let x^2 + 4xy + y^2 + 3 = 0
are there any points on the curve where the tangent is horizontal or vertical? justify your answer.
2 x dx + 4 x dy + 4 y dx + 2 y dy = 0
dx (2x+4y) + dy(2y+4x) = 0
dy/dx = -(2x+4y)/(4x+2y)
if the top is 0, tangent horizontal
if the bottom is zero, vertical
for example for yertical
4x=-2y
y = -2x
plug that back in the original equation
x^2 +4x(-2x) + (-2x)^2 + 3 = 0
x^2 -8x^2 +4x^2 = -3
-3 x^2 = -3
x = +/- 1
Be sure to check my arithmetic and do the other half of the problem for numerator = 0