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

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. And the function we're given is Y is equal to negative four tangent of one half X. Now our gra only one period. So we want to think about what is the period of this function. OK. Well, we have Y is equal to negative or tangent of one half X. Right now, the regular period of tangent just tangent of X is P, the period of the function we have, it's gonna be related to that period P through this B value A and the B value is the value that coefficient in front of the X inside of the brackets. And whatever number is multiplying that X. So in this case, B is one half. Now the period of our function T is going to be equal to that regular period pi of just tangent of X divided by that B value. So in this case, it's pi divided by one half or we can simplify this as two P. OK. So we have a period of two pie and that's what we wanna grow. Now, this blank graph that we're given that we're gonna draw on goes from negative pi all the way up to pi. OK. So that is a two pie period there. So we're gonna use that as our period. We're gonna graph from negative pi to pi. All right. Now, another thing we can notice is this negative sign out in front and we have a negative out in front of our function. What this means is that we have a vertical reflection. So our usual tangent curve we expect to do something like this. OK. So it starts at a negative value in increases throughout one period. But when we reflect that, we're gonna expect that it does the opposite. So it's gonna start positive and become negative. All right. So that gives us an idea of what the shape of the graph will look like. Now, we know that tangent has some asymptotes, OK? It has vertical asymptotes. So let's go ahead and figure out what the asymptotes of this particular function are. OK. So we're gonna go ahead and we're gonna find the vertical asymptotes or VA s. Now we have Y is equal to negative four tangent of one half X again, just rewriting that function. But recall that tangent is equal to sine divided by cosine. So we can write this function as negative for sign of one half X divided by cosine of one half X. And when we do that, it's easier to see where those vertical asymptotes come from. OK. If the denominator of our function is zero, the function will be undefined. That's where we're gonna get those vertical asymptotes. Now, the denominator is cosine of one half X. So our vertical asymptotes are going to occur when cosign of one half of X is equal to zero. Hm. All right. Well, what does cosine equals zero? Cosine equals zero if the thing inside of the cosine is some multiple of pi divided by two, an odd multiple. So the thing inside of our cosine here is one half X cosine is gonna equal zero when this is pi divided by two, OK. Plus K pie or K is an integer. OK. So at pi divided by two or three pi divided by two or five pi divided by two K as K increases, these are the possible values where cosine will be zero. Now, we wanna figure out the restriction on X, OK, not on one half of X. So we wanna isolate for X. So we're gonna do that by multiplying both sides of this equation by two. When we do that, we get that X for our vertical asymptotes is going to be equal to pi plus two K P A again K is an integer. Now we're graphing from negative pi to pi so we don't care about all of the possible values of X. We just wanna find the ones in that interval that give us vertical asymptotes. So we wanna look on our interval from negative pi up to pi, OK. And what we'll see is that we have vertical asymptotes at X equals negative pi. Can we get that if K is negative one? And if K is zero, we get that X is equal to pi. OK. So that is our next vertical Asymptote. So we have vertical asymptotes at the end of this period that we're graphing. OK. So we're gonna go ahead and add those two asymptotes to our graph. We're gonna draw them as dotted lines. OK? So an X equals negative pi is the far left of our graph. We're gonna draw a dotted vertical line and then an X equals pi is the far right of our graph and again a dotted vertical line. OK. So we know the period, we have an idea of what this function is going to look like or what the graph is going to look like based on the reflection. We know where the asymptotes are. The last thing to do is to do a little table of values. And we're gonna work out some of these values in between our two asymptotes within this period so that we can fill in the gaps. OK. So we have our X values, we have our Y values and we're gonna choose the X values that are shown on our graph. OK. So we have X equals negative pi negative pi divided by two zero pi divided by two and pi now at negative pi and pi we know that these are our vertical asymptotes. So Y is going to positive or negative infinity. And based off of the other values, we'll be able to infer which one it is. Now when X is negative pi divided by two, our function becomes negative four tangent of negative pi divided by four. Now tangent of negative pi divided by four is negative one multiplied by negative four. This gives us positive four. What about zero? When X is zero, we have negative four tangent of zero, tangent of zero is just zero. And so this entire thing works out to zero. When X is pi divided by two, we get negative four tangent of pi divided by four tangent of pi divided by four is positive one multiplied by negative four. This gives us negative four. So we have now our vertical asymptotes and these three functions or these three points that we're interested in graphing. Let's go back to our graph, add them in and will be done. So at X equals negative pi divided by two. We said that we were going to be at about at four. And so that's about four. On our graph, X equals zero, Y is equal to zero, X equals pi divided by two, Y is equal to negative four. It's about their honor graph. And we can see now that on the left, at our left Asymptote and negative pi we're gonna be approaching positive infinity. And as we approach positive pi our function is going to approach negative infinity. OK. So we are getting this vertical reflection type curve that we expected when we, when we talked about that vertical reflection. OK. So we're gonna join these points with a nice smooth line looking like a tangent curve in approaching those asymptotes. And that is the graph that we wanted. 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

Tangent Function

Transformation of Functions

Period of a Function

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