# What is the equation of the normal line of f(x)=-7x^3-x^2-x-1 at x=-8?

Dec 14, 2015

$y = \frac{1}{1329} x + 3527$.

#### Explanation:

$f \left(- 8\right) = - 7 {\left(- 8\right)}^{3} - {\left(- 8\right)}^{2} - \left(- 8\right) - 1 = 3527$.

$f ' \left(x\right) = - 21 {x}^{2} - 2 x - 1$.

$f ' \left(- 8\right) = - 21 {\left(- 8\right)}^{2} - 2 \left(- 8\right) - 1 = - 1329$.

Hence the slope of the tangent at $x = - 8$ is $- 1329$.

But the normal is perpendicular to the tangent and hence the slope of the normal at that point is $m = \frac{1}{1329}$.

But the normal line is a straight line so of form $y = m x + c$.

Substituting the point (x,y)=((-8,3527), we get for the normal :

$3527 = \frac{1}{1329} \cdot \left(- 8\right) + c$

$\therefore c = 3527$.

Thus the normal has equation $y = \frac{1}{1329} x + 3527$.