What is the equation of the line normal to #f(x)=(5-x)^2 # at #x=2#?

1 Answer
Feb 15, 2016

#x-6y=-52#

Explanation:

Given #f(x)=(5-x)^2#
#color(white)("XXX")rarr f(x)=25-10x+x^2#
Its general slope is given by the derivative:
#color(white)("XXX")f'(x) = 2x-10#

At #x=2# this slope will be
#color(white)("XXX")2(2)-10 = -6#
and a line normal to it (i.e. perpendicular to it) will have a slope
#color(white)("XXX")=-(1/(-6))=1/6#

At #x=2, f(2)= (5-2)^2=9#
giving a point #(2,9)#

Using the slope-point form for the normal line
#color(white)("XXX")y-9=1/6(x-2)#

We can simplify this as
#color(white)("XXX")6y-54=x-2#
or (in standard form)
#color(white)("XXX")x-6y=-52#
graph{(y-(5-x)^2)(x-6y+52)=0 [-7.83, 14.67, -0.585, 10.665]}