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

Apr 19, 2017

$6 x - y + 22 = 0$

#### 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,

$f ' \left(x\right) = 2 \left(x - 3\right) - 2$

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

$f ' \left(- 1\right) = - 6$

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 $f \left(x\right)$ at $x = - 1$,

$f \left(- 1\right) = 16$

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

The equation of the normal is $\left(y - 16\right) = 6 \left(x + 1\right)$

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