Question

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

Answer 1

This is a quadratic function.

Explanation:

A function `f(x) = ax^2+bx+c` with `a != 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=-b/(2a)`.
(If you forget the veertex formula, use the derivative `f'(x) = 2ax+b` to find the critical number, `x=-b/(2a)`.)

Answer 2

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 `uuu`.

So the function is convex ( concave up ) for `{x in RR}`

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 `oo`

( 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(x-h)^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=2x^2+6x+5`

Bracket of the terms containing the variable:

`(2x^2+6x)+5`

Factor out the coefficient of `x^2`:

`2(x^2+6/2x) +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(x^2+6/2x+(3/2)^2)-2(3/2)^2+5`

Convert to the square of a binomial:

`2(x+3/2)^2-18/4+5`

`2(x+3/2)^2+1/2`

So minimum value `=k=1/2`

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