Graph the functions in Exercises 23–26 in the ts-plane (t-axis...

Question

Graph the functions in Exercises 23–26 in the ts-plane (t-axis horizontal, s-axis vertical). What is the period of each function? What symmetries do the graphs have?

s = −tan πt

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Graph the function Y equals negative tangent of 2 pi X. Determine its period and the type of symmetry it exhibits. Consider only 1 period when graphing. Awesome, so it appears for this particular problem we're ultimately trying to create a graph for this function of Y. So now that we know that we're trying to graph this function of Y and we're also trying to determine its period and the type of symmetry it exhibits, and we're also told to consider only one period when we graph, so we only need to graph one period. So with that said, now that we know what we're solving for, let us quickly note that our function is in the form of Y equals. A multiplied by tangent. Of BX, where in this case A is equal to -1. And let us also note that B in this case is equal to 2 pi. Fantastic. So now, our next step is we need to note that the period in this particular case is going to be pi divided by B, so that means it's going to be equal to pi divided by 2 pi, which is going to be simply equal to 1/2. So now our next step is we need to locate the two adjacent vertical asymptotes. So we need to note that BX is equal to negative pi divided by 2. We also need to note that 2 pi X is equal to negative pi divided by 2. And when we rearrange this expression to isolate and solve for X, we will find that X is equal to. Once again, we're rearranging our expression of 2 pi X is equal to negative pi divided by 2. So when we rearrange that to isolate and solve for X, we'll find that X is simply equal to -14. Let us also note that we can go ahead and easily write that B X is equal to divided by 2. Which we can go ahead and write that 2 pi X is equal to positive pi divided by 2. So once again, rearranging this expression to isolate and solve for X, we will find that X is simply equal to 14. So thus, the two adjacent vertical ascentotes are X is equal to 1/4 and X is equal to 1/4. So if we draw those vertical asymptotes or vertical, yeah, the vertical asymptotes onto our graph, we'll graph one at 1/4 and at Positive 1/4, which we use dotted lines to denote our asymptotes. So I used red dotted lines to represent my asymptotes. So when you draw this on your graph, make sure you use a dotted line to indicate that it is an asymptote. So with that said, our next step now is we need to divide the interval formed by X equals -1/4 and X equals 1/4 into 4 equal parts. So when we do that, we will find that the period P divided by 4 is going to be equal to 1/4 minus -14, all divided by 4, which when we simplify, we will find that this is equal to 1/8. So let us also note that we can go ahead and write that X1 is equal to -14, and let us also note that X2 is going to be equal to X1 + 1/8, which is going to be equal to -18. And X3 is going to be equal to X2 plus. 18 And X3 is equal to X2 + 1/8, which is going to be simply equal to 0. And X4 is equal to X3 + 1 divided by 8. Which is gonna be equal to simply 1/8, whereas X2 is equal to -18, X4 is going to be equal to 18, and then finally, X5 is equal to X4 + 1/8, which this is going to be equal to 1/4, noting that we're taking our value for X1, which is going to be equal to 1/4, and we plug it in for the value of X1 and then plus 1/8 and then following the same pattern we plug in the value for X2, X3, X4, and we add 1/8 to get our get the resulting answer. So now, our next step that we need to take is we need to evaluate the functions at X2, X3, and X4. So at Y of -1/8, it's going to be equal to negative, tangent of 2 multiplied by negative 1/8. And this is going to be equal to when we plug it into our calculator, it's going to be simply equal to 1. And instead of rewriting the whole equation out every single time, I'll just give you the answer. So Y of 0 is going to be equal to 0 and Y of 18. is going to be equal to -1. Once again, that's when you're plugging in a value of 0 in for X, and when you're plugging in a value of 1/8 in for X. So now, Our next step is we need to plot the following points. We need to plot -18.1. We also need to plot 0.0, which we should recognize as the origin point, and we also need to plot the 018.1. And we're going to plot these 3 points onto our graph and then connect them with a smooth curve which will approach the vertical absentos, but they will not touch them. So when we graph these 3 points, we should produce the following graphs. So I've gone ahead and plugged our function into some graphing software, and when I did so, I produced the following graph. And as we can see, we have our vertical asymptotes at -1/4 and 1/4, and like I said, we have our our three coordinate points that we plotted at -1/8.1, at 0.0, and at 1/8.1, and we connected them with a smooth curve and it starts up in the second quadrant, following our vertical and then it goes dips down, then it starts to curve like as a cubic function. Almost, and curve cross through the origin point in the center which intersects both the Y and X axis. And then we'll curve down to strike our 3rd coordinate point and then dip down going into the 4th quadrant following our vertical asymptote. Note that our curve will get very, very close to our vertical asymptotes, but it will never cross or intersect our vertical asymptotes. And that's it. We've officially solved for this problem. So to quickly recap here, We need to note that once again, our period, and let's say our period P is gonna be equal to 1/2. And our symmetry is going to be symmetric about the origins. So, let's make a note of that. So symmetric. About The origin. So it's gonna be symmetric about The origin. So let's quickly recap here. To quickly note, so we know exactly what's happening here. So we plotted our points and we produced the following graph, but we need to note. Really quickly here, let us note that. And let's write this in blue so we don't get confused. That F of X is going to be equal to negative tangent of, and we're running out of room here. So let's move it down a little bit so we can see. So it's gonna be negative tangent of 22 pi X, fantastic, that's our original function. So, let us note that because F of X is equal to this value, we need to recall that tangent of negative X. So let's note that Tangent of negative X is gonna be also equal to negative tangent of X. So thus we can go ahead and write, let's scroll down, so we can go ahead and write that F of negative X is equal to negative tangent of 2 pi multiplied by negative X. And this, when we do plug this in, so 2 multiplied by negative X, we will get an answer that is equal to negative F of X, because this is also equal to tangent of 2 X. Fantastic. So, therefore, without a doubt, Y will equal negative tangent of 2 pi X, and it is an odd function. So like, I'll draw a little arrow to note that it's an odd function, so an arrow pointing to odd because we're dealing with an odd function, so. Fantastic. So it's odd and it will be symmetric about the origin like we discussed, and that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Graph the functions in Exercises 23–26 in the ts-plane (t-axis horizontal, s-axis vertical). What is the period of each function? What symmetries do the graphs have?

s = −tan πt

Key Concepts

Graphing Functions

Periodicity

Symmetry in Graphs

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