# What is the equation of the normal line of f(x)=(x-3)^3(x+2) at x=0?

Nov 19, 2015

The equation of the normal line is $y = - \frac{1}{27} x - 54$

#### Explanation:

Expanding the formula for $f$ gives:

$f \left(x\right) = \left({x}^{3} - 9 {x}^{2} + 27 x - 27\right) \left(x + 2\right)$

$= {x}^{4} - 9 {x}^{3} + 27 {x}^{2} - 27 x + 2 {x}^{3} - 18 {x}^{2} + 54 x - 54$

$= {x}^{4} - 7 {x}^{3} + 9 {x}^{2} + 27 x - 54$.

Therefore, $f ' \left(x\right) = 4 {x}^{3} - {21}^{2} + 18 x + 27$.

Also note that $f \left(0\right) = - 54$ and $f ' \left(0\right) = 27$.

The normal line at the point $\left(x , y\right) = \left(0 , f \left(0\right)\right) = \left(0 , - 54\right)$ has slope equal to the negative reciprocal of $f ' \left(0\right) = 27$, so this slope equals $- \frac{1}{27}$. Since this normal line goes through the point $\left(0 , - 54\right)$, it follows that its equation is $y = - \frac{1}{27} x - 54$.