Let us analyze the sign of f' and hence determine any maximum or minimum point and the intervals of increase and decrease. f '(x) is positive if

2 a x + b > 0

add -b to both sides of the inequality to obtain
2 a x > -b

We now need to consider two cases and continue solving the inequality above.

case 1: a > 0

We divide both sides of the inequality by 2 a and obtain

x > - b / 2a

We now use a table to analyze the sign of f ' and whether f is increasing over a given interval.

The quadratic function with a > 0 has a minimum point at (-b/2a , f(-b/2a)) and the function is decreasing on the interval (-infinity , -b / 2a) and increasing over the interval (-b / 2a , + infinity).

case 2: a < 0

We divide both sides of the inequality by 2 a but because a is less than 0, we need to change the symbol of inequality

x < - b / 2a

We now analyze the sign of f ' using the table below

The quadratic function with a < 0 has a maximum point at (-b/2a , f(-b/2a)) and the function is increasing on the interval (-infinity , -b / 2a) and decreasing over the interval (-b / 2a , + infinity).

B - Vertex form.

Quadratic functions in their vertex form are written as

f(x) = a (x - h)^{ 2} + k

where a, h and k are real numbers with a not equal to zero.

The first derivative of f is given by

f '(x) = 2 a (x - h)

We analyze the sign of f' using a table. f '(x) is positive if

a (x - h) > 0

We need to consider two cases again and continue solving the inequality above.

case 1: a > 0

We divide both sides of the inequality by a and solve the inequality

x > h

The table below is used to analyze the sign of f '.

The quadratic function with a > 0 has a minimum at the point (h , k) and it is decreasing on the interval (-infinity , h) and increasing over the interval (h , + infinity).

case 2: a < 0

We divide both sides of the inequality by a but we need to change the symbol of inequality because a is less than 0.

x < h

We analyze the sign of f ' using the table below

The quadratic function with a < 0 has a maximum point at (h , k) and the function is increasing on the interval (-infinity , h) and decreasing over the interval (h , + infinity).

Example 1: Find the extremum (minimum or maximum) of the quadratic function f given by

f(x) = 2 x^{ 2} - 8 x + 1

Solution to Example 1.

We first find the derivative

f '(x) = 4 x - 8

f '(x) changes sign at x = 8 / 4 = 2. The leading coefficient a is positive hence f has a minimum at (2 , f(2)) = (2 , -7) and f is decreasing on (-infinity , 2) and increasing on (2 , + infinity). See graph below to confirm the result obtained by calculations.

Example 2: Find the extremum (minimum or maximum) of the quadratic function f given by

f(x) = - (x + 3)^{ 2} + 1

Solution to Example 2.

The derivative is given by

f '(x) = - 2 (x + 3)

f '(x) changes sign at x = -3 . The leading coefficient a is negative hence f has a maximum at (-3 , 1) and f is increasing on (-infinity , -3) and decreasing on (-3 , + infinity). See graph below of f below.

Exercises on Properties of Quadratic Functions.

For each quadratic function below find the extremum (minimum or maximum), the interval of increase and the interval of decrease.

a) f(x) = x^{ 2} + 6 x

b) f(x) = -x^{ 2} - 2 x + 3

c) f(x) = x^{ 2} - 5

d) f(x) = -(x - 4)^{ 2} + 2

e) f(x) = -x^{ 2}

Answers to Above Exercises.

a) minimum at (-3 , -9)
decreasing on (-infinity , -3)
increasing on (-3 , + infinity)

b) maximum at (-1 , 4)
increasing on (-infinity , -1)
decreasing on (-1 , + infinity)

c) minimum at (0 , -5)
decreasing on (-infinity , 0)
increasing on (0 , + infinity)

d) maximum at (-4 , 2)
increasing on (-infinity , -4)
decreasing on (-4 , + infinity)

e) maximum at (0 , 0)
increasing on (-infinity , 0)
decreasing on (0 , + infinity)