# How do you find the maximum, minimum and inflection points and concavity for the function f(x) = 2x^2 + 6x + 5?

Dec 29, 2017

#### Explanation:

A function $f \left(x\right) = a {x}^{2} + b x + c$ with $a \ne 0$ has a graph that is a parabola.

It opens upward and is concave up if $a > 0$ and it opens downward and is concave down if $a < 0$. It has no inflection points.

The function has a minimum if $a > 0$ and a maximum if $a < 0$.

The minimum or maximum occurs at the vertex which is at $x = - \frac{b}{2 a}$.
(If you forget the veertex formula, use the derivative $f ' \left(x\right) = 2 a x + b$ to find the critical number, $x = - \frac{b}{2 a}$.)

Dec 29, 2017

See below.

#### Explanation:

This is just a standard quadratic. You do not need calculus to solve this problem. The coefficient of ${x}^{2}$ is positive, so the parabola is this way up $\bigcup$.

So the function is convex ( concave up ) for $\left\{x \in \mathbb{R}\right\}$

There are no inflection points, because the function is convex for the entire domain. Infection points only occur when the concavity changes.

Maximum value is $\infty$

( this can be deduced by the fact that the function is concave up for entire domain )

To find the minimum value, we need to arrange the function in form:

$y = a {\left(x - h\right)}^{2} + k$

Where a is the coefficient of ${x}^{2}$, h is the axis of symmetry and k is the maximum/minimum value of the function.

$y = 2 {x}^{2} + 6 x + 5$

Bracket of the terms containing the variable:

$\left(2 {x}^{2} + 6 x\right) + 5$

Factor out the coefficient of ${x}^{2}$:

$2 \left({x}^{2} + \frac{6}{2} x\right) + 5$

Add the square of half the coefficient of $x$ inside the brackets and subtract the square of half the coefficient of $x$ outside of the brackets:

$2 \left({x}^{2} + \frac{6}{2} x + {\left(\frac{3}{2}\right)}^{2}\right) - 2 {\left(\frac{3}{2}\right)}^{2} + 5$

Convert to the square of a binomial:

$2 {\left(x + \frac{3}{2}\right)}^{2} - \frac{18}{4} + 5$

$2 {\left(x + \frac{3}{2}\right)}^{2} + \frac{1}{2}$

So minimum value $= k = \frac{1}{2}$

graph{y=2x^2+6x+5 [-5, 2, -5, 7]}