how do you find which points a tangent line is horizontal/vertical for decartes folium x^3+y^3=3xy?
differentiate implicitly
3x^2 + 3y^2 dy/dx = 3xdy/dx + 3y
dy/dx(3y^2 - 3x) = 3y - 3x^2
dy/dx = 3(y - x^2)/(3(y^2 - x)
= (y-x^2)/(y^2-x) , which is the slope of the tangent
so to be horizontal, dy/dx = 0
y-x^2 = 0 -----> y = x^2
sub back in original
x^3 + x^6 = 3x^3
x^6 - 2x^3 = 0
x^3(x^3 - 2) = 0
x = 0 or x = 2^(1/3) , the cuberoot of 2
then y = 0 or y = 2^(2/3)
horizontal tangents at (0,0) and (2(1/3) , 2^(2/3))
for vertical tangent, the denominator has to be zero
y^2 - x = 0
x = y^2
sub that back in
y^6 + y^3 = 3y^3
because of the symmetry of the equation from the one above, the solutions would look the same, except everything will in in y
so vertical tangents, at (0,0) and (2^(2/3) , 2^(1/3) )