Question

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

Answer 1

`x+10y=3`

Explanation:

Given:
`color(white)("XXX")f(x)=(x-1)(x-3)(x+2)`

`f(3)=0` (since `(x-3)=0` when `x=3`)

So the point of intersection is going to be at `(x,y)=(3,0)`

Expanding the given equation we get
`color(white)("XXX")f(x)=x^3-2x^2-5x+6`

So
`color(white)("XXX")f'(x)=3x^2-4x-5`

And at `(3,0)` the tangent slope will be
`color(white)("XXX")f'(3)= 3(3^2)-4(3)-5 = 10`

Therefore the slope of the normal will be `(-1/10)`

Using the slope-point form for the normal we have
`color(white)("XXX")(y-0)=(-1/10)(x-3)`

which can be simplified as
`color(white)("XXX")x+10y=3`

(Sorry about the image quality. I've been having trouble with the graph method and this is an attempt to overlay 2 separate graphs snipped from the screen).