# How do you find the point of inflexion of y=xe^(-x)+3? Then the inflectional tangent?

## $y = x {e}^{-} x + 3$

Sep 30, 2016

The inflection point would be $2 , 2 {e}^{- 2} + 3$
The tangent line would then be
$y - \left(2 {e}^{- 2} + 3\right) = - {e}^{- 2} \left(x - 2\right)$

#### Explanation:

y'=${e}^{- x} - x {e}^{- x}$

y"=$- {e}^{- x} - \left({e}^{- x} - x {e}^{- x}\right)$

= $x {e}^{- x} - 2 {e}^{- x}$

For inflection point make y''=0. which would give x=2

The inflection point would be $2 , 2 {e}^{- 2} + 3$

The slope at this point would be ${e}^{- 2} - 2 {e}^{- 2} = - {e}^{- 2}$

The tangent line would then be
$y - \left(2 {e}^{- 2} + 3\right) = - {e}^{- 2} \left(x - 2\right)$