Lesson HOW TO complete the square to find the minimum/maximum of a quadratic function

This Lesson (HOW TO complete the square to find the minimum/maximum of a quadratic function) was created by by ikleyn(52778)

: View Source, Show
About ikleyn:

How to complete the square to find the minimum/maximum of a quadratic function

Assume that you are given a quadratic function of a general form

y = ax%5E2+%2B+bx+%2B+c

.

You know that the plot of this function is a parabola.
You also know that if the coefficient "a" at x%5E2

is positive, then the parabola has a minimum and the parabola is opened upward.
You also know that, in opposite, if the coefficient "a" at x%5E2

is negative, then the parabola has a maximum and the parabola is opened downward.

The question is: how to find that minimum or maximum?

In other words, how to find the location of the minimum/maximum in the x-axis and what is the value of the minimum/maximum?

In this lesson I will show you how to do it.

Apply "completing the square". It works in this way

y =  =  + c =  + c = 

  =  -   =    -   
                                                  ---------------    ----------------
                                                  the full square    the new constant 
                                                     term                term

Thus completing the square transforms the appearance of the quadratic function without changing its values.
It makes visible (extracts) the full square term and creates the new constant term.


The full square term has always the same sign as the coefficient "a" has.


If a > 0, then the full square term is always non-negative.
   It has the zero value at  x = , and it provides the minimal value to the quadratic function   = ,
   which is actually the value of the quadratic function at this value of x = .


If a < 0, then the full square term is always non-positive.
   It has the zero value at  x = , and it provides the maximal value to the quadratic function   = ,
   which is actually the value of the quadratic function at this value of x = .

Thus the rule for finding the minimum/maximum of a quadratic function f(x) = ax%5E2+%2B+bx+%2B+c

    If a > 0, the function has a minimum.
    If a < 0, the function has a maximum.
    The function reaches the minimum/maximum at x = .
    To find the value of the minimum/maximum, substitute the value x = into the quadratic function.

Problem 1

Find the minimum of the quadratic function f(x) = x%5E2+-+3x+%2B+2

Solution 1. (Applying the rule)

The minimum reaches at x = -%28-3%29%2F2

.
The minimum value is f(3/2) = %283%2F2%29%5E2+-+3%2A%283%2F2%29+%2B+2

.

Solution 2. (Completing the square from the scratch)

x%5E2+-+3x+%2B+2

.

The minimum reaches at x = 3%2F2

.
The minimum value is -1%2F4

graph%28+250%2C+250%2C+-3.5%2C+5.5%2C+-2.5%2C+6.5%2C%0D%0A++++++++++x%5E2+-+3x+%2B2%0D%0A%29

Plot y = x%5E2+-+3x+%2B2

Problem 2

Find the maximum of the quadratic function f(x) = -x%5E2+%2B+2x+%2B+3

Solution 1. (Applying the rule)

The maximum reaches at x = -2%2F%282%2A%28-1%29%29

= 1.
The maximum value is f(1) = -%28-1%29%5E2+%2B+2%2A1+%2B+3

= 4.

Solution 2. (Completing the square from the scratch)

-x%5E2+%2B+2x+%2B+3

.

The minimum reaches at x =

.
The minimum value is 4.

Plot y = -x%5E2+%2B+2x+%2B3

Problem 3

Prove that polynomial p(x) = x%5E2+-4x+%2B+5

doesn't have real roots.

Solution

One way to prove it is to calculate the discriminant and check that it is lesser than zero.
The discriminant of this polynomial is d = %28-4%29%5E2+-+4%2A5

= 16 - 20 = -4.
It is really lesser than zero, so the polynomial p(x) really doesn't have real roots.

But there is another way to prove it.
Complete the square, and you will get

p(x) = x%5E2+-4x+%2B+5

.

So you have the full square %28x-2%29%5E2

which is always positive, and the positive addition +1, which makes the polynomial positive for all x over the entire number line.
It is clear now that the polynomial doesn't have real roots.

Problem 4

Prove that polynomial p(x) = x%5E4+-6x%5E2+%2B+11

doesn't have real roots.

Solution

I will prove it by completing the square, again.

p(x) = x%5E4+-6x%5E2+%2B+11

.

So you have the full square %28x%5E2-3%29%5E2

which is always positive, and the positive addition +2, which makes the polynomial positive for all x over the entire number line.
It is clear now that the polynomial doesn't have real roots.

Problem 5

Prove that polynomial p(x) = x%5E6+-10x%5E3+%2B+27

doesn't have real roots.

Solution

I will prove it by completing the square, again.

p(x) = x%5E6+-10x%5E3+%2B+29

.

So you have the full square %28x%5E3-5%29%5E2

which is always positive, and the positive addition +2, which makes the polynomial positive for all x over the entire number line.
It is clear now that the polynomial doesn't have real roots.

My other lessons in this site on finding the maximum/minimum of a quadratic function are
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
    - A rectangle with a given perimeter which has the maximal area is a square
    - A farmer planning to fence a rectangular garden to enclose the maximal area
    - A farmer planning to fence a rectangular area along the river to enclose the maximal area
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Finding the maximum area of the window of a special form
    - Using quadratic functions to solve problems on maximizing revenue/profit
    - When is the best time to sell a pig ?
    - Find the point on a given straight line closest to a given point in the plane
    - Minimal distance between sailing ships in a sea
    - Advanced lesson on finding minima of (x+1)(x+2)(x+3)(x+4)
    - OVERVIEW of lessons on finding the maximum/minimum of a quadratic function

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

This lesson has been accessed 13230 times.