Question

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

Answer 1

`x-17y=880`

Explanation:

The slope of an equation is given by its derivative.

If `f(x)=x^3-7x^2-x`
then `f'(x)=3x^2-14x-1`
and
at `x=4`
the slope is `m= f'(4)=3(4^2)-14(4)-1=-17`

If a line has a slope of `m` then any line perpendicular to it (i.e. its "normal") has a slope of `-1/m`

At `x=4`
`color(white)("XXX")f(4)=4^3-7(4^2)-4 = -52`

So the required normal line
`color(white)("XXX")`has a slope of `1/17`
and
`color(white)("XXX")`passes through the point `(4,-52)`

Using the slope-point form:
`color(white)("XXX")y-(-52)=1/17(x-4)`

`color(white)("XXX")rarr 17y+884 = x-4`

or, in standard form:
`color(white)("XXX")x-17y=880`