In Exercises 18–24, graph two full periods of the given tangen...

Question

In Exercises 18–24, graph two full periods of the given tangent or cotangent function. y = −2 tan π/4 x

Accepted Answer

Hello, today we're gonna be graphing two periods of the given function. So what we are given is Y is equal to negative three tangent of pi over six X. Now, before we graph this function, let's go ahead and take a look at the parent function which is Y is equal to tangent of X. The parent function has two asymptotes for the graph. The first Asymptote is located at negative pi over two. And the second ascent to is located at positive pi over two, the graph has an X intercept of zero comma zero and the graph of tangent has a snakelike structure which roughly looks like this. Furthermore, the period of this graph is going to be pi. So now that we have the information for the parent function in order to graph our given function, let's first identify all the transformations that are occurring on the tangent graph. The first transformation occurs with the negative sign in front of the graph. The negative sign indicates that there is going to be a reflection about the X axis. Next, the second transformation occurs with the leading coefficient of negative three. Normally this would represent the amplitude of graph. So we want to take the absolute value of negative three, which is going to be three. Now again, this will normally change the amplitude of a trigonometric graph. But for tangent tangent does not have amplitude because the range of the graph is from negative infinity to positive infinity. Instead what this three tells us is that there is going to be a stretch of the graph. And this three is going to help us find two points on the tangent graph that represent this stretch. It won't tell us what the X values of these points are going to be. But it will tell us that the Y value is going to be plus or minus three. Finally, the next transformation occurs from the coefficient of pi over six for the X term. Now the coefficient of the X term is not one and this indicates that there is going to be a shape, a phase shift for the given graph. What this means is that the period of the graph is going to change. Now, the period of a tangent graph is given to us as pi over B and B represents the coefficient in front of the X term. In this case, here B is going equal to pi over six. So we can write the period as Pi over pi over six, we have a complex fraction. And in order to simplify this, we're gonna multiply the denominator by its reciprocal and multiply the numerator by the reciprocal of the denominator. So this is going to allow the denominator to simplify to one. And what we are left with is pi multiplied by six over pi we can represent pie as a fraction by putting it over one. And we can simplify both of the pies to just one that, that leaves us with one time six, which is going to give us six. So what this means is that the period is going to be six units long. Now, what we wanna do is we wanna find the locations of the new ascent totes since the period is going to change. Now, for the original tension graph, the period, the asymptotes are represented as negative pi over two and pi over two. And we know that X needs to lie between the values. Well, in this case, here we are given pi over six X. So what we wanna do is you want to put pi over six X between pi over two and negative pi over two and we want to go ahead and isolate X in the given inequality in order to do this, we're gonna multiply every term in the inequality by six over pi. And this is going to simplify the inequality to give us negative three is less than X is less than three. So what this means is that the locations of the new ascent totes are going to be negative three and positive three. Now, one more thing to note is that there is no horizontal shift of the graph. So the X intercept is going to remain the same at the origin zero comma zero. So now that we have the locations of the new asymptotes, we have the X intercept and we have the period, let's go ahead and graph this information. So we're gonna go ahead and draw an axis. Now we know that the first period or the first Asymptote is going to lie at negative three. And we know that the second Asymptote is going to lie at positive three, the X intercept is going to remain the same at the origin zero comma zero. But the first transformation is a negative sign. We're going to be flipping the shape of the graph about the X axis. And that means the new shape of the tangent graph is roughly going to look like this. Now recall that we stated that there is a stretch of the graph, the stretch or at least the leading coefficient of the amplitude is three. So we know that the Y values for two points that represent the stretch are going to be negative three and positive three. But we need to find the X values for these points. How do we do that? Well, this point is going to lie halfway between the distance of the first Asymptote and the X intercept. So what we can go ahead and do is we can take that distance, which is represented by negative three plus zero and divide that distance in half. This is going to simplify to give us negative 3/2. And that means that the first X X value for the first point is going to be negative 3/2. Now we want to repeat this process for the second point of the stretch. Now this point is gonna lie between the X intercept and the second as Tode, that distance can be represented as zero plus three. And we're going to divide that distance in half in order to get the stretch of the point. So we get zero plus three, which is going to simplify to 3/2. And that means the X value of the second point is going to be 3/2. So this is the fully labeled version of the 1st 1st rotation of the tangent graph. But note that the problem asks us to graph two periods of the tangent graph. So we're going to need to identify a third as Tode for the tangent graph. Now keep in mind the period represents the distance between one Asymptote to the next. The distance from negative three to positive three is a total distance of six. And in order to get a third ascent Tode, we just need to add another distance of 6 to 33 plus six is going to give us nine, which means that the third ascent Tode is located at nine and the second period is going to mimic the same image as the first period. So the second period is going to have the same shape and this is going to be the graph of two periods 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 18–24, graph two full periods of the given tangent or cotangent function. y = −2 tan π/4 x

Key Concepts

Period of Tangent Function

Amplitude and Vertical Stretch

Asymptotes of Tangent Function

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