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

Feb 10, 2017

$x + 2 y - 1 = 0$. See normal-inclusive Socratic graph.

#### Explanation:

graph{((2x-2)/(e^(2x-2))-y)(x+2y-1)=0 [-10, 10, -5, 5]}

$f = 2 {e}^{2} {e}^{- 2 x} \left(x - 1\right) = 0$, at x = 1.

So, the foot of the normal is P( 1, 0 ).

f'=2e^2(e^(-2x)(x-1)'+(x-1)(e^(-2x))')#

$= 2 {e}^{2} \left({e}^{- 2 x} - 2 {e}^{- 2 x} \left(x - 1\right)\right) = 2$, at x = 1.

Slope of the normal $= - \frac{1}{f '} = - \frac{1}{2}$.

So, the equation to the normal is

$y - 0 = - \frac{1}{2} \left(x - 1\right)$, giving

$x + 2 y - 1 = 0$