Quadratic Relations of the Form y = ax² + bx + c · Quadratic Expressions and Equations

Completing the Square: From Standard Form to Vertex Form

3.5About 75 minutes

Learning goals

I can explain completing the square with algebra tiles.
I can rewrite y = x² + bx + c in vertex form by completing the square.
I can complete the square when a ≠ 1, including when a is negative.
I can state the vertex and the maximum or minimum value from the vertex form.

Words to know

completing the square
Rewriting a quadratic expression so that it contains a perfect square trinomial, to change y = ax² + bx + c into y = a(x − h)² + k.

vertex form
A quadratic relation written as y = a(x − h)² + k. Its vertex is (h, k).

perfect square trinomial
A trinomial such as x² + 6x + 9 that equals the square of a binomial, (x + 3)².

equivalent expressions
Expressions that have the same value for every value of the variable.

minimum / maximum value
The y-coordinate of the vertex: the lowest value of y when a > 0, or the highest value when a < 0.

Why change forms?

In Quadratic Relations and Their Graphs you saw that vertex form, y = a(x − h)² + k, shows the vertex (h, k) at a glance. But many relations arrive in standard form, like y = 2x² − 12x + 7. Where is its vertex? You cannot tell by looking.

Completing the square is a way to rewrite y = ax² + bx + c as y = a(x − h)² + k. The name is literal: you will complete an incomplete square of algebra tiles.

Algebra tiles for x² + 6x. Half of the x-tiles go on each side of the x² tile. The square is complete only when you add 9 unit tiles in the corner: 9 = (6 ÷ 2)².

The key idea: add and subtract (b ÷ 2)²

For x² + bx, the number that makes a perfect square is (b ÷ 2)², because

(x + b/2)² = x² + bx + (b/2)².

You cannot just add a number to an expression; that would change its value. So you add it and subtract it in the same step. Adding and subtracting the same number is adding zero, so the expression stays equivalent.

In this course, b ÷ 2 will always be an integer after any common factor is taken out (no fractions).

Worked example

Example 1: a = 1

Write y = x² + 6x + 5 in vertex form, and state the vertex.

Half of 6 is 3, and 3² = 9.
Add and subtract 9: y = x² + 6x + 9 − 9 + 5.
The first three terms are a perfect square: y = (x + 3)² − 9 + 5.
Simplify: y = (x + 3)² − 4.
Check by expanding: x² + 6x + 9 − 4 = x² + 6x + 5 ✓.

y = (x + 3)² − 4, vertex (−3, −4), a minimum because a = 1 > 0.

When a is not 1

Factor a out of the first two terms only: y = a(x² + __x) + c.
Inside the bracket, add and subtract (half the x-coefficient)².
Move the subtracted number out of the bracket. It must be multiplied by a when it leaves.
Write the perfect square and simplify the constants.

Completing the square when a = 2. The most common mistake is in step 3: the −9 is inside a bracket multiplied by 2, so it becomes −18 when it leaves.

Worked example

Example 2: a = 2

Write y = 2x² − 12x + 7 in vertex form.

Factor 2 from the first two terms: y = 2(x² − 6x) + 7.
Half of −6 is −3, and (−3)² = 9: y = 2(x² − 6x + 9 − 9) + 7.
Move −9 out of the bracket, multiplying by 2: y = 2(x² − 6x + 9) − 18 + 7.
y = 2(x − 3)² − 11.
Check: 2(x² − 6x + 9) − 11 = 2x² − 12x + 18 − 11 = 2x² − 12x + 7 ✓.

y = 2(x − 3)² − 11; vertex (3, −11), minimum value −11.

y = 2x² − 12x + 7 and y = 2(x − 3)² − 11 are the same parabola. Completing the square found the vertex (3, −11). The points (0, 7) and (6, 7) are mirror images across x = 3.

Worked example

Example 3: A negative a

Write y = −x² + 4x + 1 in vertex form. What is the maximum value?

Factor −1 from the first two terms: y = −(x² − 4x) + 1.
Half of −4 is −2; (−2)² = 4: y = −(x² − 4x + 4 − 4) + 1.
Move −4 out, multiplying by −1: y = −(x² − 4x + 4) + 4 + 1.
y = −(x − 2)² + 5.

y = −(x − 2)² + 5. It opens down, so the maximum value is 5, at x = 2.

