Question

What is the equation of the normal line of `f(x)=-2x^3+4x^2+2x` at `x=-1`?

Answer 1

Equation of normal is `x-12y+49=0`

Explanation:

We are seeking equation at `x=-1` on the curve `f(x)=-2x^3+4x^2+2x` i.e. at `(-1,f(-1))` i.e. `(-1,4)`

as `f(-1)=-2(-1)^3+4(-1)^2+2(-1)=2+4-2=4`

As slope of tangent at `f(-1)` is `f'(-1)` and as

`f'(x)=-6x^2+8x+2` and hence slope of tangent is `-6(-1)^2+8(-1)+2=-6-8+2=-12`

As normal is perpendicular to tangent, its slope is `(-1)/(-12)=1/12`

and equation of normal is `y-4=1/12(x+1)` or `x-12y+49=0`

graph{(x-12y+49)(y+2x^3-4x^2-2x)=0 [-9.71, 10.29, -1.28, 8.72]}