Question

What is the equation of the normal line of `f(x)=x^3-4x^2+7x` at `x=-2`?

Answer 1

`y=-1/35x-1332/35`

Explanation:

From the given equation
`f(x)=x^3-4x^2+7x` at the given abscissa `x=-2`

We need the negative reciprocal of slope m and the point

Solve for the slope m
`m=f' (-2)`

`f' (x)=d/dx(x^3-4x^2+7x)`

`f' (x)=3x^2-8x+7`

`m=f' (-2)=3(-2)^2-8(-2)+7`

`m=f' (-2)=`3(4)+16+7#

`m=f' (-2)=35`

Solve for `f(-2)`

`f(x)=x^3-4x^2+7x`
`f(-2)=(-2)^3-4(-2)^2+7(-2)`

`f(-2)=-8-16-14`
`f(-2)=-38`

Our needed point `(x_1, y_1)=(-2, -38)`

Let us now solve for the normal line

`y-y_1=-1/m(x-x_1)`

`y--38=-1/35(x--2)`

`y+38=-1/35(x+2)`

#y=-1/35(x+2)-38

`y=-1/35x-2/35-38`

`y=-1/35x-1332/35`

Kindly see the graph of `f(x)=x^3-4x^2+7x` and the normal line
`y=-1/35x-1332/35`
graph{(y+1/35x+1332/35)(y-x^3+4x^2-7x)=0[-90,90,-45,45]}

God bless....I hope the explanation is useful.