if you have the equation of a graph how do you find a point on curve where the tangent is horizontal/vertical?

Question

the graph is x^3+y^3 = 3xy here,,just the process in general is fine

Answer 1

derivative is 3x^2 + 3y^2(dy) = 3(dy)
3(dy) - 3y^2(dy) = 3x^2
(dy)(3-3y^2) = 3x^2
dy = (3x^2)/(3-3y^2)

Graph is horizontal (numerator of dy is zero) when x = 0. Graph is vertical (denominator dy is zero) when y = 1.

Answer 2

For horizontal tangent, look for y' = 0 or dy/dx=0.
For vertical tangent, look for x' = 0 or dx/dy =0.

Differentiating implicitly, x^3+y^3 = 3xy becomes
3x^2 + 3y^2*dy/dx = 3x*dy/dx + 3y.
Simplify and solve for dy/dx:
(3y-3x^2)/(3y^2-3x) = dy/dx
dy/dx = (y-x^2)/(y^2-x)
Set (y-x^2)/(y^2-x) = 0 for horizontal tangents. (y=x^2)
Substitute into your original equation to get
x^3+x^6 = 3x^3 => x^6= 2x^3 => x^6-2x^3=0 =>x^3(x^3-2)=0 => x=0 and you get (0,0) or x = cube root of 2 and you get
(2^(1/3), 2^(2/3)).

Set (y^2-x)/(y-x^2) = 0 for vertical tangents> (x=y^2).

Answer 3

differentiate the equation and solve for 0. this will give you the point where the slope is 0, ie horizontal

Answer 4

Differentiate implicitly

giving

3x^2 + 3y^2*dy/dx = 3y+3x*dy/dx

Then group together dy/dx

dy/dx(3y^2-3x)=3y-3x

dy/dx=(3y-3x)/(3y^2-3x)

Then solve for dy/dx=0

Hoped it helped

Answer 5

dy/dx=0 horizontal
dy/dx=infinity vertical.