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

Jan 10, 2018

$2 y + x = 1$

#### Explanation:

$f \left(x\right) = \left(2 x - 2\right) {e}^{2 x - 2}$
$\implies f ' \left(x\right) = \left(4 x - 2\right) \cdot {e}^{2 x - 2}$

$\implies f ' \left(1\right) = 2$
$\implies {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\text{at x =1}} = 2$

$\implies {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}_{\text{at x =1}} = - \frac{1}{2}$

Hence we get the slope of the normal: $m = - \frac{1}{2}$

Also, $f \left(1\right) = 0$, hnce f(x) passes through $\left(1 , 0\right)$
Using point slope form of line,

$\left(y - 0\right) = m \left(x - 1\right)$
$\implies y = - \frac{1}{2} \left(x - 1\right)$
$\implies 2 y + x = 1$