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

The Normal Line is

$y = \frac{x}{12} - \frac{49}{6}$ or $x - 12 y = 98$

#### Explanation:

the given: $f \left(x\right) = {\left(x - 2\right)}^{\frac{3}{2}} - {x}^{3}$ and $x = 2$

$f \left(2\right) = {\left(2 - 2\right)}^{\frac{3}{2}} - {\left(2\right)}^{3}$

$f \left(2\right) = - 8$

the point on the curve : $\left(2 , - 8\right)$

compute slope $m$ then use $- \frac{1}{m}$ for the normal line

$f ' \left(x\right) = \left(\frac{3}{2}\right) {\left(x - 2\right)}^{\frac{1}{2}} - 3 {x}^{2}$

$f ' \left(2\right) = \left(\frac{3}{2}\right) {\left(2 - 2\right)}^{\frac{1}{2}} - 3 {\left(2\right)}^{2}$

$f ' \left(2\right) = - 12$

The slope to be used to find the normal line is $- \frac{1}{m}$

$- \frac{1}{m} = - \frac{1}{-} 12 = \frac{1}{12}$

Determine now the Normal Line using

$y - {y}_{1} = \left(\frac{1}{12}\right) \left(x - {x}_{1}\right)$

$y - \left(- 8\right) = \left(\frac{1}{12}\right) \left(x - 2\right)$

$x - 12 y = 98$ or $y = \frac{x}{12} - \frac{49}{6}$