Maximum and Minimum using Quadratic Functions and not too much algebra!

These slides were prepared for management mathematics students, some of whom had no algebra at the start of the course.

-4 -3 -2 -1 0 1 2 3 40123456789

10

y = x²

-4 -3 -2 -1 0 1 2 3 40

2

4

6

8

10

y = x²

-4 -3 -2 -1 0 1 2 3 402468

10

y = x²

Quadratic Functions The graphs of many equations are called functions.

In economics and finance we are often interested in where the minimum value (turning point) is (the X ordinate) and what the minimum value is (the Y ordinate).

If the graph is inverted then we are looking for the maximum value.

The basic equation for a quadratic function is y = x2. The minimum value is found at x = 0 and the minimum value is y = 0

-4 -3 -2 -1 0 1 2 3 40

1

2

3

4

5

6

7

8

9

10

y = x²

A negative in front of the x2 inverts the graph. The maximum value is found at x = 0 and the maximum value is y = 0.

-4 -3 -2 -1 0 1 2 3 4

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

y = -x²

A constant (number) added to the end of the equation shifts the graph up the Y axis. The minimum value is found at x = 0 and the minimum value is y = 4.

-4 -3 -2 -1 0 1 2 3 4

-6

-4

-2

0

2

4

6

y = x² - 4

A constant inserted inside the bracket slides the graph along the X axis in the Opposite direction. The minimum value is found at x = 2 and the minimum value is y = 0.

-2 -1 0 1 2 3 4 5 60

1

2

3

4

5

6

7

8

9

10

y = (x - 2)²

We can have combinations of these. The minimum value is found at x = 3 and the minimum value is y = 2

-7 -6 -5 -4 -3 -2 -1 0 1 2 3

-4

-2

0

2

4

6

8

y = (x + 3)² - 2

Quadratic equations do not always come in this convenient form. They could be:

y = 3x2 4x + 5

or y = (x + 4)(x -5)

By a process called “completing the square” these equations can be written in the form

a(x + d)2 + e

which is the form of the last graph in the last slide above.

To rewrite the equation in completed square form we use the theory that

ax2 + bx + c = a + c

which is equivalent to a(x + d)2 + e

where d =

and e = + c

e.g. If y = 3x2 4x + 5, write this in completed square form:

a = 3, b = 4, c = 5

substituting: d = = = or - 0.67

e = + 5 = + 5 = + 5 = 3.67

The equation can be written as

y = 3(x 0.67)2

The equation can be written asy = 3(x 0.67)2

Find the x value for which this is a minimum and state the minimum value.

This is a minimum at x = 0.67 (note opposite sign) and the minimum value is 3.67.

Some problems are already in completed square form. You do not need to recalculate d and e. Just read your answers off the given form:

e.g. if y = (x + 4)(x -5) can be written as (x 0.5)2 20.25, find the X value for which this is a minimum and state the minimum value.

This is a minimum at x = 0.5 and the minimum value is y = 20.25.

Some problems require a little algebra manipulation before you can calculate d and e.e.g. If P = Q(Q 6) find the value of Q for which P is a maximum and what is the maximum value of P?

Expand the bracket P = Q2 + 6QUse the formula to write in completed square form:

a = 1, b = 6, c = 0d = = = 3e = + c = + 0 = 9

so P = 1(Q 3)2 + 9The maximum value of P is at Q = 3 and it is P = $9