40:15.976 --> 00:40:22.522
for the sine. And then because A is negative, remember it's going to go down first, so at negative 3 pi over 2
00:40:22.524 --> 00:40:36.766
it's down at negative 3, 0, up to 3, back down to 0. Alright, so there's one cycle for the sine. Let's go ahead and
00:40:36.768 --> 00:40:57.523
sketch another one. So this is negative pi, so it's going to go up to 3, back to 0, down to negative 3, back to 0.
00:40:57.525 --> 00:41:09.060
Ok, now we have two periods for the reciprocal function negative 3 sine of (x minus pi). We want to get the
00:41:09.062 --> 00:41:17.570
graph of the cosecant. If the sine is 0, then we know we're going to have vertical asymptotes. So in this case,
00:41:17.572 --> 00:41:27.833
every time the sine is 0, here at negative pi, 0, pi, 2 pi, 3 pi, that's where your asymptotes are going to be. So
00:41:27.835 --> 00:41:43.605
we have an asymptote here, we have one on the y, we have one at pi, 2 pi, and 3 pi. Ok, so now that we know
00:41:43.607 --> 00:41:55.364
where the asymptotes are, we want to figure out the two points on each cycle. This point right here, this
00:41:55.366 --> 00:42:04.617
(pi over 2, 3), we can see is a relative max for the sine function. That's going to be a relative min for the cosecant.
00:42:04.619 --> 00:42:20.894
And this point, which is, let's go ahead and label it. It was 3 pi over 2, it's down at negative 3. That minimum
00:42:20.896 --> 00:42:36.404
point for the sine becomes a maximum point for the cosecant. Let's draw a graph. Ok. So we've got our points
00:42:36.406 --> 00:42:58.679
labeled. The next thing we want to know is what are the relative maximum points for the cosecant? Well, let's see
00:42:58.681 --> 00:43:05.678
where they are. This point is a relative maximum point and this is a relative maximum point. What point is that?
00:43:05.680 --> 00:43:29.177
It occurs at (negative pi over 2, -3), and then this one is at (3 pi over 2, -3). Ok and for these two periods those are
00:43:29.179 --> 00:43:35.672
the only two relative maximum points. Now what are the relative minimum points? Well for these two cycles
00:43:35.674 --> 00:43:56.423
there are two relative minimum points. Let's go ahead and list those. This one is at (pi over 2, 3) and this one
00:43:56.425 --> 00:44:13.176
is at (5 pi over 2, 3). So we have determined the easiest way to graph the cosecant is always to graph its
00:44:13.178 --> 00:44:18.995
reciprocal of sine which you should be good at. And then once you get that, it's very easy.
00:44:18.997 --> 00:44:25.992
Every time the sine is 0, that's where your vertical asymptotes are. Label, find your two points for each cycle,
00:44:25.994 --> 00:44:32.995
and then sketch your graph. Remember you approach the asymptotes. Once you have that, you can easily find
00:44:32.997 --> 00:44:38.025
your relative maximum and relative minimum points.
00:44:38.027 --> 00:44:43.528
You need to practice working these examples until you have mastered the skills and concepts involved.
00:44:46.030 --> 00:44:51.789
Let's take a look at the secant function, y equals secant x. Recall that the secant function is the reciprocal
00:44:51.791 --> 00:44:59.289
of the cosine. I'm going to write this as 1 over the cosine of x, which means the secant will be undefined
00:44:59.291 --> 00:45:14.811
when the cosine is 0. And when is that? Let's see, undefined. The cosine is 0 at odd multiples of pi over 2,
00:45:14.813 --> 00:45:27.576
which is a way of saying 2n plus 1 over 2 times pi, where n is an integer. What's going to happen for the
00:45:27.578 --> 00:45:34.080
secant is that's where the vertical asymptotes will be. First let's do this. Let's sketch our basic function
00:45:34.082 --> 00:45:43.831
y equals cosine x that you should be very familiar with. So let's see, remember cosine (0, 1), (pi over 2, 0),
00:45:43.833 --> 00:46:00.360
(pi, -1), (3 pi over 2, 0), (2 pi, back up to 1). And then let's draw another period to the left, so (negative pi over 2, 0),
00:46:00.362 --> 00:46:12.608
(negative pi, -1), negative 3 pi over 2, back up. Alright, so here's our basic cosine function.
