Hey, everyone. In this problem, we're asked to graph the equation r 2 = 25 ∙ cos 2 · θ . Now here, this is the equation for a lemniscate because I have this r 2 , which I know is only something that happens when dealing with equations of lemniscates. Now, lemniscates are arguably our easiest shape that we know how to graph, so let's dive right in here. Now looking at our first step, we want to plot our first petal here, which is going to be located at 25 for our r value, and then we need to determine what θ is based on the trigonometric function that we're dealing with. Now, remember that in our equation, this value at the beginning is actually a squared. So, we need to make sure that we take the square root of that to get a. Now the square root of 25 is equal to 5, so this gives me my value of a that I will be plotting for that first petal. Now, we need to determine what θ is. Here, I have 25 times the cosine of 2θ. So, I'm dealing with a positive r² and a cosine, which tells me that θ will be located at 0. So, I can plot my very first petal at 5, which will end up being right here on my polar axis. Now, all that's left to do is reflect this petal over the pole. So, doing that here, reflecting this over to the other side, landing me right here. I have my second petal, and I can move on to my final step and just connect these with a smooth and continuous curve. Now we know that the general shape of our lemniscate is a sort of infinity symbol or a propeller or whatever you want to think of this as, but remember that we're always going to go to that hole in there with these sorts of petals or propellers or whatever you want to call these. Now, we've graphed this lemniscate at r 2 = 25 ∙ cos 2 · θ . Let me know if you have any questions.