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

Sep 5, 2016

$\textcolor{b l u e}{y = \frac{1}{9} x + \frac{1}{9}}$

#### Explanation:

$\textcolor{b l u e}{\text{determine the gradient of } f \left(x\right)}$

Expanding the brackets

$f \left(x\right) \equiv \left(x - 1\right) \left({x}^{2} + 4 x + 4\right)$

$f \left(x\right) \equiv {x}^{3} + 4 {x}^{2} + 4 x \text{ } - {x}^{2} - 4 x - 4$

$= {x}^{3} + 3 {x}^{2} - 4$

$f ' = 3 {x}^{2} + 6 x$

$f ' \left(1\right) = 3 {\left(1\right)}^{2} + 6 \left(1\right) = 9$

$\text{The above value of 9 is the gradient of " f(x)" at } x = 1$

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$\textcolor{b l u e}{\text{Determine the equation of the straight line}}$

So the gradient of the straight line normal to this is $- \frac{1}{9}$ and it will pass through the point ${P}_{1} \to \left({x}_{1} , {y}_{1}\right)$ where ${x}_{1} = 1 \leftarrow \text{ given}$

So at ${P}_{1}$

$f \left(1\right) = {y}_{1} = {\left({x}_{1}\right)}^{3} + 3 {\left({x}_{1}\right)}^{2} - 4$

${y}_{1} = 1 + 3 - 4 = 0$

Giving $y = m x + c \text{ " ->" } {y}_{1} = - \frac{1}{9} {x}_{1} + c$

$c = 0 + \frac{1}{9} \left(1\right) = + \frac{1}{9}$

$\textcolor{b l u e}{y = \frac{1}{9} x + \frac{1}{9}}$