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 = − 1/2 cot π/2 x

Accepted Answer

Hello, today we're gonna be graphing two periods of the given function. So what we are given is Y is equal to negative 1/ cotangent of pi over three X. Now, before we graph this function, let's go ahead and take a look at the parent function which is just cotangent of X. The parent function of contingent of X has two asy totes that are located at zero and at pi the function has an X intercept at pi over two. And the shape of the cotangent graph roughly looks like this. Furthermore, the period of the original cotangent graph is given to us as pie. So in order to graph the function that is given to us what we want to go ahead and do is identify all the transformations that are occurring on the cotangent graph. The first transformation occurs with the negative sign in front of the function. A negative sign indicates that we are reflecting the graph about the X axis. So we are going to flip the graph around the X axis. The second transformation comes from the coefficient in front of the cotangent graph. The coefficient is negative 1/4 which normally indicates that there is a change in the graph's amplitude. So what we wanna do is you want to take the absolute value of this coefficient which is going to be 1/4. And because this value is less than one, this indicates that there is going to be a shrink in the graph. Now, normally the amplitude would change because of this coefficient. But notice that cotangent doesn't have an amplitude because its ranges from negative infinity to positive infinity. Instead, what this value allows us to do is find two points on the cotangent graph that indicate the shrink of the graph. Now this value won't give us the X value of these two points. But it will tell us that the Y values are going to be plus or minus 1/4. Now the third transformation occurs from the coefficient in front of the X term. The coefficient is given to us as pi over three. And this is going to indicate a phase shift for the given Cotangent graph. So the phase shift means that there is going to be a change in the graphs period. In order to find a new period, the period is defined as pi over B and B represents the coefficient that is in front of the X term of the cotangent function. So in this case, B is going to equal the pi over three. So we can rewrite the period as pi over the quantity pi over three. Now, in order to simplify the period, we're gonna multiply the denominator by its reciprocal and we're gonna multiply the numerator by the reciprocal of the denominator. The denominator is going to simplify to just one. And what we are left with is pi times three over pi which is going to leave us the value of three. So that means that the phase shift is now going to be three or in other words, the new period for this graph is going to be three. Now we need to find the X intercept. Now the X intercept for graph is defined as the period divided by two. Earlier, we found that the period is three. So if we plug in the period for three, we get 3/2, which means that the X intercept for this function is going to be 3/2 comma zero. So now that we have the period, we have the phase shift and we have the different transformations. Let's go ahead and plot our given function. So the function of the cotangent graph is still going to start at zero. That is going to be the first Asymptote. But the first, the second ascent tote is not going to be pi since the period is now three, that means the second aceto lies at the location of three. Now we indicated earlier that the X intercept is going to be in the middle at 3/2 comma zero. And because we are reflecting the cotangent graph about the X axis, we're going to be flipping the sign or flipping the direction of the coon graph. So the new shape of this cotangent graph is roughly gonna look like this. And I'll keep in mind that we indicated that there is a shrink of the graph. We wanna go ahead and find two points that represent the shrinking of the graph. Now, we know from earlier that the Y values are going to be positive 1/4, a negative 1/4. But we need to go ahead and get the X values for this point. So for the first point, how do we get the X value? Well, this point is going to lie between the first Asymptote and the X intercept. So what we can go ahead and do is we can take the distance from the first Asymptote to the X intercept, which can be represented as zero plus 3/2. Then we can divide that distance in half. So zero plus 3/2 is going to give us 3/ and 3/2, divided by two is going to give us 3/4. So the X value for the first point is going to be 3/4. And now we want to get the X value for the second point while the second point lies between the X intercept and the second Asymptote. So we can take that distance which can be represented as three, 3/2 plus three and divide that distance by two, 3/2 plus three is going to give us 9/ and 9/2, divided by two is going to give us the value of 9/4. So the X value for the second point is going to be 9/4. So this is the fully labeled version of the first period of the given function. But recall that the instructions ask us to graph two periods of our function. So we need to find the location of a third Asymptote. Well, we know that one period has the distance of three on the graph. So in order to get the location of the third Asymptote, we just need to add three to the second Asymptote, which is three, three plus three is going to give us six. So that means that the second, as I mean, the third ascent tote is going to be located at six and the second period is just going to mirror the image of the first period. So this is going to be the graph of the second period. And with that being said, this is going to be the entirety of the graph of 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 18–24, graph two full periods of the given tangent or cotangent function. y = − 1/2 cot π/2 x