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

May 27, 2016

$y = - \frac{1}{11} x - 18 \frac{2}{11}$

#### Explanation:

Given -

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

The first derivative gives the slope at any given point

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4 x + 3$

At $x = - 2$ its slope is

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4 \left(- 2\right) + 3 = 8 + 3 = 11$

It can be taken as the slope of the Tangent, drawn to that point on the curve.

Y- co-ordinate of the point -

$y = - 2 {\left(- 2\right)}^{2} + 3 \left(- 2\right) - 4$
$y = - 8 - 6 - 4 = - 18$

The point on the curve, the normal passing through is $\left(- 2 , - 18\right)$

If the two lines are perpendicular then -

${m}_{2} \times {m}_{2} = - 1$
$11 \times {m}_{2} = - 1$
${m}_{2} = \frac{- 1}{11}$

The equation of the normal is -

$m x + c = y$
$- \frac{1}{11} \left(- 2\right) + c = - 18$
$\frac{2}{11} + c = - 18$
$c = - 18 - \frac{2}{11} = 18 \frac{2}{11}$

Equation is -

$y = - \frac{1}{11} x - 18 \frac{2}{11}$