 So let's continue to try and graph our implicit function based on calculus. Well, let's use a few other things that are at our disposal. If the only tool you use is a hammer, you treat every problem like a nail, whether it's cutting down a tree or opening a jar of pickles. And using a hammer to cut down a tree or open a jar of pickles is either difficult or messy, and so we might want to use some other tools. And one of the important tools is symmetry, and so to sketch this graph we'll note several things about the function. First, suppose I have a point on the curve. Then, because it's a point on the curve, I know that the x and y values will make the equation defining the curve a true statement. But if this statement is true, notice that this statement is also true, and so that means that this point x0 negative y0 is also on the curve. And if we put that together, it tells us the following. If x0 y0 is a point on the curve, so is x0 negative y0, which means that the graph is symmetric about the x-axis. Now remember, we've already found the critical points for this function, and these will give us a starting point for the graph. First, we know that at the critical points 1, 0, negative 2.302, 8, 0, and 1.302, 8, 0, the tangent line is vertical. So let's graph those points and a short section of the tangent line, and we'll end up with a graph that may look something like this. We also found some other critical points where the derivative was 0, so again, plotting those points and a short section of the nearby tangent line, which gives us a slightly better idea of what the graph looks like. But remember, we also found that the graph is rising to this point and falling after it, so we can include that in our sketch. And so our graph might look something like this, and remember we did a similar analysis for this point, where we found the graph was falling to that point and rising after it. And so we can include that information in our graph as well. There's one other feature that came in our analysis of the critical points. Remember that when we tried to find the point where x equals square root of 4 thirds, we found that the equation had no solutions, and that means there is no point on the graph where x equals square root of 4 thirds. What this means is that a curve must break into two parts, one to the left of square root of 4 thirds and one to the right of square root of 4 thirds. So what we'll do is we'll put in a barrier line there, a do not cross this line indicator. And it's important to understand that this vertical line here doesn't really mean anything other than it's a barrier we should not cross. It doesn't imply that this is a vertical asymptote. It just indicates that there's a break in the graph. There's a discontinuity here. Because the graph is symmetric about the x-axis, we can focus on what happens above the x-axis. And because we have this do not cross line, we can focus on what happens to the left of that do not cross line and to the right of that do not cross line. On the left of the do not cross line, we can just connect the dots, keeping in mind that at the beginning and the end, our tangent line should be vertical. And one possibility is to draw something like this. To the right of the do not cross line, it's a little bit more difficult to see what's going on. But one thing we do know is that when the graph hits this point, the tangent line is vertical. And so we might continue it this way. And finally, remember that the graph is symmetric about the x-axis. And so once we know what the top part looks like, we know the bottom part looks the same. So reflecting the top portion across the x-axis is a graph that looks like this.