What is the equation of the line normal to f(x)=6x^3-8 at x=7?

1 Answer
Dec 4, 2016

The equation of the line normal to f(x) = 6x^3-8 at x=7 is:

y-2050=-1/882(x-7)

Explanation:

The equation of the tangent line to a curve y=f(x) at x=barx is:

y-f(bar x) =f'(barx) (x -bar x)

The normal to this line is:

y-f(bar x) =-1/(f'(barx)) (x -bar x).

Calculate:

f'(x) =18x^2

and for bar x =7

f(barx) = 2050

f'(bar x) = 882

So the equation of the normal line is:

y-2050=-1/882(x-7)