Worked example

Example 4: Getting ready to sketch

Write y = 3x² − 12x + 1 in vertex form, then describe how it relates to y = x².

y = 3(x² − 4x) + 1.
y = 3(x² − 4x + 4 − 4) + 1 = 3(x² − 4x + 4) − 12 + 1.
y = 3(x − 2)² − 11.
Transformations of y = x²: a vertical stretch by a factor of 3, a translation 2 units right and 11 units down.

y = 3(x − 2)² − 11, vertex (2, −11). You will sketch relations like this in the next lesson.

Watch out

Factor a out of the x² and x terms only, not the constant.
Use half of the coefficient inside the bracket, after factoring out a.
When the subtracted number leaves the bracket, multiply it by a.
Check by expanding your vertex form. It must give back the original.

Hands-on activity

Complete the square with paper tiles

You need: paper tiles from earlier lessons (squares for x², long strips for x, small squares for 1).

Lay out 1 x²-tile and 8 x-tiles.
Put half the x-tiles on the right of the x² tile and half underneath, standing and lying as in the lesson picture.
Count how many unit tiles fill the empty corner.
Write x² + 8x + __ = (x + __)².
Repeat with 1 x²-tile and 4 x-tiles, then 10 x-tiles. What is the pattern?

Model answers: x² + 8x + 16 = (x + 4)²; x² + 4x + 4 = (x + 2)²; x² + 10x + 25 = (x + 5)². The corner always needs (half the number of x-tiles)² unit tiles.

Just for fun

Completing the square is one of the few times in math when adding zero is a brilliant move. You add 9 and subtract 9, change nothing at all, and suddenly the vertex appears. It's like tidying your room by moving everything one centimetre to the left, except it actually works.

Your turn

What number completes the square x² + 10x + __? Write the result as a square.

Write y = x² + 8x + 3 in vertex form and give the vertex.

Write y = x² − 6x + 11 in vertex form and give the vertex.

Write y = x² + 2x − 8 in vertex form and give the vertex.

Write y = 2x² + 8x + 5 in vertex form.

Write y = −x² − 6x + 2 in vertex form and state the maximum value.

Write y = 3x² − 18x + 20 in vertex form.

More practice: warm-up

What number completes x² − 14x + __?

Write x² + 12x + 36 as a square.

Write y = x² + 4x + 1 in vertex form.

Write y = x² − 2x + 5 in vertex form.

State the vertex of y = (x − 7)² + 3.

More practice: on your own

Write y = 4x² + 16x + 9 in vertex form and give the minimum value.

Write y = −2x² + 12x − 10 in vertex form and give the maximum value.

Write y = 5x² − 10x + 1 in vertex form.

Use completing the square to show that y = x² + 6x + 10 has no zeros.

A student wrote 2x² + 12x + 1 = 2(x + 3)² − 8. Find and fix the mistake.

Challenge

Complete the square for y = x² + 2bx + c, where b and c are constants.

Write y = −3x² − 24x − 50 in vertex form. Does the graph have any zeros? Explain.

Solve x² + 6x − 16 = 0 by completing the square (not by factoring).

Check your understanding

Quiz

Which number completes the square x² + 12x + __?

Write y = x² − 4x + 7 in vertex form.

What is the vertex of y = x² + 10x + 21?

Which is the vertex form of y = 2x² − 8x + 3?

What is the maximum value of y = −x² + 2x + 5?

When completing the square, why do we subtract the same number we add?

For parents and teachers

Big idea: Expectation 3.5 asks students to express y = ax² + bx + c in the form y = a(x − h)² + k by completing the square, in situations involving no fractions, using tools such as algebra tiles and diagrams. The vertex form then gives the vertex and the maximum or minimum value.

How to help: Ask your teen to explain the tile picture: why does the corner need (b ÷ 2)² tiles? Then have them check every answer by expanding the vertex form. If the check fails, the error is almost always in moving the subtracted number out of the bracket.

Common difficulties:

Factoring a out of the constant term as well.
Forgetting to multiply by a when the subtracted number leaves the bracket.
Sign of h: (x + 3)² means h = −3, so the vertex is at x = −3.