# How do you find points of inflection and determine the intervals of concavity given y=(3x+2)/(x-2)?

Nov 17, 2017

There are no points of inflection.
The interval of concavity is $\left(- \infty , 2\right)$ and the interval of convexity is $\left(2 , + \infty\right)$

#### Explanation:

Calculate the first derivative

$y = \frac{3 x + 2}{x - 2}$

This is a quotient $\frac{u}{v}$ and the derivative is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$u = 3 x + 2$, $\implies$, $u ' = 3$

$v = x - 2$, $\implies$, $v ' = 1$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(x - 2\right) - 1 \left(3 x + 2\right)}{x - 2} ^ 2 = \frac{3 x - 6 - 3 x - 2}{x - 2} ^ 2$

$= - \frac{8}{x - 2} ^ 2$

And the second derivative is

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(- 8 {\left(x - 2\right)}^{-} 2\right) ' = - 8 \cdot - \frac{2}{x - 2} ^ 3 = \frac{16}{x - 2} ^ 3$

The points of inflection are when $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0$

Here, the second derivative is $\ne 0$

So, there are no points of inflection

Build a sign chart for the concavity

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , 2\right)$$\textcolor{w h i t e}{a a a a}$$\left(2 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$S i g n \frac{{d}^{2} y}{\mathrm{dx}} ^ 2$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a}$$\cup$

The interval of concavity is $\left(- \infty , 2\right)$ and the interval of convexity is $\left(2 , + \infty\right)$

graph{(3x+2)/(x-2) [-41.1, 41.08, -20.56, 20.56]}