What is the equation of the normal line of #f(x)=x^3-x^2+8x-9# at #x=-1#?

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Answer 1 · 2016-09-01T20:23:12

#x+13y+248=0#.

#f(x)=x^3-x^2+8x-9#

rArr f(-1)=-1-1-8-9=-19#.

Therefore, we require the eqn. of normal to the curve

# C : y=f(x)=x^3-x^2+8x-9# at the pt.#A(-1,-19)#.

We have, #f'(x)=3x^2-2x+8 rArr f'(-1)=3+2+8=13#.

But, we know that, #f"(-1)# gives the slope of tgt. line to the Curve

#C# at the pt.#A#.

#:." the slope of tgt. at "A" is "13#.

Normal is #bot# to tgt., so, its slope is #-1/13#, it passes thro. #A#.

Hence, eqn. of normal is # : y+19=-1/13(x+1)#, or,

#13y+247+x+1=0, i.e., x+13y+248=0#.

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