In Exercises 17–24, graph two periods of the given cotangent func...

Question

In Exercises 17–24, graph two periods of the given cotangent function.y = −3 cot π/2 x

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Hello, today we're gonna be drawing the graph of two periods for the given function. So in order to draw the graph, we're going to need to first identify all the transformations that are happening to the Cotangent graph. The first transformation occurs with the negative sign in front of the Cotangent graph. The negative sign indicates that the graph is being reflected about the X axis. So that means that we are flipping the graph about the X axis. The next thing we want to identify is the fact that the coefficient in front of X is not one, it's pi over three. Since the coefficient in front of our X term is not one that means that there is going to be a phase shift for our given graph. And what a phase shift means is that the period of the cotangent graph is going to be changed. So how do we identify the new period? Well, the period is defined as Pi over B and B represents the coefficient in front of the X term. In this case, here B is going to be pi over three. So we can write the period as Pi over pi over three. In order to simplify this, we're gonna multiply the numerator and denominator by the reciprocal of the denominator. This is going to allow the denominator to simplify to one. And in the numerator, we have pi times three over pi both of the pies are going to simplify to just one and that leaves us with three times one which is three. So the period for the given graph is going to be three units long. Now, another transformation we can identify is going to be the coefficient of four in front of the Cotangent graph. Now typically when you have a coefficient of not one in front of any trigonometric function, this indicates that there is a change in amplitude, but cotangent doesn't have an amplitude instead what this means is that we can use this value to locate two points on the graph itself. We need to identify the X values for these points. But we know that the Y value is gonna be plus or minus four. And finally, we need to identify the X intercept for the given cotangent graph. In order to get the X intercept, we're going to be taking our period and dividing it by two, our period was three. And if we plug this in, we get an X intercept of 3/2, so the X intercept is gonna be located at 3/2 comma zero. So now we have a couple of pieces of information to graph for a cotangent graph. Let's go ahead and take a look at the parent function for cotangent, the parent function of cotangent has two asymptotes located at zero. And that pie, we also know that the X intercept is located at pi over two comma zero. And the shape of cotangent roughly looks like this. So if we draw our new graphic cotangent, let's go ahead and start by labeling our new asy totes. So we identified that the asymptotes or at least the period of the graph is going to be three. And that means that the Asymptote is going to start at zero. But the second Asymptote is going to be located at three instead of pi. Now again, keep in mind that we want two periods for the graph of the function. So we're going to need to identify a second ascent tode for the functional cotangent and that's going to be located at six. Now, if you're wondering how we got this value, the distance for the period between zero and three is three, and if we add another distance of three to the ascent tode of three, we get a new ascent tode at six. Now let's go ahead and label the X intercept for the new cotangent graph. The X intercept is located at 3/ comma zero. And using our first transformation, we know that we are reflecting the shape of the cotangent graph about the X axis. So if we are reflecting the shape of the graph about the X axis. The new shape of the cotangent graph is roughly gonna look like this. Now, what we wanna do is we want to identify two points on the co tangent graph and this is where the transformation of four comes into play. See we just need to identify what the X values of these points are going to be because we know that the Y, the Y values of these points are going to be a negative four and positive for. So how do we get the X values for these two points? While we know that this point here is located some distance between zero and the X intercept 3/2. So what we can do is we can take zero and add 3/2 to this distance and divide the distance in half to find the X value zero plus 3/2 is going to give us 3/2 and that's gonna be divided by two. And if we simplify this value, we get an X value of 3/4, that means that the first point on the cotangent graph is gonna be located at 3/4 comma negative four. Now let's go ahead and identify the point, the second point on the cotangent graph. Again, this point is located between the X intercept and the second asy tote of positive three. So we can add the distance between 3/2 and three and divide that distance by two to get that X value 3/2 plus three is going to give us 9/2 divided by two. And simplifying this is going to give us the value of 9/4. So that means that the second point on the cotangent graph is gonna be located at 9/4 comma positive four. And now that we know the shape of the first period of the function, the second period is just going to mirror that same shape. And this is going to be the graph for the given co Taio function. So I hope this video helps you in in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 17–24, graph two periods of the given cotangent function.y = −3 cot π/2 x

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