Given x^2+4xy+y^2=-12, find the equations of all horizontal tangent lines. I already found that y'= -x/2+y

Question

across · Answer

Implicit differentiation yields$$x^2+4xy+y^2=-12,$$$$2x+4y+4x\frac{dy}{dx}+2y\frac{dy}{dx}=0,$$$$2x+4y+\frac{dy}{dx}\left(4x+2y\right)=0,$$$$\frac{dy}{dx}\left(4x+2y\right)=-2x-4y,$$$$\frac{dy}{dx}=-\frac{x+2y}{2x+y},$$$$-\frac{x}{2}-y=0.$$

anonymous · Answer

oh wow i forgot to use chain rule thanks

anonymous · Answer

wait so how do i find the horizontal tangent lines?

mr.math · Answer

The tangent line is horizontal when of a function $y=f(x)$ is horizontal if $\frac{dy}{dx}=0$, and that's what across did when she set up $\frac{dy}{dx}=-\frac{x+2y}{y+2x}=0$. This expression is zero if, and only if the numerator is zero, thus the line $x+2y=0$ represents all horizontal tangent lines to $x^2+4xy+y^2=-12$.

mr.math · Answer

Sorry, The tangent line of a function $y=f(x)$ is horizontal if ...