00:46:12.610 --> 00:46:19.122
So since the secant is the reciprocal of the cosine, we're going to be able to get some points. First of all, remember
00:46:19.124 --> 00:46:27.636
we said that wherever the cosine is 0 the secant will be undefined. That's where the vertical asymptotes will be.
00:46:27.636 --> 00:46:39.389
So I can see that there's going to be a vertical asymptote at pi over 2, 3 pi over 2, negative pi over 2,
00:46:39.391 --> 00:46:52.910
negative 3 pi over 2. Next, this point here on the cosine (0, 1), well the reciprocal of 1 is 1, so this point will also
00:46:52.912 --> 00:46:59.660
be a point on the secant function. And all these values of cosine between 0 and 1, the reciprocal of those values
00:46:59.662 --> 00:47:07.420
are numbers bigger than one, and it approaches the asymptotes. So this is what that portion of the graph looks like.
00:47:07.422 --> 00:47:16.687
Here we have a point for cosine (pi, -1). Well, the reciprocal of negative 1 is of course negative 1, so that will be a
00:47:16.687 --> 00:47:24.194
point on the secant. And all these values for the cosine between negative 1 and 0, the reciprocal of those values are
00:47:24.194 --> 00:47:32.702
numbers less than negative 1, and it approaches the vertical asymptotes as well. Here's one period for the secant.
00:47:32.704 --> 00:47:41.711
Let's draw another. Oh I forgot this portion. Of course reciprocal of 1 is 1, and then it approaches the asymptote.
00:47:41.723 --> 00:47:52.207
So here are two cycles for the secant functions. You need to learn this graph and learn where the asymptotes are
00:47:52.209 --> 00:47:56.714
and those two points for each cycle.
00:47:58.212 --> 00:48:07.464
Let's look at characteristics of the secant function. First the domain, values of x are all real numbers except
00:48:07.466 --> 00:48:16.241
when x is equal to 2n plus 1 over 2 times pi, where n is an integer. That's another way of saying odd multiples
00:48:16.243 --> 00:48:25.003
of pi over 2 are not in the domain. The range would be the y-values. If we look at the graph we can see the
00:48:25.005 --> 00:48:34.253
y-values go from negative infinity up to and including negative 1. And then it jumps up. It goes from 1 to infinity.
00:48:34.255 --> 00:48:43.003
So that would be the range. The period is the same as the period for cosine which is 2 pi. It has infinitely many
00:48:43.005 --> 00:48:51.527
vertical asymptotes, and they occur where it's undefined which is x equals 2n plus 1 over 2 times pi. Again
00:48:51.529 --> 00:48:57.283
same thing as odd multiples of pi over 2. So we can see our asymptotes are at negative 3 pi over 2,
00:48:57.285 --> 00:49:09.298
negative pi over 2, pi over 2, and 3 pi over 2. The next one would be at 5 pi over 2. For each cycle for the
00:49:09.300 --> 00:49:19.303
secant function, there are two important points. One is where a relative max occurs, and it occurs at the x-value pi.
00:49:19.305 --> 00:49:29.301
The next one would occur 2 pi away times n, again n is an integer, and the maximum value is negative 1.
00:49:29.303 --> 00:49:38.826
Again, in this region this is the highest y-coordinate. Here we have a relative minimum point. It occurs
00:49:38.828 --> 00:49:47.584
at x equals all multiples of 2 pi n. So one is at 0. The next would be at 2 pi, negative 2 pi. Again, n is an integer,
00:49:47.586 --> 00:49:58.594
and the minimum value is 1. The function is even, same as the cosine, which means the secant of negative x
00:49:58.596 --> 00:50:08.607
is the secant of x. That's important. Let's underline relative max and relative min. And because it's even,
00:50:08.609 --> 00:50:15.112
that means that the graph is symmetric about the y-axis. It's important that you know how to graph the secant,
00:50:15.114 --> 00:50:21.110
and know where the asymptotes are and where those two points are.