Converting quadratic forms | Year 11

01 / LEARN

The idea, step by step

Three views of one function

Expanded form exposes the y-intercept. Factored form exposes the roots. Vertex form exposes the turning point and axis. All three describe the same function when the conversions are correct.

Choose the form that makes the requested feature easy to see. Expanding before finding roots may waste effort, whereas completing the square is useful for finding a minimum.

a x^{2} + b x + c = a (x - r) (x - s) = a {(x - h)}^{2} + k

Conversion checks

Distribute to get expanded form. Factor or solve for roots to get factored form. Complete the square to get vertex form. Keep the leading coefficient at every stage.

A quadratic with negative discriminant cannot be written using real linear factors. Expanded and vertex forms still work, so do not force a real factorisation that does not exist.

A common trap

Dropping the leading coefficient in a form change.

Instead: The coefficient of x squared stays a. Keep it multiplying the squared bracket or factors.

02 / WORKED EXAMPLES

See the thinking

Example 1

One rule, three forms

Convert $x^{2} - 6 x + 8$ to factored and vertex forms.

A pair with sum -6 and product 8 is -2,-4.
Complete $x^{2} - 6 x + 9 - 1$ .

(x - 2) (x - 4) = {(x - 3)}^{2} - 1

Example 2

Retain the coefficient

Expand $2 (x + 1) (x - 3)$ , then find vertex form.

The bracket product is $x^{2} - 2 x - 3$ .
Distribute 2 to obtain $2 x^{2} - 4 x - 6$ .
Complete the square inside $2 (x^{2} - 2 x) - 6$ .

2 x^{2} - 4 x - 6 = 2 {(x - 1)}^{2} - 8

Check your understanding

Which form directly gives the roots of $y = 3 (x + 2) (x - 5)$ ?

Show answer

Factored form; roots -2 and 5.

Each factor is zero at one root.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Expand $(x - 4) (x + 2)$ .
2 marks
Factorise $x^{2} + 2 x - 8$ .
2 marks
Write $x^{2} + 2 x - 8$ in vertex form.
2 marks
Convert $- 2 {(x - 3)}^{2} + 8$ to expanded/factored forms.
3 marks
Roots are -1 and 3 and the graph passes through (0,-6). Find the quadratic.
3 marks
Can $x^{2} + 4$ factorise into real linear factors? Explain.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Expand $(x - 4) (x + 2)$ .
2 marks
Multiply and collect the middle terms.
$x^{2} - 2 x - 8$
Factorise $x^{2} + 2 x - 8$ .
2 marks
Use product -8 and sum 2.
$(x + 4) (x - 2)$
Write $x^{2} + 2 x - 8$ in vertex form.
2 marks
Add/subtract 1.
${(x + 1)}^{2} - 9$
Convert $- 2 {(x - 3)}^{2} + 8$ to expanded/factored forms.
3 marks
Expand carefully.
Solve ${(x - 3)}^{2} = 4$ for roots 1,5.
$- 2 x^{2} + 12 x - 10 = - 2 (x - 1) (x - 5)$
Roots are -1 and 3 and the graph passes through (0,-6). Find the quadratic.
3 marks
Use $y = a (x + 1) (x - 3)$ .
Substitute (0,-6) to get a=2.
$y = 2 (x + 1) (x - 3)$
Can $x^{2} + 4$ factorise into real linear factors? Explain.
2 marks
Its discriminant is -16, so there are no real roots.
A real linear factor would supply a real root.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

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