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

1 Answer
May 7, 2018

The normal line is given by #y=-x-4#

Explanation:

Rewrite #f(x)=(2x^2+1)/x# to #2x+1/x# to make differentiation simpler.

Then, using the power rule, #f'(x)=2-1/x^2#.

When #x=-1#, the y-value is #f(-1)=2(-1)+1/-1=-3#. Thus, we know that the normal line passes through #(-1,-3)#, which we will use later.

Also, when #x=-1#, instantaneous slope is #f'(-1)=2-1/(-1)^2=1#. This is also the slope of the tangent line.

If we have the slope to the tangent #m#, we can find the slope to the normal via #-1/m#. Substitute #m=1# to get #-1#.

Therefore, we know that the normal line is of the form
#y=-x+b#

We know that the normal line passes through #(-1,-3)#. Substitute this in:
#-3=-(-1)+b#
#therefore b=-4#

Substitute #b# back in to get our final answer:
#y=-x-4#

You can verify this on a graph:
graph{(y-(2x^2+1)/x)(y+x+4)((y+3)^2+(x+1)^2-0.01)=0 [-10, 10, -5, 5]}