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

1 Answer
Oct 21, 2016

Let's start by simplifying the function.

#f(x) = 1/sqrt((x - 1)(x - 1))#

#f(x) = 1/(x- 1)#

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

#f(2) = 1/(2 - 1) = 1/1 = 1#

Next, let's differentiate the function.

#f(x) = 1/(x - 1)#

By the quotient rule:

#f'(x) = (0 xx (x - 1) - 1(1))/(x- 1)^2#

#f'(x) = -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'(2) = -1/(2 - 1)^2 = -1/1 = -1#

The normal line is perpendicular to the tangent. So, the slope of the normal line is #-1/(-1) = 1#

We now know the normal line's point of contact on the function #(2, 1)# 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(x- x_1)#

#y - 1 = 1(x- 2)#

#y - 1 = x - 2#

#y = x - 1#

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

Hopefully this helps!