Find all points where the tangent line is horizontal or vertical

Question

Find all the points where the tangent line to (y^3)-xy=-6 either horizontal or vertical

Answer 1

Hi John,

 

Using implicit differentiation find y', so that you have a formula for slopes of tangent lines to the graph of the original function.

 

3y²y' -xy' -y = 0 (implicit differentiation)

y'(3y² - x) = y    

y' = y/(3y² - x)

 

For horizontal tangent lines we want to know when y' = 0. That will only happen when the numerator has a value of 0, which means when y=0. If you plug 0 into the original function for y, you will find that there is no corresponding x value to make the equation true. So there are no horizontal tangent lines.

 

When will there be a vertical tangent line? When the denominator of y' = 0 (perhaps).

 

That means when x = 3y².

 

Substitute 3y² for x in the original function to determine if in fact there is an x value that makes that true:

 

y³ - (3y²)y = -6

-2y³ = -6

y³ = 3

 

y = (3)^1/3 (or cube root of 3)

 

When y = 3^1/3, solve for x.

 

(3^1/3)³ - x(3^1/3) = -6

 

3 - x(3^1/3) = -6

 

x = 9/(3^1/3)

 

So, the point on the graph of the original function where there is a vertical tangent line is:

 

(9/3^1/3, 3^1/3)

 

This graph confirms the above:

 

https://www.desmos.com/calculator/c9dqzv67cx