What is the equation of the line normal to #f(x)=(x-1)^2-2x+5# at #x=-2#?

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Answer 1 · 2018-07-18T15:19:06

#x-8y+146=0#

The slope #f'(x)# of tangent to the curve: #f(x)=(x-1)^2-2x+5# is given by differentiating given function w.r.t. #x# as follows

#f'(x)=d/dx((x-1)^2-2x+5)#

#=2(x-1)-2#

#=2x-4#

hence the slope of tangent at #x=-2# is

#f'(-2)=2(-2)-4#

#=-8#

hence the slope of normal to the curve at the same point #x=-2# is

#=-1/{-8}#

#=1/8#

Now, substituting #x=-2# in the given function: #f(x)=(x-1)^2-2x+5# we get #y# coordinate of the point as follows

#f(-2)=(-2-1)^2-2(-2)+5#

#=18#

Hence, the equation of normal to the curve at #(-2, 18)\equiv(x_1, y_1)# & having a slope #m=1/8# is given as

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

#y-18=1/8(x-(-2))#

#8y-144=x+2#

#x-8y+146=0#