I have an assignment for tomorrow - and this topic has been puzzling me for a while. I hope someone out there can help me :)

Consider the function $f(x)=x^3$ and the point $P=(a,a^3)$, where $a≠0$, which is a point on the graph for the function. The tangent to the graph, through the point P, intersects the graph one other place than in the point P. We will call the other intersection $Q$. Now, my task is to determine the coordinates for the point Q - and show that the slope of the tangent line for the function through Q is 4x as steep as through P.

Currently, I know that the equation of the tangent through P must be: $$y'=f'(x_{0})*(x-x_{0})+f(x_{0})$$ $$3a^2x-3a^3+a^3$$

$$3a^2x-2a^3$$ Do I set the above $=x^3$ and then calculate? I tried doing that but ended up with something I couldn't calculate.

Can anybody help me please. Thank you all very much.

EDIT 1: Added a missed calculation.

  • $\begingroup$ @Peter I'll just clarify. I just skipped a calculation :) $\endgroup$ – MathyFoxy Oct 18 '15 at 22:03
  • $\begingroup$ You may find this question of mine helpful. $\endgroup$ – Chappers Oct 18 '15 at 22:04
  • $\begingroup$ Now you set $x^3=3a^2x-2a^3$. You know that $x=a$ is a solution, do polynomial division : $(x^3-3a^2x+2a^3):(x-a)=x^2+ax-2a^2=(x-a)(x+2a)$ $\endgroup$ – Peter Oct 18 '15 at 22:07
  • $\begingroup$ So, $Q(-2a/-8a^3)$ and the tangent has slope $3\times (-2a)^2=12a^2=4\times 3a^2$. $\endgroup$ – Peter Oct 18 '15 at 22:13
  • $\begingroup$ @MathyFoxy, if my answer was useful, consider accepting it by clicking the gray tick on the left of it. $\endgroup$ – dbanet Oct 19 '15 at 18:10

As you have already observed, $P\in\{(x,y)\in\mathbb{R}^2\mid y=f(x)\}$, therefore the slope of a tangent through $P$ is simply $f'(a)$ where $f'(x)=\frac{\mathrm df(x)}{\mathrm dx}=3x^2$.

Now, engaging your equation for tangent at $a$, let us find the intersection point $Q$: \begin{cases} y=f(x)=x^3,\\ y=f'(a)(x-a)+f(a)=3a^2(x-a)+a^3=3xa^2-3a^3+a^3=3xa^2-2a^3; \end{cases} $$x^3=3xa^2-2a^3;$$ $$0=x^3-3xa^2+2a^3;$$ this is a depressed cubic, so substituting (see Vieta's Substitution) $x=w+\dfrac{a^2}{w}$: $$0=\left(w+\frac{a^2}{w}\right)^3-3\left(w+\frac{a^2}{w}\right)a^2+2a^3;$$ expanding yields: $$0=\frac{a^6}{w^3}+2a^3+w^3;$$ substitute $u=w^3$ and multiply through by $u$ to have a quadratic: $$0=u^2+2ua^3+a^6;$$ which is a full square: $$0=(u+a^3)^2$$ therefore the solution for $u$ is $-a^3$, hence we have: $$w^3=-a^3;$$ $$w=-a;$$ returning back to the terms of $x$, we have: $$x=w+\frac{a^2}{w}=-a-\frac{a^2}{a}=-2a;$$ that is, the tangent to $y=f(x)$ at $a$ also intersects the curve at the point $-2a$, which has slope $f'(-2a)=3(-2a)^2=12a^2$, which is exactly four times more than $f'(a)=3a^2$, which is what had to be demonstrated.

  • $\begingroup$ Great! Makes sense :) Thank you so much! $\endgroup$ – MathyFoxy Oct 20 '15 at 21:27

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