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

Oct 21, 2016

Let's start by simplifying the function.

$f \left(x\right) = \frac{1}{\sqrt{\left(x - 1\right) \left(x - 1\right)}}$

$f \left(x\right) = \frac{1}{x - 1}$

Now, let's start by finding the corresponding y-coordinate to $x = 2$.

$f \left(2\right) = \frac{1}{2 - 1} = \frac{1}{1} = 1$

Next, let's differentiate the function.

$f \left(x\right) = \frac{1}{x - 1}$

By the quotient rule:

$f ' \left(x\right) = \frac{0 \times \left(x - 1\right) - 1 \left(1\right)}{x - 1} ^ 2$

$f ' \left(x\right) = - \frac{1}{x - 1} ^ 2$

The slope of the tangent (the line that touches the graph at the point $x = a$) at the point $x = 2$ is

$f ' \left(2\right) = - \frac{1}{2 - 1} ^ 2 = - \frac{1}{1} = - 1$

The normal line is perpendicular to the tangent. So, the slope of the normal line is $- \frac{1}{- 1} = 1$

We now know the normal line's point of contact on the function $\left(2 , 1\right)$ and the slope of the normal line. We can finally use point-slope form to determine the equation of the normal line.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = 1 \left(x - 2\right)$

$y - 1 = x - 2$

$y = x - 1$

Hence, the equation of the normal line is $y = x - 1$.

Hopefully this helps!