Question

What is the equation of the normal line of `f(x)=(x-3)^3(x+2)` at `x=0`?

Answer 1

The equation of the normal line is `y=-1/27 x-54`

Explanation:

Expanding the formula for `f` gives:

`f(x)=(x^3-9x^2+27x-27)(x+2)`

`=x^4-9x^3+27x^2-27x+2x^3-18x^2+54x-54`

`=x^4-7x^3+9x^2+27x-54`.

Therefore, `f'(x)=4x^3-21^2+18x+27`.

Also note that `f(0)=-54` and `f'(0)=27`.

The normal line at the point `(x,y)=(0,f(0))=(0,-54)` has slope equal to the negative reciprocal of `f'(0)=27`, so this slope equals `-1/27`. Since this normal line goes through the point `(0,-54)`, it follows that its equation is `y=-1/27x-54`.