A2.8 Complete the square to convert standard form to vertex form

Ontario Grade 11 Mathematics

Quadratic Functions

Convert a quadratic from standard form to vertex form

A quadratic can be written in different forms. Standard form makes the coefficients easy to see. Vertex form makes the vertex easy to see. Completing the square is a method for changing standard form into vertex form without changing the quadratic. This lesson reviews the needed algebra, explains the square-making step, and works through one conversion.

What you will learn

Recognize standard form and vertex form for a quadratic.
Complete a square by using half the coefficient of the linear term.
Convert a quadratic with a leading coefficient other than 1 into vertex form.
Check a conversion by expanding the vertex form.

1. Prerequisite bridge: perfect-square trinomials

A quadratic expression has a squared variable, such as

x^2

. A coefficient is a number multiplying a variable or expression. In standard form, a quadratic is written as

ax^2+bx+c

, where

a

,

b

, and

c

are numbers and

a\ne 0

. The term

bx

is the linear term because the variable has exponent 1.

You may already know how to expand two matching brackets. For example, multiplying

(x+4)(x+4)

gives

x^2+8x+16

. This result is a perfect-square trinomial: three terms that come from squaring a binomial. A binomial is an expression with two terms.

The key pattern is that the number beside

x

inside the brackets is half the coefficient of

x

in the trinomial. Half of 8 is 4, and the last term is

4^2=16

. Completing the square uses this pattern in reverse. It creates a perfect-square trinomial from an expression that does not yet have the needed last term.

x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2

Standard form is $ax^2+bx+c$ , with $a\ne 0$ .
A perfect-square trinomial can be written as the square of a binomial.
For $x^2+bx$ , the number to add to make a perfect square is $\left(\frac{b}{2}\right)^2$ .

2. What completing the square does

Vertex form is written as

a(x-h)^2+k

. The vertex is the turning point of the graph of a quadratic. In vertex form, its coordinates are

(h,k)

. For this lesson, the goal is to rewrite the same quadratic in that form.

Think of

x^2+6x

as a nearly complete square. Half of 6 is 3, and

3^2=9

. Adding 9 makes the expression a perfect square:

x^2+6x+9=(x+3)^2

. But adding 9 changes the value of an expression. To keep the original quadratic unchanged, subtract 9 as well. The added and subtracted amounts cancel.

When the coefficient of

x^2

is not 1, first factor that coefficient from the

x^2

and

x

terms. Then complete the square inside the brackets. The added amount inside the brackets is multiplied by the outside coefficient, so keep track of that change. Any constant already outside the brackets stays in the expression.

a(x^2+bx)+c=a\left[\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2\right]+c

Take half the coefficient of $x$ inside the group, then square that result.
Add and subtract the same amount to keep the expression equal to its original value.
If a number is factored outside the brackets, account for it when balancing the expression.

3. A reliable conversion process

Start with the standard-form quadratic and group the squared and linear terms. If the leading coefficient is not 1, factor it from those two terms only. Do not factor the constant into the group unless you also account for it correctly.

Next, use the coefficient of

x

inside the brackets. Divide it by 2 and square the result. Add that square inside the brackets, and subtract the same square there so the expression remains equal. Because the group may have an outside coefficient, the subtraction inside is also multiplied by that coefficient.

Rewrite the three terms inside the brackets as a squared binomial. Then simplify the remaining constants. The result should have a squared binomial multiplied by the original leading coefficient, followed by a constant. That is vertex form.

A quick check is to expand the squared binomial and combine the constants. The result should return to the original standard form. This check can catch a sign error or a missed multiplication by the outside coefficient.

ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}

Factor the leading coefficient from the first two terms when needed.
Balance the added square and simplify all constants.
Expand the final form to check that it matches the starting quadratic.

4. Use the form to read the vertex

The completed form displays the squared binomial and the remaining constant. Compare it with

a(x-h)^2+k

. Be careful with signs: the value inside the brackets is written as

x-h

. For example,

(x+3)^2

is the same as

(x-(-3))^2

, so its horizontal coordinate is negative 3.

The value outside the squared binomial is the vertical coordinate. The coefficient multiplying the square remains the leading coefficient. Completing the square does not change the quadratic; it only changes how the same expression is written.

