What is the equation of the normal line of #f(x)=(x-3)^3(x+2)# at #x=0#?

1 Answer
Nov 19, 2015

The equation of the normal line is #y=-1/27 x-54#

Explanation:

Expanding the formula for #f# gives:

#f(x)=(x^3-9x^2+27x-27)(x+2)#

#=x^4-9x^3+27x^2-27x+2x^3-18x^2+54x-54#

#=x^4-7x^3+9x^2+27x-54#.

Therefore, #f'(x)=4x^3-21^2+18x+27#.

Also note that #f(0)=-54# and #f'(0)=27#.

The normal line at the point #(x,y)=(0,f(0))=(0,-54)# has slope equal to the negative reciprocal of #f'(0)=27#, so this slope equals #-1/27#. Since this normal line goes through the point #(0,-54)#, it follows that its equation is #y=-1/27x-54#.