# What is the equation of the normal line of f(x)=x^3-4x^2+7x at x=-2?

$y = - \frac{1}{35} x - \frac{1332}{35}$

#### Explanation:

From the given equation
$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 7 x$ at the given abscissa $x = - 2$

We need the negative reciprocal of slope m and the point

Solve for the slope m
$m = f ' \left(- 2\right)$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} - 4 {x}^{2} + 7 x\right)$

$f ' \left(x\right) = 3 {x}^{2} - 8 x + 7$

$m = f ' \left(- 2\right) = 3 {\left(- 2\right)}^{2} - 8 \left(- 2\right) + 7$

$m = f ' \left(- 2\right) =$3(4)+16+7

$m = f ' \left(- 2\right) = 35$

Solve for $f \left(- 2\right)$

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 7 x$
$f \left(- 2\right) = {\left(- 2\right)}^{3} - 4 {\left(- 2\right)}^{2} + 7 \left(- 2\right)$

$f \left(- 2\right) = - 8 - 16 - 14$
$f \left(- 2\right) = - 38$

Our needed point $\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - 38\right)$

Let us now solve for the normal line

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

$y - - 38 = - \frac{1}{35} \left(x - - 2\right)$

$y + 38 = - \frac{1}{35} \left(x + 2\right)$

y=-1/35(x+2)-38

$y = - \frac{1}{35} x - \frac{2}{35} - 38$

$y = - \frac{1}{35} x - \frac{1332}{35}$

Kindly see the graph of $f \left(x\right) = {x}^{3} - 4 {x}^{2} + 7 x$ and the normal line
$y = - \frac{1}{35} x - \frac{1332}{35}$
graph{(y+1/35x+1332/35)(y-x^3+4x^2-7x)=0[-90,90,-45,45]}

God bless....I hope the explanation is useful.