Horizontal tangent to implicit curve | AP Calculus AB | Khan Academy

We're told to consider the curve given by the equation that gives this equation. It can be shown that the derivative of y with respect to x is equal to this expression, and you could figure that out with just some implicit differentiation and then solving for the derivative of y with respect to x. We've done that in other videos.

Write the equation of the horizontal line that is tangent to the curve and is above the x-axis. Pause this video and see if you can have a go at it.

So let's just make sure we're visualizing this right. So let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there, and then another one might be maybe right over here.

They want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know? What is true if this tangent line is horizontal? Well, that tells us that at this point, dy/dx is equal to zero. In fact, that would be true at both of these points. We know what dy/dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y.

So when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero.

So what is going to be the corresponding y value when x is equal to negative three? And if we know that, well, this equation is just going to be y is equal to something—it's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y.

So let's do that. It's going to be negative three squared plus y to the fourth plus 6 times negative three is equal to 7. This is 9. This is negative 18. And so we're going to get y to the fourth minus 9 is equal to 7, or adding 9 to both sides, we get y to the fourth power is equal to 16.

And this would tell us that y is going to be equal to plus or minus 2. Well, there would be then two horizontal lines: one would be y is equal to 2, the other is y is equal to negative two. But they want us the equation of the horizontal line that is tangent to the curve and is above the x-axis. So only this one is going to be above the x-axis and we're done. It's going to be y is equal to 2.