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

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Answer 1 · 2016-09-05T11:48:00

#color(blue)(y=1/9x+1/9)#

#color(blue)("determine the gradient of "f(x))#

Expanding the brackets

#f(x)-=(x-1)(x^2+4x+4)#

#f(x)-=x^3+4x^2+4x" "-x^2-4x-4#

#=x^3+3x^2-4#

#f'=3x^2+6x#

#f'(1)=3(1)^2+6(1) = 9 #

#"The above value of 9 is the gradient of " f(x)" at "x=1#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the equation of the straight line")#

So the gradient of the straight line normal to this is #-1/9# and it will pass through the point #P_1->(x_1,y_1)# where #x_1=1 larr" given"#

So at #P_1#

#f(1)=y_1=(x_1)^3+3(x_1)^2-4#

#y_1=1+3-4=0#

Giving #y=mx+c" " ->" "y_1=-1/9x_1+c#

#c=0+1/9(1) = +1/9#

#color(blue)(y=1/9x+1/9)#