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

May 23, 2017

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

#### Explanation:

We are seeking equation at $x = - 1$ on the curve $f \left(x\right) = - 2 {x}^{3} + 4 {x}^{2} + 2 x$ i.e. at $\left(- 1 , f \left(- 1\right)\right)$ i.e. $\left(- 1 , 4\right)$

as $f \left(- 1\right) = - 2 {\left(- 1\right)}^{3} + 4 {\left(- 1\right)}^{2} + 2 \left(- 1\right) = 2 + 4 - 2 = 4$

As slope of tangent at $f \left(- 1\right)$ is $f ' \left(- 1\right)$ and as

$f ' \left(x\right) = - 6 {x}^{2} + 8 x + 2$ and hence slope of tangent is $- 6 {\left(- 1\right)}^{2} + 8 \left(- 1\right) + 2 = - 6 - 8 + 2 = - 12$

As normal is perpendicular to tangent, its slope is $\frac{- 1}{- 12} = \frac{1}{12}$

and equation of normal is $y - 4 = \frac{1}{12} \left(x + 1\right)$ or $x - 12 y + 49 = 0$

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