# Finding where tangent line cuts through the graph

$$\ L$$ is the tangent line at $$\ x = x_0$$ to the graph of $$\ y= ax^3 +bx$$. I need to find the $$\ x$$ value of the second point where $$\ L$$ cuts through the graph.

I tried to define $$\ (x_0, ax_0^3+bx_0)$$ and then just find the equation of the tangent line at $$\ x_0$$ which is $$\ y = 3ax_0^2 -2ax_0^3 + bx$$ and then just compare it to the graph so I get $$\ 3ax_0^2-2ax_0^3 +bx = ax^3 +bx$$ but all I get is $$\ x^3 = 3x_0^2x-2x_0^3$$ and I can not get much out of it.

$$\begin{eqnarray} x^3-3x_0^2x+2x_0^3 &=&x^3-x_0^2x-2x_0^2x+2x_0^3\\ &=& x(x^2-x_0^2)-2x_0^2(x-x_0)\\ & =& (x-x_0)\Big(x(x+x_0)-2x_0^2\Big)\\ &=&(x-x_0)\Big(x^2+xx_0-2x_0^2\Big)\\ &=&(x-x_0)(x+2x_0)(x-x_0)\\ &=&(x-x_0)^2(x+2x_0) \end{eqnarray}$$ So it intersect the graph at $$(-2x_0,...)$$.
You have to solve the equation $$4x^{3}-3x_0^{2}x-2x_0^{3}=0$$ and you already know that $$x=x_0$$ Is one solution. The equation can be written as $$(x-x_0)^{2}(x+2x_0)=0$$. Hence $$x=-2x_0$$is the answer.