# How do you find all points of inflection given y=x^3-2x^2+1?

Apr 17, 2017

$\left(\frac{2}{3} , 0.41\right)$

#### Explanation:

Inflection points occur when the second derivative is equal to $0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 4 x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 x - 4$

Let $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$0 = 6 x - 4$

$6 x = 4$

$x = \frac{4}{6} = \frac{2}{3}$

Solve for y-cordinate,

$y = {\left(\frac{2}{3}\right)}^{3} - 2 {\left(\frac{2}{3}\right)}^{2} + 1$

$y = \frac{8}{27} - 2 \left(\frac{4}{9}\right) + 1$

$y = \frac{8}{27} - \frac{8}{9} + 1$

$y = \frac{11}{27}$ or $0.41$

Therefore the point of inflection for the function $y = {x}^{3} - 2 {x}^{2} + 1$ is $\left(\frac{2}{3} , 0.41\right)$