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

1 Answer
Apr 19, 2017

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