Question

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

Answer 1

`color(blue)(y=-9x-82/3)`

Explanation:

The line that is normal to `f(x)=-1/x` at `x=-3` is perpendicular to the tangent line at `x=-3`.

We first need to find the gradient of the tangent line at `x=-3`. We now find the derivative of `f(x)=-1/x`.

`dy/dx(-1/x)=1/x^2`

Plugging in `x=-3`:

`1/(-3)^2=1/9`

We know that if two lines are perpendicular, then the product of their gradients is `-1`

Let gradient of the normal be `bbm`, then:

`m*1/9=-1=>m=-9`

The normal passes through the point with x coordinate `-3`. We now need a corresponding `y` coordinate. Plugging in `x=-3` into `f(x)=-1/x`:

`-1/(-3)=1/3`

Using point slope form of a line:

`(y_2-y_1)=m(x_2-x_1)`

`y-(1/3)=-9(x-(-3))`

`color(blue)(y=-9x-82/3)`

Graph: