# What is the equation of the line normal to f(x)=9x^2 - 28x - 34  at x=-1?

Dec 29, 2016

$46 y - x - 139 = 0$

#### Explanation:

let $\text{ } y = f \left(x\right)$

1) find the gradient of the normal

This is done in two parts

(i) find gradient of tangent ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x = - 1}$

$y = 9 {x}^{2} - 28 x - 34 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 18 x - 28$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x = - 1} = 18 \times - 1 - 28 = - 18 - 28 = - 46$

(ii) the normal is perpendicular to the tangent .

${m}_{t} \times {m}_{n} = - 1$

$- 46 \times {m}_{n} = - 1 \implies {m}_{n} = \frac{1}{46}$

2) find the eqn. of normal

use $\text{ } y - {y}_{1} = {m}_{n} \left(x - {x}_{1}\right)$

need to find $\text{ " y_1" }$first

$y \left(- 1\right) = 9 \times {\left(- 1\right)}^{2} - 28 \times \left(- 1\right) - 34$

$y \left(- 1\right) = 9 + 28 - 34 = 3$

$\left({x}_{1} , {y}_{1}\right) = \left(- 1 , 3\right)$

$\therefore y - 3 = \frac{1}{46} \left(x - - 1\right)$

$46 y - 138 = x + 1$

$46 y - x - 139 = 0$