Find Turning Point by Completing the Square

In this revision note, you will learn the steps to find turning point by completing the square of Quadratic
Graphs.

3 Forms of Quadratic Expressions

A quadratic expression can be written in 3 main forms:

(i) General form: ax2 + bx + c

(ii) Factorised form: (x + p)(x +q)

(iii) “Completed square” form: a(x + h)2 + k

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completing the square, o level math notes

 

Completing the Square

Apply the following steps to “complete the square”.

Step 1:  

Observe the general form. If the coefficient of x2 is not 1, factorise the coefficient of x2 for the terms with x2 and x only.

 

Example 1

 

 2x2 + 16x – 1 = 2( x2 + 8x ) – 1

Step 2: Apply the identity: 

a2 + 2ab + b2 = (a + b)2

a2 + 2ab = (a + b)– b2

With reference to Example 1:

·    a2 is represented by x2,

·     a is represented by x,

·     2ab is represented by 8x, and

·     b is represented by 4. 

**In general, b is always half the coefficient of the term with x (not x2). 

  2x2 + 16x – 1 = 2( x2 + 8x ) – 1

= 2[(x + 4)2 – 42] – 1

This is the “completed square”.

Step 3:

Simplify the constants to obtain the “completed square” form: a(x + h)2 + k

 2x2 + 16x – 1 = 2( x2 + 8x ) – 1

= 2[(x + 4)2 – 42] – 1

= 2(x + 4)2 – 2(16) – 1

= 2(x + 4)2 – 33

 

Find the Turning Point by Completing the Square

Completing the Square Example 1

State the minimum value of = x2 − 4x + 10 and the corresponding value of x.

 

Solution

= x2 − 4x + 10

= (x − 2) 2 − (−2) 2 + 10

= (x− 2) 2 + 6

 

Since ( x − 2) 2 ≥ 0 for all real values of x, then ( x − 2) 2 + 6 ≥ 6 for all real values of x.

The minimum value of y is 6, when x = 2. (ans)

*Tip: To find the turning point from the “completed square” form, we let the expression inside the bracket be zero.

Note: Since the graph is a U-shaped graph, the turning point is a minimum point and the coordinates are (2 , 6).

 

Completing the Square Example 2

State the maximum value of = −2x 2 − 10x + 5 and the corresponding value of x.

 

Solution

y = −2x2 − 10x + 5

= −2 ( x + 5/2 )2 + 35/2

 

Since ( x + 5/2 )2 ≥ 0 for all real values of x, then −2 ( x + 5/2 )2 ≤ 0 for all real values of x.

And −2 ( x + 5/2 )2 + 35/2 ≤ 35/2

The maximum value of y is 35/2 , when x = − 5/2 . (ans)

Note: Since the graph is an inverted-U graph, the turning point is a maximum point and the coordinates are (− 5/2 , 35/2 ).

 

Summary of how to find a turning point by completing the square

 

In general, the coordinates of the turning point of a quadratic graph after completing the square, y = a(x + h)2 + k is always given by ( −h, k). This information is very useful for graph sketching.

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