# What are the points of inflection, if any, of f(x)=2x^3 + 8x^2 ?

Jan 28, 2017

At $x = - \frac{4}{3}$ there is point of inflection

#### Explanation:

Given -

$y = 2 {x}^{3} + 8 {x}^{2}$

To find the point of inflection put send derivative equal to zero and find the value of $x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} + 16 x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 12 x + 16$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0 \implies 12 x + 16 = 0$

$x = \frac{- 16}{12} = - \frac{4}{3}$

graph{2x^3+8x^2 [-10, 10, -5, 5]}