# How do you find the inflection points of the graph of the function: f(x) = (x+2) (x-4)^2?

$x = 2$

#### Explanation:

Given function,

$f \left(x\right) = \left(x + 2\right) {\left(x - 4\right)}^{2}$

$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 32$

$f ' \left(x\right) = 3 {x}^{2} - 12 x$

$f ' ' \left(x\right) = 6 x - 12$

For, inflection point $f ' ' \left(x\right) = 0$

$6 x - 12 = 0$

$x = 2$