What is the equation of the normal line of #f(x)=x^3+6x^2-3x# at #x=-1#?

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Answer 1 · 2018-03-29T07:51:40

#12y-x+14=0#

#f(x)=x^3+6x^2-3x#

to find the normal at #x=-1#

so the #y# value

#y=f(-1)=(-1)^3+6(-1)^2-3(-1)#

#y=-1+6-3=2#

#:. (x_1,y_1)=(-1,2)---(1)#

we need the gradient at #x=-1#

the tangent gradient is

#f'(-1)#

and the normal gradient will be calculated from

#m_tm_n=-1#

#f(x)=x^3+6x^2-3x#

#f'(x)=3x^2+12x-3#

#m_t=f'(-1)=3(-1)^2+12(-1)-3#

#m_t=3-12-3=-12#

#=>m_n= -1/m_t=1/12 " from "(1)#

eqn of normal

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

#(y- -1)=1/12(x-2)#

#12(y+1)=x-2#

#=>12y-x+14=0#