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

##### 1 Answer
May 18, 2016

I found: $y = - \frac{1}{3} x + \frac{4}{3}$

#### Explanation:

We can first find the coordinates of the point of intersection: knowing that $x = 1$ we substitute into our function and it gives us:
$y = f \left(1\right) = 1 \cdot {e}^{2 \cdot 1 - 2} = 1 \cdot {e}^{0} = 1$
So we get $\left(1 , 1\right)$.

Next we try to evaluate the GRADIENT of the line perpendicular (normal).
We evaluate the derivative of the function:
$f ' \left(x\right) = 1 \cdot {e}^{2 x - 2} + 2 x {e}^{2 x - x} = {e}^{2 x - x} \left[2 x + 1\right]$
Evaluate the derivative at $x = 1$
$f ' \left(1\right) = 1 \left[2 + 1\right] = 3$
This is the slope ${m}_{T}$ of the TANGENT to your curve at your point. To finde the slope of the normal ${m}_{N}$ we need:
${m}_{N} = - \frac{1}{m} _ T = - \frac{1}{3}$

Finally we use the general equation of a line through a point and having slope equal to ${m}_{T}$:
$y - {y}_{0} = {m}_{T} \left(x - {x}_{0}\right)$
$y - 1 = - \frac{1}{3} \left(x - 1\right)$
$y = 1 - \frac{1}{3} x + \frac{1}{3}$
$y = - \frac{1}{3} x + \frac{4}{3}$

Graphically: