 So let's see if we can use calculus to help us graph implicit functions. So let's try and find all critical points for the function defined implicitly by y squared equals x cubed minus 4x plus 3. So first we'll find the derivative. Now remember the critical points are going to be where the derivative is zero or undefined. And since our derivative is a quotient, it's possible for it to be undefined. And so one place where critical points occur is going to be wherever y equals zero. To find this point we need to solve our equation for x. Now this is a cubic equation and most of us were never taught how to solve cubic equations, but that's okay. We have plenty of options. We can use trial and error or a computer algebra system to find a solution x equal to 1. And just as a quick note, we see that as x approaches 1 and y approaches zero, our derivative is going to approach plus or minus infinity. Depending on whether y is approaching zero from above or approaching zero from below, and that suggests that our tangent line is vertical at our point x equal to 1, y equal to zero. Are there any other such critical points? By the root theorem, we know that if x equals 1 is a solution to x cubed minus 4x plus 3 equals zero, then x minus 1 is a factor. And so we can remove that factor and additional solutions to y equals zero will be found by solving the quadratic x squared plus x minus 3 equals zero. And so we'll use the quadratic formula, which gives us two additional solutions. Alternatively, if we'd used a computer algebra system, we would have found all three solutions to begin with. By a similar analysis, we see that as x approaches either of these values and y approaches zero, our derivative is going to approach plus or minus infinity. And again, what that tells us is that at these points, our tangent line is going to be vertical. Where else do we get critical points? We'll get additional critical points where the derivative is zero. And because this is a quotient, the only way a quotient can be zero is if the numerator is zero. So we'll set numerator equal to zero and solve. And this gives us x equals plus or minus square root of 4 over 3. Again, we have our x value. Now we want to look for our y value. If x equals square root of 4 over 3, then our equation tells us that y squared equals x cubed minus 4x plus 3. But then an interesting thing happens. When we evaluate this right hand side, we get negative 0.0792. And it's impossible for the square of a real number to be negative. So there is no solution. And what this means also is that there is no point on the graph where x equals square root of 4 thirds. The graph somehow skips over this x value. What about x equals negative square root of 4 thirds? Here, if we substitute our x value into our equation, we find our y value. And it's important to notice that there are two possible y values for this one x value. And so we get two additional critical points. And we'll make one quick note about mathematical style. It's not considered good form to mix dialects. And what that means is that we have our y value in decimal form. We should give our x value in decimal form as well. And so we'll convert this x equals negative square root of 4 thirds into its decimal equivalent. And our critical points are going to be given in decimal form. Now, in order to determine whether these are local maximum or local minimum values, we might find the second derivative. But that's going to be a nightmare because this is an implicit function, which is a quotient. Instead, we'll do a first derivative test. Since our derivative is in terms of both x and y, then we have to evaluate our critical points based on both their x and y values. So let's consider this first critical point. So remember our x coordinate here was the decimal equivalent of negative square root of 4 thirds. So now let's consider if we're to the left of this point, our x values will be less than negative square root of 4 thirds, and our y values will be positive. And so this means our derivative will be positive. And that's because in our derivative the denominator is going to be a positive number and the numerator is also going to be a positive number. And that means the graph is rising up to meet that point. Now, once we go to the right of the point, the x values will be greater than negative square root of 4 thirds and the y values will still be positive, so our derivative is going to be negative. And again, that's because our denominator is going to be positive and our numerator is going to be negative. And so the graph is falling from this point. And that means that this point will be a local maximum. Similarly, if we're to the left of this other point, the x values will be less than negative square root of 4 thirds and the y values will be negative, so our derivative is going to be negative. And again, that's because our denominator is a negative number and our numerator is a positive number. And if we're to the right of this point, the x values will be more than negative square root of 4 thirds and the y values will still be negative, so our derivative will be positive. So the graph is falling to the point and rising after it, so this point is a local minimum.