This conversion is useful when a question asks for vertex form or for the vertex represented by that form. The method in this lesson is limited to converting a quadratic from standard form to vertex form. Keep the algebra focused on making and balancing the perfect-square trinomial.

a(x-h)^2+k\quad\longrightarrow\quad(h,k)

In $a(x-h)^2+k$ , the vertex is $(h,k)$ .
Read the horizontal coordinate with care because the bracket shows $x-h$ .
The coefficient of the squared bracket remains $a$ .

Worked example

Convert standard form to vertex form

Convert

2x^2+12x+11

to vertex form, then state the vertex.

Group and factor
Factor 2 from the squared and linear terms. Keep the constant 11 outside the brackets.
$2(x^2+6x)+11$
Make a perfect square
Inside the brackets, the coefficient of $x$ is 6. Half of 6 is 3, and its square is 9. Add and subtract 9 inside the brackets so the value stays unchanged.
$2(x^2+6x+9-9)+11$
Rewrite the square
The first three terms inside the brackets form the square $(x+3)^2$ . Take care: the subtracted 9 is still multiplied by the outside factor 2.
$2\bigl((x+3)^2-9\bigr)+11$
Simplify and identify
Multiply 2 by negative 9, then add 11. The result is vertex form. Since the bracket is $(x-(-3))^2$ and the remaining constant is negative 7, the vertex is $(-3,-7)$ .
$2(x+3)^2-7$

Answer: The vertex form is

2(x+3)^2-7

. The vertex is

(-3,-7)

.

Check: Expanding gives

2(x^2+6x+9)-7=2x^2+12x+18-7=2x^2+12x+11

, which matches the original expression.

Common mistakes and how to avoid them

Using the coefficient of

x

without first factoring out the leading coefficient.

Correction: If the leading coefficient is not 1, factor it from the squared and linear terms first. Then use the coefficient of

x

inside the brackets.

Adding the square but not subtracting it.

Correction: Add and subtract the same amount inside the brackets. The two changes cancel, so the expression keeps its original value.

Forgetting that the outside coefficient multiplies the subtracted square.

Correction: Simplify the subtracted term together with the outside coefficient before combining it with the constant.

Reading the sign inside the bracket as the vertex's horizontal coordinate.

Correction: Compare the bracket with

x-h

. For example,

(x+3)^2

corresponds to

h=-3

.

Lesson summary

Standard form is $ax^2+bx+c$ ; vertex form is $a(x-h)^2+k$ .
To complete the square, take half the coefficient of $x$ in the grouped expression and square it.
Balance the added square by subtracting the same amount, while accounting for any outside coefficient.
Check the conversion by expanding the result back to standard form.

Check your understanding

Question 1

Which number completes

x^2+10x

as a perfect-square trinomial?

5
10
25
100

Show answer and explanation

25

Half of 10 is 5, and

5^2=25

. Thus,

x^2+10x+25=(x+5)^2

.

Question 2

Which expression is vertex form for

x^2-8x+3

?

$(x-4)^2-13$
$(x+4)^2-13$
$(x-4)^2+3$
$(x-8)^2-61$

Show answer and explanation

(x-4)^2-13

Half of negative 8 is negative 4, and its square is 16. So

x^2-8x+3=(x-4)^2-16+3=(x-4)^2-13

.

Question 3

When rewriting

3(x^2+4x)+2

, what must be included to complete the square inside the brackets?

Add and subtract 2 inside the brackets.
Add and subtract 4 inside the brackets.
Add and subtract 8 inside the brackets.
Add and subtract 16 inside the brackets.

Show answer and explanation

Add and subtract 4 inside the brackets.

Half of the coefficient 4 is 2, and

2^2=4

. Add and subtract 4 inside the brackets; the outside factor 3 still multiplies both terms.

Key terms

Quadratic: An expression or function whose highest power of the variable is 2.
Standard form: The form $ax^2+bx+c$ for a quadratic, where $a\ne 0$ .
Vertex form: The form $a(x-h)^2+k$ , which shows the vertex as $(h,k)$ .
Perfect-square trinomial: A three-term expression that can be written as the square of a binomial.
Coefficient: A number that multiplies a variable or expression.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation A2.8. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

A2.8 · Complete the square to convert standard form to vertex form

Ontario Grade 11 Mathematics

Convert a quadratic from standard form to vertex form

What you will learn

1. Prerequisite bridge: perfect-square trinomials

2. What completing the square does

3. A reliable conversion process

4. Use the form to read the vertex

Convert standard form to vertex form

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MCF3M

About this lesson