How to find unkown tangent point

Question

As part of my homework i've the following question:

The tangent line $ L $ is crossing the graph of $ y = ax^3 + bx $ at point $ x = x_0 $, find another point where the tangent-line $L$ is crossing the graph. Define $ a = 1$ and $b = 0$.

Second part of the question is to graph $y = x^3$ and show the corssing points.

I was able to find find the equation of $L$ using the derivative of the function and the point: $$ y - x_0^3 = 3x_0^2(x - x_0) \\ y = 3x_0^2x - 2x_0^3 $$

I'm unable to find another point the tangent $L$ cross the graph (which is assume is $y = x^3$). Should i resolve the value of $x_0$?

Answer 1

Why not solve the set of equation? $$\begin{cases} y = ax^3 + bx \\ y - x_0^3 = (3ax_0^2+b)(x - x_0) \end{cases}$$ The solution of the set of equations is the intersection points of the tangent line and the curve.

Answer 2

Just solve it in general: $$y=ax^3+bx \text{ and } y-y_0=m(x-x_0)$$ where $y_0=ax_0^3+bx_0 $ and $m=3ax_0^2+b$.

As you know already one solution of this cubic equation, namely $x=x_0$, perform a long division to get a very nice result: the other solution (besides $x=x_0$ of course again) is $$x=-2x_0,$$ independent from $a$ and $b$.

Well, I didn't knew that before, thanks for that interesting question.