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

Aug 1, 2017

$y = - x + 3$

#### Explanation:

First find the equation of the line TANGENT to the function at the point; then find the equation of the line that passes through that point but has a perpendicular slope.

$f \left(x\right) = {x}^{2} - x + 2$

$f \left(1\right) = {1}^{2} - 1 + 2 = 2$

$f ' \left(x\right) = 2 x - 1$

$f ' \left(1\right) = 2 \left(1\right) - 1$

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

We now have the point and the slope of the line tangent to $f \left(x\right)$ at $x = 1$.
$\textcolor{b l u e}{\text{Tangent line: } y = 1 \left(x - 1\right) + 2}$

Simplified tangent line: $y = x + 1$
Since the slope is $1$, the slope of the normal line will be $\frac{- 1}{1} = - 1$

$\textcolor{b l u e}{\text{Normal line: } y = - 1 \left(x - 1\right) + 2}$

Simplified normal line:
$y = - x + 1 + 2$

$y = - x + 3$