What is the slope of the line normal to the tangent line of f(x) = x^2e^(x-1)+2x-2  at  x= 1 ?

Jul 18, 2018

$- \frac{1}{4}$

Explanation:

We know that a normal line is orthogonal to a tangent line, so we should find the slope of the tangent line first, then find the slope orthogonal to it.

The slope of the tangent line is just the derivative:
$f ' \left(x\right) = 2 x {e}^{x - 1} + {x}^{2} {e}^{x - 1} + 2$
$f ' \left(1\right) = 2 {e}^{0} + 1 {e}^{0} + 2 = 4$

Two orthogonal lines have
${m}_{1} {m}_{2} = - 1$

so
m_(norm) * 4 = -1 implies m_(norm) = -1/4