Converting quadratic forms | Year 11

01 / LEARN

Understand the idea.

Three views of one function

The same quadratic has useful different forms. Expanded form shows its vertical-axis intercept; factored form shows real roots; vertex form shows the turning point and symmetry axis. Choose the form that directly exposes the feature you need.

A form change must preserve the entire function, including its leading coefficient. Factoring and completing the square rearrange its structure; they do not change its values or turn it into a new quadratic.

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

Conversion checks

Use distribution to expand, factorisation or known roots to form linear factors, and completing the square to obtain vertex form. Check conversions by expanding back to the same coefficients; checking only one numerical input cannot establish an identity.

A quadratic with negative discriminant has no real linear factors, though its expanded and vertex forms still exist. In that case, do not force a real factorisation. A positive or zero discriminant permits the corresponding real factors.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

One rule, three forms

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

Worked solution

1
Find the factor constants.
$(- 2) + (- 4) = - 6, (- 2) (- 4) = 8$
For a monic quadratic, the factor constants must reproduce the middle coefficient and constant term.
2
Write factored form.
$x^{2} - 6 x + 8 = (x - 2) (x - 4)$
Expanding the factors gives the required sum and product.
3
Complete the square.
$\begin{aligned} x^{2} - 6 x + 8 & = (x^{2} - 6 x + 9) - 1 \\ = {(x - 3)}^{2} - 1 \end{aligned}$
Half the linear coefficient is negative three; adding nine inside the square requires a compensating negative one outside.
4
Connect the forms’ features.
$x = 2, 4, V = (3, - 1)$
The roots from factored form are symmetric about the vertex input shown in square form.

Answer

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

EXAMPLE 02

Retain the coefficient

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

Worked solution

1
Expand the two brackets before distributing two.
$(x + 1) (x - 3) = x^{2} - 3 x + x - 3 = x^{2} - 2 x - 3$
Every term in the first bracket multiplies every term in the second.
2
Write the expanded quadratic.
$2 (x + 1) (x - 3) = 2 x^{2} - 4 x - 6$
The leading coefficient two multiplies all three resulting terms.
3
Complete the square without losing the coefficient.
$\begin{aligned} 2 x^{2} - 4 x - 6 & = 2 (x^{2} - 2 x) - 6 \\ = 2 [{(x - 1)}^{2} - 1] - 6 \\ = 2 {(x - 1)}^{2} - 8 \end{aligned}$
The compensation inside the bracket is multiplied by two before combining constants.
4
Check the final form by expansion.
$2 {(x - 1)}^{2} - 8 = 2 x^{2} - 4 x + 2 - 8 = 2 x^{2} - 4 x - 6$
The final form reproduces every coefficient of the expanded quadratic.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Factored form; roots -2 and 5.

Factored form exposes the zeros directly: $x + 2 = 0$ or $x - 5 = 0$ . The nonzero coefficient three does not change the roots, which are $- 2$ and $5$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Three views of one function

Conversion checks

See the method in action.

One rule, three forms

Find the factor constants.

Write factored form.

Complete the square.

Connect the forms’ features.

Retain the coefficient

Expand the two brackets before distributing two.

Write the expanded quadratic.

Complete the square without losing the coefficient.

Check the final form by expansion.

Try it yourself.

Check your understanding

Keep practising