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

2 Answers
Dec 29, 2017

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)#.)

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 #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]}