In Exercises 5–12, graph two periods of the given tangent functio...

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In Exercises 5–12, graph two periods of the given tangent function.y = 3 tan x/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 two tangent of X over six. Now, I wanna do a quick rewrite of this function. If we take a look inside the argument of tangent, we have X over six and X over six is the equivalent of 1/6 times X. So we can rewrite the given function as Y is equal to two tangent of 1/6 X. Now, before we identify the graph for the given function, let's take a look at the parent function. The pair function is going to be tangent of X. The parent function is going to have a period of pi and it contains two ascent totes. The first ascent tode is located at negative pi over two. And the second aceto is located at positive pi over two. The pair function contains an X intercept at the origin zero comma zero and tangent contains a shape like structure which roughly looks like this. So now that we know the shape of the original tangent function, let's go ahead and identify all the transformations that are happening in our given function, the first transformation occurs from the coefficient of two two represents a horizontal stretch of the given tangent graph. Now, normally what this means is that the amplitude is going to be stretched. But tangent does not have an amplitude because the range of tangent is from negative infinity to positive infinity. So instead we can use this value of two to find two points on the tangent function that represent the horizontal stretch. This value won't tell us what the X values of the point are going to be. But it will tell us that the Y values of the points are going to be plus or minus two. The next transformation occurs from the coefficient of 1/6 in front of the X term. No, the coefficient in front of X is not one, it's 1/6. And this indicates that there is a phase shift for the tangent graph. What this means is that the period of the graph is changing. So how do we identify the new period while the period is given to us as pi over B and B represents the coefficient in front of the X term. So in this case here, the period can be rewritten as pi over 1/6. In order to simplify this, we multiply both the numerator and denominator by the reciprocal of the denominator, the denominator is going to simplify to just one. And what we are left with is pi times 6/1, which is going to give us six pi. So this is going to be the distance of the new period. Now, what we need to do is we need to identify the locations of the new ascent totes. Now, for the original tangent function, the ascent totes are given to us as negative pi over two or I'm sorry, X is between negative pi over two and positive pi over two. But in this case here we are given positive 1/6. So we can rewrite the asymptotes as 1/6 X between negative pi over two and pie over two. And what we wanna do now is you wanna suffer X for the given inequality in order to do this, we're gonna multiply every term in this inequality by six. That's going to leave us with just X in the middle of the inequality. And X is going to be between to be between the values of negative three pi and positive three pi. So that means that the asymptotes for the first period of the function are going to be identified as negative three pi and positive three pi. And since there is no horizontal shift or vertical shift to the graph, that means the X intercept is still going to be at the origin zero comma zero. So now that we have the locations of the first set of asymptotes and we have the X intercept, we can go ahead and sketch the first period of the function. So let's go ahead and draw an axis. Recall that the location of the first ascent tode is going to be at negative three pi and the location of the second as is going to be a positive three pi the X intercept is going to be located at the origin zero comma zero. And because we are not doing any reflections about any axis, the shape of the function is gonna remain the same. So we still have this snakelike structure. Now we need to identify the points that show that there is a horizontal stretch. Now recall from earlier that we know that the Y values of these points are gonna be plus or minus two, but we need to find the X values of these points. So how do we go about doing that? Well, we know that this point has to lie some distance between the first Asymptote and the X intercept. So what we can go ahead and do in order to get the first X value is we can take negative three pi and add the distance of zero to it and divide this distance by two, negative three pi plus zero is going to give us negative three pi and negative three pi over two is in its simplest form. So the location of the first point is going to be at negative three pi over two. Now we want to find the second X value for the second stretch point. Well, that's gonna be between zero and the second ascent to at three pi. So we can add that distance zero plus three pi and divide that distance by two. And that is going to leave us with positive three pi over two. So the X value for the second point is going to be three pi over two. Now this is the fully labeled version of the first period of the graph. But recall that the problem wants us to graph two periods of the function. So we need to identify a third ascent tote for the graph. Now keep in mind that the period is six pi. In order to get the third ascent tode, we're going to be adding six pie to three pi and that is going to give us the value of nine pi the shape of the graph of the second period is gonna mimic that of the first period. And this is going to be the complete graph for the given 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.

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In Exercises 5–12, graph two periods of the given tangent function.y = 3 tan x/4

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