In Exercises 29–44, graph two periods of the given cosecant or se...

Question

In Exercises 29–44, graph two periods of the given cosecant or secant function.y = −2 csc πx

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Welcome back. I am so glad you're here. We are asked to sketch the graph of the following co can function using two periods. Our function is Y equals negative three co can of two pi X. And then we have a graph, we have a vertical Y axis, a horizontal X axis. They come together the origin in the middle. The domain of our given graph is from negative 1 to 3 and the range is from negative 10 to 10. All right. So what do we do with the coy can function? Well, what we want to do is start by graphing the reciprocal function. The reciprocal function of Cosic can is sign. So we can just set this equal to Y equals negative three sign of two pi X. And we'll work for a while on graphing the sine function, we recognize that this sine function is in the format of Y equals a sign of B X A is what's being multiplied by the sign. So our A is negative three and B is what's being multiplied by the X and so B is two pi and recalling from previous lessons on graphing the sign function we're going to start off by finding our amplitude in our period. We know that when it's in this format that it starts at the origin. So we know that for our amplitude, that's going to be the absolute value of A and A here is negative three. So the absolute value of negative three which is three, our amplitude is three. Now, for the period, we know that that is two pi divided by B and so that will be two pi divided by B is two pi so two pi divided by two pi which is one. So our period here is one great. We've got that information. Now, step two in graphing the sign function is that we are going to identify our five key X values which are going to be our X intercepts the maximum and the minimum. So as we said before this starts at the origin, so the X sub one, the first X value for our key point is at zero. And now what we do is we add to each of these X values, we're going to add the period divided by four to get each subsequent X value. And so the period here, as we said is 11 divided by four, we're going to add 1/4 for each one. So X up two, that is zero plus 1/4. Well, that is 1/4 for our X sub three, we'll take 1/4 plus 1/4. That's 2/4 which is one half for our X sub four, we'll take one half plus 1/4 that's 3/4. And for our X sub five, that fifth key X value 3/4 plus 1/4 is one. And that makes sense. We've gone one period and we're going from 0 to 1. Now, we do have to graph two periods so we can keep going a little bit more and we'll go one more period. So X sub six, adding one plus 1/4 is 5/ for X sub seven, adding 5/4 plus 1/4 will simplify to be three halves, four, X sub eight, three halves plus 1/4 is 7/ N for X sub nine 7/4 plus 1/4 is a positive two. So there's are two periods worth of X values. Now, we need to first step three discern the corresponding Y values. And to do this, there are a couple ways to do it. We can take our X values and plug those in to our sine function. So the first one, we would have Y equals, then you plug into your calculator negative three multiplied by the sign of two pi multiplied by zero. So all in the parentheses, two pi multiplied by zero and you would get a Y sub one value of zero. But we also know this because our starting point is at the origin for this sine function which will give us X and Y values of 00. Now, we can go through this and plugging in all the X values into the function to get the corresponding Y values. But we can discern a lot of this without that information. So we know that our 1st 3rd and 5th key X values are all X intercepts. So that means that our Y sub one, Y sub three and Y sub five are all going to be zero because those are X intercepts now with the Y sub two and the Y sub 41 of those is at a maximum and one of them is at a minimum. And you can plug the values into the calculator to figure out which one is the maximum and the minimum for the X values. But you can also just look at our function in front of that sign our a value this negative three that's being multiplied. Well, that's a negative. And that tells us that our function is reflected over the X axis. So typically, if it was being multiplied by a positive number there, the first, our Y sub two, the first point that would be a maximum. But because it's negative, that's going to be our minimum and our minimum is the amplitude distance from our X axis. So for the minimum, that's going to be a negative three. So our Y sub two value is negative three and Y sub four, that's going to be the amplitude above the X axis. So that's going to be a positive three. Now, for our Y sub six, Y sub seven, Y sub eight and Y sub nine is just going to follow the pattern. So Y sub six goes back to what Y sub two is, which is a negative three, Y sub seven is back to zero, Y sub eight is up to three and Y sub nine is back down to zero. Now, this is our sign function. We can plot these points and graph it. But don't forget we're aiming for the coy camp function. So we're not done once we've finished this. All right, let's start off by plotting our points. So the first point is 00. That's right at the origin. The second point is 1/4 negative three. So from the origin, we go to the right 1/4 of a unit and we go down three units for the 0.1 half zero. From the origin, we go to the right one half of a unit for the point 3/4 3, we go from the origin to the right 3/4 of a unit and up three units for the 0.10. From the origin, we go to the right one unit. Now for the point 5/4 negative three from the origin, we go to the right 5/4 of a unit and go down three units for the 0.3 halves zero, we go from the origin to the right three halves of a unit. For the point 7/4 3 from the origin, we go to the right 7/4 and we go up three units. And for the 0.20 from the origin, we go to the right two units. OK. So that is our sign function, we can connect these as smoothly as possible, but this isn't our goal. So for the CO C can function a couple of things, every X intercept from our sign function becomes a vertical as some tote. So we can draw, we can draw some dotted lines along all the places where there's an X A X intercept. So where X equals zero, there is an X intercept and at X equals one half that becomes a vertical as tote at X equals one that becomes a vertical Asymptote at X equals three halves that becomes a vertical Asymptote. And at X equals two because we only need two periods worth here. All right. So there's all our vertical asom totes and we could label them X equals zero, X equals one half, X equals one, X equals three halves and X equals two. Every minimum of our sign function becomes a maximum. And we are going to draw our lines that will be hugging up against our vertical asymptotes. They hit our sign functions minimums and then head back down toward negative infinity for the Y values. So to the right of our vertical asom tote X equals zero, we have a line going down to negative infinity for the Y values it hugs up along our vertical Asymptote X equals zero. It touches our minimum from the sine function at 1/4 negative three and then turns around and then heads back to a negative infinity for the Y values along the vertical Asymptote X equals one half. So we're going to continue this now, for the next one, we know that every maximum from our sine function becomes a minimum. That's not how you spell minimum, minimum of our co C can't function. So to the right of the vertical Asymptote, X equals one half here, our line will be heading up toward positive infinity for the Y values hugging along our vertical Asymptote X equals one half. It's going to touch our sine functions maximum at 343, which will become the minimum for this. Cos you can't function, it turns around and heads back up to positive infinity for the Y values hugging along the vertical Asymptote of X equals one. Now, we've got two more lines to draw for our two periods. So to the right of our vertical Asymptote, X equals one, we are back down in the negatives for our Y values. So heading toward negative infinity for the Y values hugging along our vertical Asymptote, X equals one to the right of it. Then we are going to encounter our sine functions minimum at 5/4 negative three. It touches, that turns around hugs up along our vertical Asymptote, X equals three halves and heads toward negative infinity for the Y values last line to draw to the right of our vertical Asymptote X equals three halves. We're now in the positive part of the Y axis heading up toward positive infinity, to the right of that vertical asom tote, it's going to come down along that line and it's going to touch our sign functions maximum at 7/4 3, which becomes the minimum for this CO C can function then turns around heads up along the vertical asom tode X equals two toward positive infinity for the Y values. And that's it. That's our CO C can't function well done. We'll catch you on the next one.

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In Exercises 29–44, graph two periods of the given cosecant or secant function.y = −2 csc πx

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