Hey, everyone. So let's see how we can go about solving this example. So here we are asked to graph the function \( y = -2 \cdot \cot\left(\frac{1}{4}x\right) \).
Now what I noticed in this problem is we have quite a few things going on. We have a negative sign here, which is going to cause our graph to be flipped or reflected over the x axis. I also notice we have this 2 here, which is going to change the amplitude or tallness of our graph. And we have this 1/4 on the inside of our cotangent, which is going to change the period. So where do we even get started with this?
Well, what I like to do is start with what I'm familiar with. And what I'm going to do is I'm going to ignore this amplitude and negative sign out front. I'm just going to focus on \( y = \cot\left(\frac{1}{4}x\right) \). And the reason that I'm focusing on this is because I know how to deal with the period of a cotangent, and I know what the general cotangent graph looks like, so I can build off of something.
Now the period for a cotangent graph is going to be \( \frac{\pi}{b} \), where \( b \) is the number that is in front of the \( x \). Since I can see that that's 1/4, I get that we have \( \frac{\pi}{\frac{1}{4}} \). You can flip this fraction and bring it to the top, which is going to give you \( \frac{4\pi}{1} \), which is just \( 4\pi \). So the period is \( 4\pi \).
Something that I know about the cotangent graph is the cotangent graph has an asymptote that starts at an \( x \) value of 0. So since we start at an \( x \) value of 0 and our period is \( 4\pi \) when we have this \( b \) value in front of \( x \), that means that starting at 0, we're going to repeat every \( 4\pi \) units.
So I can draw another asymptote here at \( 4\pi \), and I can draw another asymptote over there at \( 8\pi \). Now something I know about the cotangent graph is that it goes up into the left, and then goes down into the right, and it reaches 0 on the \( x \) axis right in between the two asymptotes in the middle.
Now what you should typically see with a cotangent graph is you're going to see points that are in between these two values, and in between those two values. So this over here should be \( \pi \), and this should be \( 3\pi \), and that makes sense do we have \( \pi, 2\pi, 3\pi, 4\pi \), and at these values you should see an output of 1. So what we should have is an output right about there, which would be at 1, and we should see an output right here, which would typically be at negative one.
So adjusting our curve a little bit, this is what the function would look like for the cotangent of \( \frac{1}{4}x \), and then you would get this curve continuously repeating as we go along.
But this is not what we're trying to find. We're not trying to find the cotangent of \( \frac{1}{4}x \). We're trying to find \( -2 \times \cot\left(\frac{1}{4}x\right) \).
Now this number out in front here, the 2, this is going to change the amplitude or basically vertically stretch our graph. So rather than having an output of 1 when you're at \( \pi \), you should actually have an output of 2 at this value. And then rather than having an output at negative one for \( 3\pi \), you should actually have an output at negative 2.
So adjusting our curve, the graph should look a bit more stretched vertically when we incorporate this 2. But there's one more thing we need to change about our graph, and that's the negative sign. This negative sign is going to take our whole graph, and it's going to flip it over the x-axis. So doing this, we actually need to, in essence, reverse these two points on the y axis. So we should have a point right here at \( \pi \) negative 2, and a point right here at \( 3\pi \) positive 2.
So for the graph \( y = -2 \times \cot\left(\frac{1}{4}x\right) \), it should look something like this where we start at the lower left, and then we go ahead and go up reach a 0 on the y axis at \( 2\pi \), and then we come up here and go to the top right. So this is what our curve should look like, and because the cotangent is a repeating graph as we've learned in previous videos, this curve should continue happening between each of these asymptotes. So this would be the graph of our function and the solution to the problem. So hope you found this video helpful, thanks for watching.
