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

- [Instructor] We're told to consider

the curve given by the equation.

They give 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.

And 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.

And 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.

And 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.

So it's going to be negative three squared

plus y to the fourth

plus six times negative
three is equal to seven.

This is nine.

This is negative 18.

And so we're going to get y to the fourth

minus nine is equal to seven,

or, adding nine 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 two.

Well, there would be then
two horizontal lines.

One would be y is equal to two.

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 two.