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

Question

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

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Hello, today we're gonna be graphing two periods of the given function. So what we are given is Y is equal to negative tangent of X minus pi over three. Now, before we graph this function, let's go ahead and take a look at the parent function which is just tangent of X. The graph of tangent of X has two asymptotes. The first Asymptote is located at negative pi over two and the second Asymptote is located at pi over two. The graph has an X intercept at the origin zero comma zero. And the shape of the tangent graph roughly looks like this. Furthermore, the period of the tangent graph is going to be pi so this is going to be the parent function of our given function. So in order to graph the given function, let's go ahead and identify all the transformations that are occurring on the tangent graph. The first transformation comes from the negative sign in front of the tangent graph. The negative sign indicates that we are going to reflect the graph about the X axis. Next, let's take a look at the insight argument of tangent. So we are given the quantity X minus pi over three. This represents a horizontal shift in the graph as it is represented by the quantity X minus H. Now X minus pi over three is in the standard form of X minus H. This is where H is going to equal to positive pi over three. And that means we have a horizontal shift of the graph pi over three units to the right. So what this means is that the original X intercept is going to be shifted pi over three units to the right. But if we are shifting the X intercept to the right, we must also shift the two asy totes the same distance as well. So that means the measures of the asymptotes are going to change along with the X intercept. So we need to identify the new uh the new asymptotes. So the original ascent totes for the parent function are given to us as negative pi over two and positive pi over two, we are shifting these ascent totes pi over three units to the right. So we're going to be adding pi over three to both of the ascent totes. So the first Asymptote can be represented as pi over two plus pi over three. And the second ascent tote can be represented as positive pi over two plus pi over three. Now, we just need to add these values together negative pi over two plus pi over three is going to give us the value of negative pi over six and pi over two plus pi over three is going to give us the value of five pi over six. So these are gonna be the locations of the new ascent totes for the tangent function. Now that we have the locations of the asymptotes and we have the location of the X intercept. Let's go ahead and graph the function. So we're going to draw an axis. Now, the first asom tote is located at negative pi over six. And the second Asymptote is gonna be located at five pi over six. Now we shifted the X intercept pi over three units to the right, which means the new X intercept is going to be located at pi over three comma zero. And one more thing to note is that we are reflecting the original tangent graph along the X axis. So that means we're going to be flipping the shape of the tangent graph and the new shape of the tangent graph is roughly gonna look like this. So this is going to be the first, the first period of the tangent graph. But recall that the problem asks us to graph two periods of the function. So we need to identify a third Asymptote for the function. So that way we can graph a second period of the function. Now what is the location of this third cento while the period of the function never changed, the period is still going to be P. So since the period is pi, that means the length from one as to to the next is gonna be a total length of pi. So the location of the third Asymptote is going to be five pi over six plus another distance of pi and five pi over six plus pi is going to give us 11 pi over six. This is going to be the location of the third Asy Tode and the second period is going to I reflect the image of the first period. So this is going to be the shape of the second period of the tangent graph. And this is going to be the complete graph of two periods of the given tangent graph. 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 = −tan(x − π/4)

Key Concepts

Tangent Function Properties

Transformations of Functions

Graphing Periodic Functions

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