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

Dec 4, 2016

The equation of the line normal to $f \left(x\right) = 6 {x}^{3} - 8$ at $x = 7$ is:

$y - 2050 = - \frac{1}{882} \left(x - 7\right)$

#### Explanation:

The equation of the tangent line to a curve $y = f \left(x\right)$ at $x = \overline{x}$ is:

$y - f \left(\overline{x}\right) = f ' \left(\overline{x}\right) \left(x - \overline{x}\right)$

The normal to this line is:

$y - f \left(\overline{x}\right) = - \frac{1}{f ' \left(\overline{x}\right)} \left(x - \overline{x}\right)$.

Calculate:

$f ' \left(x\right) = 18 {x}^{2}$

and for $\overline{x} = 7$

$f \left(\overline{x}\right) = 2050$

$f ' \left(\overline{x}\right) = 882$

So the equation of the normal line is:

$y - 2050 = - \frac{1}{882} \left(x - 7\right)$