# What is the equation of the normal line of f(x)=(x-4)e^(x-2) at x=0?

Sep 22, 2017

$y = 2.463 x + 0.5413$

#### Explanation:

Equation of the line will equal
$y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right)$

Differentiate to find the gradient function using the product rule
$\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

$u = x - 4$
$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

$v = {e}^{x - 2}$
$\frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x - 2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(x - 4\right) {e}^{x - 2} + {e}^{x - 2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(x - 4\right) {e}^{x - 2} + {e}^{x - 2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = x {e}^{x - 2} - 3 {e}^{x - 2}$

At $x = 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {e}^{- 2} = {m}_{1}$

The normal (${m}_{2}$) is perpendicular to the tangent (${m}_{1}$) so:
${m}_{2} = \frac{- 1}{{m}_{1}}$

${m}_{2} = \frac{- 1}{- 3 {e}^{- 2}}$
${m}_{2} = \frac{1}{3 {e}^{- 2}}$
${m}_{2} = 2.463$

At $x = 0$,
$y = \left(0 - 4\right) {e}^{0 - 2}$
$y = \left(4\right) {e}^{- 2}$
$y = 0.5413$

So the equation of the normal to the curve at x=0 is,
$y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right)$
$y - 0.5413 = 2.463 \left(x - 0\right)$
$y = 2.463 x + 0.5413$