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

$y + 5 = - \frac{1}{5} \left(x + 1\right)$
$f ' \left(x\right) = - 2 x + 3$
slope of tangent line when x = -1 is $5$ so slope of normal is $- \frac{1}{5}$.
When $x = - 1 , y = - 5$
Equation of normal: $y + 5 = - \frac{1}{5} \left(x + 1\right)$