What is the equation of the normal line of #f(x)=x^3-x^2+6x-1# at #x=5#?

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Answer 1 · 2017-01-17T11:54:13

#x+71y-9154=0#. See the normal-inclusive Socratic second graph.

The first is a not-to-scale graph.

graph{x^3-x^2+6x-1 [-20, 20, -150, 150]} At x = 5, y = 129.

#f'=3x^2-2x+6=71#, at #x = 5.#

Slope of the normal is #-1/f'=-1/71.

The equation to the nthe normal is

y-129=-1/71(x-5), giving

#x+71y-9154=0#

The normal=inclusive-graph below. is for uniform scale near the foot

of the normal (5, 129), shown as a small circle.

graph{(x^3-x^2+6x-1-y)(x+71y-9154)((x-5)^2+(y-129)^2-.1)=0 [-20, 20,, 120, 140]}