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?
Best Answer
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!