In this problem, we're asked to graph \( r = 2 + 2\sin(\theta) \). Now I know that this is the graph of a cardioid because I have addition happening and both my \( a \) and \( b \) values are equal to each other. So, I already know the general shape of this graph, and it remains to just determine the orientation and the exact positioning of this on my graph. So let's go ahead and get started with step 1, first looking at symmetry. Now, since our equation contains a sine function, that tells me that my graph is going to be symmetric about the line \( \theta = \frac{\pi}{2} \), which we can keep in mind while graphing our points here.

Now we want to go ahead and find and plot some points, and we're specifically going to use our quadrantal angles starting, of course, with 0. So plugging in 0 for \( \theta \), I get \( 2+2\sin(0) \), which the sine of 0 is just 0, so this gives me 2. So I can plot that first point at \( (2,0) \), which will end up being right here. Now \( \frac{\pi}{2} \), plugging that in for \( \theta \), I have \( 2 + 2 \times \sin(\frac{\pi}{2}) \). Now the sine of \( \frac{\pi}{2} \) is 1, so this ends up being 2 + 2 or 4. So I can plot my second point at \( (4,\frac{\pi}{2}) \). Now that will end up being right here at the top of that axis.

Now I want to take a moment to remember my symmetry here because since I'm symmetric about this line \( \theta = \frac{\pi}{2} \), I can go ahead and just reflect this point over onto my other quadrantal angle. So I have this other point at \( 2\pi \). Then I have just one final point to find at \( 3\frac{\pi}{2} \), so \( 2+2\sin(3\frac{\pi}{2}) \). The sine of \( 3\frac{\pi}{2} \) is negative 1. So this ends up being \( 2 - 2 \), which is just 0. So here, my final point is actually going to be located at my pole.

Now from here, we want to connect this with a smooth continuous curve. Now we remember the shape of a cardioid, but remember that the orientation of that can change. Your furthest point from your pole will always be the sort of bottom of your cardioid that's very round. Then right here, we're going to have the sort of dimple heart shape that happens with our cardioid. And you want to make sure and round this around. Now it's okay if you don't get super precise. And you can always choose to plot some more points here if you want to get a bit more exact or if your professor asks you to get more exact. But this is the general shape and positioning of my cardioid for \( r = 2+2\sin(\theta) \). Thanks for watching, and let me know if you have questions.