Question

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

Answer 1

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

The first is a not-to-scale graph.

Explanation:

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]}