Question

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

Answer 1

`6x-y+22=0`

Explanation:

Lets differentiate the function to find the slope of the tangent at `x=-1`, then we can find slope of the normal and find the equation of it.

Differentiating the function w.r.t `x`, we get,

`f'(x)=2(x-3)-2`

We find the slope of the tangent at `x=-1`, we get

`f'(-1)=-6`

Slope of the tangent is -6, hence slope of the normal is 6 , this because the product of slope of tangent and slope of normal at a point is `-1`.

Before finding the equation, we need the value of `f(x)` at `x=-1`,

`f(-1)=16`

Now we have the point `(-1,16)` and slope = `6`, which enough data to find the equation of the normal.

The equation of the normal is `(y-16)=6(x+1)`

On simplification, we get `6x-y+22=0`