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

Jul 8, 2016

It is $y = - \frac{1}{2} x + \frac{3}{2}$.

#### Explanation:

We can first find the tangent to that curve in the point $x = 1$.
The slope of the tangent is given by the derivative

${m}_{t} = \frac{d}{\mathrm{dx}} {e}^{2 x - 2}$ when $x = 1$

${m}_{t} = {e}^{2 x - 2} \cdot \frac{d}{\mathrm{dx}} \left(2 x - 2\right) = 2 {e}^{2 x - 2}$ and when $x = 1$ it is

${m}_{t} = 2 {e}^{2} - 2 = 2 \cdot {e}^{0} = 2$.

This is the slope of the tangent line. The normal line is orthogonal to this line. The slope of the orthogonal line is ${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{2}$.
Then the normal has slope $- \frac{1}{2}$.
The equation of the line is

$y = - \frac{1}{2} x + q$ and we have to find $q$. To do this we impose the passage from the point $x = 1$ and we have to evaluate the $y$ calculating $y = {e}^{2 x - 2} = {e}^{2 \cdot 1 - 2} = {e}^{0} = 1$. Then the line passes for the point $\left(1 , 1\right)$.

I substitute this in the equation of the normal line

$y = - \frac{1}{2} x + q$

$1 = - \frac{1}{2} \cdot 1 + q$

$q = 1 + \frac{1}{2} = \frac{3}{2}$

Then the equation of the normal is

$y = - \frac{1}{2} x + \frac{3}{2}$. 