Question

What is the equation of the line normal to `f(x)=6x^2 -x - 9` at `x=1`?

Answer 1

`y=-x/11-43/11`

Explanation:

`f(x)=6x^2-x-9`

We can find the gradient of this function `m` at `x=1` by finding the 1st derivative, then we can get the gradient of the normal `m'` using the fact that `m.m'=-1`.

`f'(x)=12x-1=m`

So at `x=1:`

`m=(12xx1)-1=11`

`:.m'=-1/11`

`f(1)=6-1-9=-4`. This is the `y` value at `x=1`

The equation of the normal line is of the form:

`y=m'x+c'`

`:.-4=-1/11+c'`

`:.c'=1/11-4=1/11-44/11=-43/11`

So the equation of the normal`rArr`

`y=-x/11-43/11`

The situation looks like this:

graph{(-x/11-43/11-y)(6x^2-x-9-y)=0 [-20, 20, -10, 10]}