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

Jan 31, 2016

$- \frac{1}{6}$

#### Explanation:

To find the slope of the normal line, first find the slope of the tangent line at $x = 1$. This can be done through finding the derivative of the function.

To find the derivative, you will have to use the product rule.

$f ' \left(x\right) = {e}^{x - 1} \frac{d}{\mathrm{dx}} \left(2 x\right) + 2 x \frac{d}{\mathrm{dx}} \left({e}^{x - 1}\right) + 2$

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

Note that $\frac{d}{\mathrm{dx}} \left({e}^{x - 1}\right) = {e}^{x - 1} \frac{d}{\mathrm{dx}} \left(x - 1\right) = {e}^{x - 1}$.

The slope of the tangent line is the value of the derivative at a certain point. In this case it is

$f ' \left(1\right) = 2 {e}^{0} + 2 \left(1\right) {e}^{0} + 2 = 6$

Recall that ${e}^{0} = 1$.

Since the slope of the tangent line is $6$, we can find the slope of the line normal to it with the knowledge that the tangent line and normal line are perpendicular. Since perpendicular lines have opposite reciprocal slopes, the line perpendicular to a line with slope $6$ has a slope of $- \frac{1}{6}$.