What is the equation of the normal line of #f(x)=(x-9)^2/(x-1)# at #x=-2#?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T09:15:03

The normal line

#y=(9x)/55-6601/165#

The given curve is

#y=(x-9)^2/(x-1)#
at the point #x=-2#

Let us solve for the point #(x_1, y_1)# first

Let #x_1=-2#

#y_1=(x_1-9)^2/(x_1-1)=(-2-9)^2/(-2-1)=121/-3=-121/3#

Our point #(x_1, y_1)=(-2, -121/3)#

Solve for the slope of the normal line #m_n=-1/m#

#m=d/dx((x-9)^2/(x-1))=((x-1)*2(x-9)-(x-9)^2*1)/(x-1)^2#

substitute #x=-2#

#m=((-2-1)*2(-2-9)-(-2-9)^2*1)/(-2-1)^2=-55/9#

#m_n=-1/m#

#m_n=-1/(-55/9)#

#m_n=9/55#

Let us solve for the normal line using #m_n# and #(x_1, y_1)#

#y-y_1=m_n(x-x_1)#

#y-(-121/3)=9/55(x-(-2))#

#y=(9x)/55-6601/165#

Kindly see the graphs of #y=(x-9)^2/(x-1)# and the normal line #y=(9x)/55-6601/165#

Desmos

God bless....I hope the explanation is useful.