# What is the equation of the normal line of f(x)=x^3+6x^2-3x at x=-1?

Mar 29, 2018

$12 y - x + 14 = 0$

#### Explanation:

$f \left(x\right) = {x}^{3} + 6 {x}^{2} - 3 x$

to find the normal at $x = - 1$

so the $y$ value

$y = f \left(- 1\right) = {\left(- 1\right)}^{3} + 6 {\left(- 1\right)}^{2} - 3 \left(- 1\right)$

$y = - 1 + 6 - 3 = 2$

$\therefore \left({x}_{1} , {y}_{1}\right) = \left(- 1 , 2\right) - - - \left(1\right)$

we need the gradient at $x = - 1$

$f ' \left(- 1\right)$

and the normal gradient will be calculated from

${m}_{t} {m}_{n} = - 1$

$f \left(x\right) = {x}^{3} + 6 {x}^{2} - 3 x$

$f ' \left(x\right) = 3 {x}^{2} + 12 x - 3$

${m}_{t} = f ' \left(- 1\right) = 3 {\left(- 1\right)}^{2} + 12 \left(- 1\right) - 3$

${m}_{t} = 3 - 12 - 3 = - 12$

$\implies {m}_{n} = - \frac{1}{m} _ t = \frac{1}{12} \text{ from } \left(1\right)$

eqn of normal

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

$\left(y - - 1\right) = \frac{1}{12} \left(x - 2\right)$

$12 \left(y + 1\right) = x - 2$

$\implies 12 y - x + 14 = 0$