The equation of a curve is $y=x^3-8$. Find the equation of the normal to the curve at the point where the curve crosses the x-axis. 
The equation of a curve is $y=x^3-8$. Find the equation of the normal to the curve at the point where the curve crosses the x-axis.

So first I factorised the function:
$$x^3-8= (x-2)(x^2+2x+4)$$
Therefore it crosses the x-axis at $(2,0)$.
Secondly, I found the derivative of the function:
$$y' = 3x^2$$
Substituting $f '(2)=12$
Tangent's equation:
$$y=12x-24$$
So the normal equation would be:
$$y=-\frac 1{12}x+\frac 16$$
Is this answer correct?
 A: Yes. Since your normal passes through the point $(2,0)$ it is correct.
A: You've done just fine! Your solutions are spot on.  Was there any point in your work that you were/are unsure about?  
If you need to hand in your work to justify your solutions, I'd make sure to mention that the point of intersection of $y= x^3-8$ with $y=0$ is the solution to $$x^3-8=(x−2)(x^2+2x+4)= 0.$$  You are correct that the point of intersection is $(2, 0)$.
From there, clear sailing!
A: Alternate solution for those knowing about the gradient of a function:
Let $f(x,y)=x^3-y$. The given curve is a level curve of $f$ ($f=8$).   Then $\nabla f(x,y) = 3x^2 \vec{i}-\vec{j}$ so the normal direction at the $x$-intercept $(2,0)$ is $\nabla f(2,0) = 12 \vec{i}-\vec{j}$.  So the line is parameterized as $\mathbf{r}(t)= 2\vec{i} + t\ (12 \vec{i}-\vec{j})=(2+12t) \vec{i}+(-t) \vec{j}$. Set $y=-t$ and thus $x=2-12y$ or (in standard form) $x+12y=2$. Solving for $y$ gives the slope-intercept form: $y=-\frac{1}{12}x+\frac{1}{6}$.
