In Exercises 45–52, graph two periods of each function.

y = 2 ta...

Question

In Exercises 45–52, graph two periods of each function.

y = 2 tan(x − π/6) + 1

Accepted Answer

Hello. Today we're gonna be drawing the graph of two periods of the function given below. So what we are given is Y is equal to tangent of X minus pi over six. Now, before we take a look at the graph of this function, let's take a look at the pair function tangent of X. So the pair function is given to us like this, we have two asymptotes for tangent of X. The first ascent tote is gonna be located at negative pi over two. And the second one is located at positive pi over two. The function has a period of pi and it has an X intercept at the origin zero comma zero. The tangent function also has a snakelike structure which roughly looks like this. So in order to graph the given function, we're gonna go ahead and identify all of the transformations that occur on the given function. So taking a look at the pair function, the difference between the pair function and the given function is the quantity X minus pi over six. So we have X minus pi over six. This is in the standard form of X minus H. And what this means is that we have a horizontal shift for the graph. Now because X minus pi over six is in the standard form of X minus H. This means that H is equal to pi over six. And what this means is that the graph is going to be shifting pi over six units to the right. So we're going to be taking our X intercept and shifting it pi over six units to the right. But if we are shifting the X intercept, we must also shift the two asy totes. Now keep in mind that the two asymptotes are given to us as negative pi over two comma pi over two. So in order to identify the new asymptotes, we're going to be adding pi over six to negative pi over two and pi over two. So negative pi over two plus pi over six is going to give us the value of positive I'm sorry, negative pi over three and pi over two plus pi over six is going to give us the value of positive two pi over three. So that means that the new asymptotes are gonna be located at negative pi over three and positive two pi over three. And since we are shifting the X intercept, pi over six units to the right, the new X intercept is gonna be located at pi over six comma zero. So now that we have the locations of the asymptotes and we have the location of the X intercept, let's go ahead and draw the first period. So we know that the location of the first ascent tote is going to be at negative pi over three. And we know the location of the second ascent tote is going to be at two pi over three. The X intercept for the first period is gonna be located at pi over six comma zero. And since we are not having any reflections of the graph, the shape of the tangent graph is going to remain the same. So this is going to be the rough shape of the first period of the tangent graph. But keep in mind that the problem asks us to graph two periods of the function. So we need to identify where a third ASOM code exists. On the on the graph will recall that the period is pi, the distance between negative pi over three and two pi over three is pi. So in order to get a third ascent to, we're going to be adding the value of pi to two pi over three, two pi over three plus Pi is going to give us the value of five pi over three. So the location of the third ascent tode is going to be at five pi over three and the second period is going to mirror the first period. So the second period is gonna roughly look like this and this is going to be the graph of the given tangent function. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 45–52, graph two periods of each function. y = 2 tan(x − π/6) + 1

Key Concepts

Tangent Function

Transformations of Functions

Graphing Periodic Functions

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