Question

What is the equation of the line normal to `f(x)=(3x-1)(x+4)` at `x=-3`?

Answer 1

`7y=x-67`

Explanation:

Take the derivative of the equation to find the slope of the tangent, and then the slope of the normal is the negative inverse.

First expand the equation to simplify getting the derivative:
`f(x)=(3x^2 +11x - 4)`
`f'(x) = 6x+11`

At the point `x=-3` the slope is `6(-3)+11 =-7`
The slope is `-7`, therefore the slope of the normal is `1/7`

The equation of the normal is therefore `y=1/7x +c`

At the point `x=-3` the original expression gives`y=27-33-4 = -10`

Plugging this back into the normal gives `c =-10-1/7*(-3) = -67/7`
The normal equation is therefore `y=1/7x -67/7` or
`7y=x-67`