# What is the equation of the normal line of f(x)=5x^3-2x^2-3x-1 at x=-8?

Jul 30, 2017

Equation of normal is $x + 989 y + 2635693 = 0$

#### Explanation:

At $x = - 8$, $f \left(x\right) = 5 {\left(- 8\right)}^{3} - 2 {\left(- 8\right)}^{2} - 3 \left(- 8\right) - 1$

$= - 2560 - 128 + 24 - 1 = - 2665$

Hencewe are seeking normal at $\left(- 8 , - 2665\right)$

Slope of tangent is given by $f ' \left(- 8\right)$ and as $f ' \left(x\right) = 15 {x}^{2} - 4 x - 3$

slope of tangent is $15 {\left(- 8\right)}^{2} - 4 \left(- 8\right) - 3 = 989$

and hence slope of normal is $- \frac{1}{989}$ and

equation of normal is $y + 2665 = - \frac{1}{989} \left(x + 8\right)$

or $x + 989 y + 2635693 = 0$