# What is the equation of the line normal to  f(x)=-sqrt((x+1)(x-2) at  x=6?

Feb 9, 2018

$y = \frac{4 \sqrt{7} x}{11} - \frac{46 \sqrt{7}}{11}$

$4 \sqrt{7} x - 11 y - 46 \sqrt{7} = 0$

#### Explanation:

We are given:
$f \left(x\right) = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{d}{\mathrm{dx}} \left[- {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{\frac{1}{2}}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{d}{\mathrm{dx}} \left[{\left(\left(x + 1\right) \left(x - 2\right)\right)}^{\frac{1}{2}}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{\frac{1}{2} - 1} \cdot \frac{1}{2} \cdot \frac{d}{\mathrm{dx}} \left[\left(x + 1\right) \left(x - 2\right)\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{\frac{1}{2} - 1} \cdot \frac{1}{2} \cdot \left(\left(x + 1\right) \frac{d}{\mathrm{dx}} \left[x - 2\right] + \left(x - 2\right) \frac{d}{\mathrm{dx}} \left[x + 1\right]\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{- \frac{1}{2}} / 2 \cdot \left(\left(x + 1\right) \left(1\right) + \left(x - 2\right) \left(1\right)\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{- \frac{1}{2}} / 2 \cdot \left(x + 1 + x - 2\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{- \frac{1}{2}} / 2 \cdot \left(2 x - 1\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{\left(2 x - 1\right) {\left(\left(x + 1\right) \left(x - 2\right)\right)}^{- \frac{1}{2}}}{2}$

$f ' \left(6\right) = - \frac{\left(2 \left(6\right) - 1\right) {\left(\left(6 + 1\right) \left(6 - 2\right)\right)}^{- \frac{1}{2}}}{2} = - \frac{11 \sqrt{7}}{28}$

However, for the normal, $\text{gradient} = - \frac{1}{f ' \left(x\right)} = - \frac{1}{- \frac{11 \sqrt{7}}{28}} = \frac{28}{11 \sqrt{7}}$

The normal will be in the form of $y = m x + c$,

$m = \frac{28}{11 \sqrt{7}}$
$x = 6$
$y = - {\left(\left(6 + 1\right) \left(6 - 2\right)\right)}^{\frac{1}{2}} = - \sqrt{7 \cdot 4} = - 2 \sqrt{7}$

$- 2 \sqrt{7} = 6 \cdot \frac{28}{11 \sqrt{7}} + c$

$c = - 2 \sqrt{7} - 6 \cdot \frac{28}{11 \sqrt{7}}$

$\textcolor{w h i t e}{c} = - \frac{46 \sqrt{7}}{11}$

$y = \frac{28 x}{11 \sqrt{7}} - \frac{46 \sqrt{7}}{11}$

$y = \frac{4 \sqrt{7} x}{11} - \frac{46 \sqrt{7}}{11}$

$y = \frac{4 \sqrt{7} x - 46 \sqrt{7}}{11}$

$11 y = 4 \sqrt{7} x - 46 \sqrt{7}$

$4 \sqrt{7} x - 11 y - 46 \sqrt{7} = 0$