Graph each function over a one-period interval.
y = 2 ta...

Question

Graph each function over a one-period interval.

y = 2 tan (¼ x)

Accepted Answer

Hey, everyone in this problem, we're asked to graph the following function and to consider only one period. So the function we're given is Y is equal to five tangent of one half X. We're given this buoyant coordinate plane to draw our graph on. So first thing we want to think about is what is the period of this function? OK. We're told to only graph one period. What is that period? So we're gonna rewrite our function Y is equal to five multiplied by a tangent of one half of X. And when we have our function written in this form and we're interested in the period, we want to think about the B value and the B value comes from the term that is multiplying the X inside of the brackets. OK. So we want that constant term multiplying the X which in this case is one half. Now how we get the period from that B value is by looking at the period of just the regular tangent function, OK. The period of the regular tangent function is just pi and so the period of our new function which we're gonna call T is going to be that regular period pi divided by our B value. So in this case, it's going to be pie divided by one half, which gives us a period of two pi. OK. So we know that our function has a period of two pi the blank graph we're given goes from negative pi to pi. So that's what we're gonna do. We're gonna graph it from negative pi to pi. All right. Now, the next thing we wanna do is look at any asymptotes. OK? We're dealing with the tangent function. So we know that there will be vertical asymptotes. So let's find our vertical asymptotes and we're just gonna short form that as VA S. Now we have tangent, OK? Um Five, sorry five tangent of one half X now recall that tangent is equal to sine divided by cosine. So we can write this function as five sign of one half X divided by cosine of one halfway. And when we do that, it's easier to see where those asymptotes come from. And remember that our vertical asymptotes are gonna occur if the denominator is zero, maybe because if the denominator is zero, our function becomes undefined. So in this case, we have a vertical Asymptote when cosine of one half of X is equal to zero, which tells us that one half of X is equal to. Well, one is cosine zero, cosine is zero at pi divided by two. Yeah, pi divided by two plus K pi K where K is an integer. So pi divided by 23, pi divided by two, et cetera, et cetera. Yeah. So we wanna isolate for X. These are gonna be the X values where we have vertical asymptotes for our function. Instead of the regular tangent function, we multiply both sides by two. And we get that X is going to be pi who was two IK or K again is an integer. So we're graphing from negative pi to pi. OK. So on that interval from negative pi to pi, we're gonna have vertical asymptotes at well, X equals negative pi. OK. We get that if K is equal to negative one and then if K is equal to zero, we get a vertical Asymptote at X equals pi. OK. All right. So these are the two vertical asymptotes we have for the interval that we're graphing. Let's go up and add those to our diagram. Yes, we have our graph and we have these vertical asymptotes and we're gonna draw them just as dotted lines at negative pi and that pie. Mhm All right. Now, let's take a look at our function again. Now, the term out in front, we have a five out in front but it's positive. OK. So this function hasn't been reflected um vertically from the regular tangent function. We're not adding anything at the end and we're not adding anything inside the brackets. So we're not having any vertical or horizontal shifting. OK. So what we expect is that this function is gonna take the typical shape of a tangent function. It's just gonna be stretched and compressed a little bit. It's not gonna be reflected, it's not going to be shifted. Now, how we can find that out exactly is by looking at a table of values. So let's go ahead and do that. We have X values and we have Y values and we're just gonna choose the X values that are shown on our graphs. So we have negative pi negative pi divided by 20, pi divided by two and pi now at negative pi and that pi that's where we have our vertical asymptotes. OK? Function is undefined there. So we don't need to worry about those points at negative pi divided by two. We have five tangent of negative pi divided by four. Hey, this is gonna give us five multiplied by well, tangent of negative pi divided by four is negative one. So we get five multiplied by negative one which gives us negative five. Now tangent at zero, that's gonna be a zero. OK. And then a pi over 25 tangent of pi divided by four is what our function becomes. And this is gonna be equal to five multiplied by positive one. And so we get five there. OK. So we have our vertical asymptotes, we have these three points in between and that's gonna be enough to graph this. So let's go back to our diagram. We're in a graph in green just so that we can see it. So we have our 0.00. And then at negative pi divided by two, we have negative five, uh positive pi divided by two, we have positive five. And we're gonna connect these in a smooth way like our tangent function where that function is going to approach those vertical asymptotes. And so our function is going to look something like this. All right, thanks everyone for watching. I hope this video helped see you in the next one.

Graph each function over a one-period interval.
y = 2 tan (¼ x)

Key Concepts

Period of a Trigonometric Function

Transformation of Functions

Graphing the Tangent Function

Watch next