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

Question

1486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-14T22:45:39

$y=1/1329x+3527$ .

$f(-8)=-7(-8)^3-(-8)^2-(-8)-1=3527$ .

$f'(x)=-21x^2-2x-1$ .

$f'(-8)=-21(-8)^2-2(-8)-1=-1329$ .

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 $m=1/1329$ .

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

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

$3527=1/1329*(-8)+c$

$therefore c=3527$ .

Thus the normal has equation $y=1/1329x+3527$ .

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