# What is the equation of the normal line of f(x)=(x-1)(x+4)(x-2)  at x=3 ?

Feb 3, 2016

23y + x - 325 = 0

#### Explanation:

before differentiating , distribute the brackets.

start with (x-1)(x+4) $= {x}^{2} + 3 x - 4$

$\left({x}^{2} + 3 x - 4\right) \left(x - 2\right) = {x}^{3} - 2 {x}^{2} + 3 {x}^{2} - 6 x - 4 x + 8$

f(x) $= {x}^{3} + {x}^{2} - 10 x + 8$

The derivative of f(x) is the gradient of the tangent and f'(3)
the value of the tangent.

$f ' \left(x\right) = 3 {x}^{2} + 2 x - 10$

and f'(3)$= 3 {\left(3\right)}^{2} + 2 \left(3\right) - 10 = 27 + 6 - 10 = 23 = m$

For 2 perpendicular lines , the product of their gradients
is -1

let ${m}_{1} = \textcolor{b l a c k}{\text{ gradient of normal }}$

then  m.m_1 = -1 → 23.m_1 = -1 rArr m_1 = -1/23

f(3) = $\left(2\right) \left(7\right) \left(1\right) = 14 \Rightarrow \left(3 , 14\right) i s \textcolor{b l a c k}{\text{ normal point}}$

equation of normal is y-b = m(x-a) $m = - \frac{1}{23} , \left(a , b\right) = \left(3 , 14\right)$

so y-14 = $- \frac{1}{23} \left(x - 3\right)$

(multiply by 23 to eliminate fraction )

23y - 322 = - x + 3 $\Rightarrow 23 y + x - 325 = 0$