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# Horizontal tangent to implicit curve

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.D (LO)
,
FUN‑4.D.1 (EK)
,
FUN‑4.E (LO)
,
FUN‑4.E.1 (EK)
,
FUN‑4.E.2 (EK)

## Video transcript

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 ax 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 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 equals 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 eight 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 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 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
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