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# Plotting maxima, minima and midline intersections of trig function

## Video transcript

Plot points corresponding to the maxima, minima, and intersections with the midline in the graph of y is equal to 2 sine of 1/4 x minus 1. And they give us this neat tool here where we can sit here and pick as many points just by clicking. We can move the points around, drag them around. If we want to get rid of a point, we just have to drag them to the trash. Now to work through the math, I've copied and pasted this problem right here on my scratch pad, and I rewrote the function. And I want to think about, when does this hit its maximum points, its minimum points, and its intersections with the midline? Well this is going to hit a maximum point whenever sine of 1/4 x is maximized. And when is sine of 1/4 x maximized? Well the maximum value that the sine function can take on is 1. So we're dealing with maximum points whenever sine of 1/4 x is equal to 1. Now what about minimum points? Well we're dealing with minimum points whenever this thing-- the minimum value that the sine function can take on is negative 1. So we're hitting minimum points whenever-- let me write it here-- so minimum points whenever sine 1/4 x hits the minimum value that the sine function can take on, hits the point negative 1. And of course, when it hits this point, we're then going to multiply it times this 2, so the maximum value of this part of expression is going to be 2 times 1. The maximum value of the whole function is going to be 2 times 1 minus 1. The minimum value is going to be 2 times negative 1 minus 1. But we'll get to that. Now what about when we intersect the midline? We intersect the midline whenever this part of the function is equal to 0. And we're essentially just left with that part right over there. So let me write this-- midline. And this is going to be 0 when sine of 1/4 x is equal to 0. So when do these things happen? When does the sine function hit 1? When does the sine function hit 0? When does the sine function hit negative 1? And to think about that, I will draw ourselves a little unit circle to help us visualize things. So a little unit circle, I don't want to make it too little so that we can-- so a little unit circle right over there, that's our y-axis. That is our x-axis. And it's a unit circle. It's centered at the origin, has a radius 1-- so just like this. And this, of course, is the point negative 1. Sorry, this is the point 0, negative 1. This is the point 1 comma 0. This is the point 0 comma 1. This is the point negative 1 comma 0. Now let's start with an angle of 0 radians. So we're going to intersect the unit circle right at that point, at the point 1 comma 0. And remember, the sine evaluated there by the unit circle definition of trig functions is going to be the y value. So we see sine of 0 radians is equal to 0-- and not just 0 radians, any integer multiple of 2 pi plus 0 radians. So we could add 2 pi. We would go around again. We would get to the same point. Add 2 pi again-- well now we've added 4 pi. We go again. We could subtract 2 pi, and we're back there again. So this is going to happen whenever-- whatever we input into the sine function is equal to 0 radians plus some multiple of 2 pi radians. So when is this going to be equal to 0? Well, when 1/4 x meets that constraint. So that's going to happen whenever 1/4 x is equal to 0 plus or minus-- I guess I could say-- some multiple of 2 pi. Note and remember, all I'm saying here is, if you input anything like this into the sine function, you're going to be at this point on the unit circle. And when you evaluate the sine of the function at any of those points, you are going to get 0. Why did I write k here? Well this is just for some integer k. So this is just saying 0 plus 2 pi, 0 plus 4 pi, 0 plus 6 pi, 0 minus 2 pi. This encapsulates all of that. So let's solve for x. Well to solve for x, we just multiply both sides by 4. We get x is equal to 4 times 0 is 0, plus or minus 4 times some multiple of 2 pi would get us-- or 4 times 2 pi k is going to give us 8 pi k. So when x is 0 plus or minus any integer multiple of 8 pi, this whole function is going to intersect the midline. Why? Because when x is any of these things, when you input any of these x values there, you multiply it by 1/4, you evaluate sine there, then this whole expression is going to be 0. We're going to hit the midline, y equals negative 1. So let's just go with the simplest one. Let's just go with x equals 0. When x equals 0, all of the stuff is 0. y is equal to negative 1. So we have the midline right over here. x equals 0. y is equal to negative 1. And we're also hitting the midline right over there. Now let's think about a maximum value. So if we increase our angle right over here all the way to pi/2 2 radians, we get to this point right over here. Sine hits its maximum value whenever we're taking the sine of pi/2 plus or minus any integer multiple of 2 pi. So this is going to hit a maximum whenever what we input into the function is equal to pi/2 plus or minus some integer multiple of 2 pi. Well once again, multiply both sides by 4. You get x is equal to pi/2 times 4 is 2 pi plus or minus 8 pi k. Now which x value would we want to do? Well and actually, even in the last one, I picked x equals 0 for simplicity. But if I did x equals 8 pi or 16 pi, those aren't even on this graph. Or if I want to do x is equal to negative 8 pi, those aren't even on the graph. So this is actually the only one that sits on this graph. Likewise over here, x equals 2 pi is the only one that's actually on this graph. If I added 8 pi, I would go off the graph. If I did 2 pi minus 8 pi at negative 6 pi, that's off the graph. So I'm going to do x equals 2 pi. When x equals 2 pi, 1/4 of 2 pi is pi/2. Sine of pi.2 we've already saw is 1. 2 times 1 is 2, minus 1 is 1. So when x is 2 pi, we're at the point y is equal to 1. That is a maximum point. And then finally-- same logic for the minimum point. We need to see when this 1/4 x hits not pi/2, but now negative pi/2. We could have also picked 3 pi over 2 if we wanted. So negative pi/2 is equal to negative pi/2 plus or minus any integer multiple of 2 pi. So once again, multiply both sides times 4. x is equal to negative 2 pi plus or minus 8 pi-- some integer multiple of 8 pi. So once again, the only of these that's actually on our graph is x is equal to negative 2 pi. When x is equal to negative 2 pi, what does our function equal? Well, when x is equal to negative 2 pi, 1/4 x is going to be equal to negative pi/2. Sine of negative pi/2 is negative 1. 2 times negative 1 is negative 2, plus negative 1 gets us to negative 3. So it's that point right over there. So those are our points. And if we actually wanted to graph-- they just ask us to graph these points-- but if we wanted to give a little bit more texture here, we could draw the midline, right over here, at y is equal to negative 1. And if we wanted to actually try to plot this function, or at least the part that we can graph, it would look something like this. But actually, let's just get the tool out, and it would keep going if we had more real estate. We're not even plotting a complete period here, but let's go back to our tool and plot these points and make sure that we got it right. So we want a point-- that's where we intersect at the midline. This is our minimum point. That's our maximum point based on what I just did. Check the answer. We got it right.