Expanded, factored and vertex forms | Year 10

01 / LEARN

The idea, step by step

One function, three useful views

Expanded form makes the y-intercept easy to read. Factored form makes the roots easy to read. Vertex form makes the turning point and direction easy to read. These forms describe the same function when correctly converted.

The leading coefficient a controls the direction and vertical scale in all three forms.

Here $a \neq 0$ . A factored form using real root factors exists only when the quadratic has real roots. If its discriminant is negative, use expanded or vertex form over the real numbers instead.

y = a x^{2} + b x + c = a (x - r_{1}) (x - r_{2}) = a {(x - h)}^{2} + k

Choose the form before doing extra work

If the question gives two roots, start with a times the two root factors and use a further point to find a. If it gives a vertex, begin with vertex form.

An expanded answer is not automatically better. State the form that answers the question most clearly, then expand or complete the square only when needed.

A common trap

Reading the root of the factor x+3 as positive 3.

Instead: A root makes the factor zero: x+3=0 gives x=−3.

02 / WORKED EXAMPLES

See the thinking

Example 1

Connect all three forms

Write $y = 2 x^{2} - 8 x + 6$ in factored and vertex forms.

Factorise: $2 (x^{2} - 4 x + 3) = 2 (x - 1) (x - 3)$ .
Complete the square: $2 [{(x - 2)}^{2} - 4] + 6 = 2 {(x - 2)}^{2} - 2$ .
The roots are $1, 3$ and the vertex is $(2, - 2)$ .

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

Example 2

Build a model from roots

A parabola has roots $- 2$ and $4$ and passes through $(0, - 16)$ . Find its rule.

Write $y = a (x + 2) (x - 4)$ .
Substitute $(0, - 16)$ : $- 16 = - 8 a$ , so $a = 2$ .
Expand if required: $y = 2 x^{2} - 4 x - 16$ .

y = 2 (x + 2) (x - 4)

Check your understanding

Which form shows the vertex directly: $x^{2} - 6 x + 5$ or ${(x - 3)}^{2} - 4$ ?

Show answer

${(x - 3)}^{2} - 4$ , giving vertex $(3, - 4)$ .

Both expressions are equal, but square form reveals the vertex.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Expand $y = (x - 2) (x + 5)$ .
2 marks
Write $y = x^{2} - 4 x - 5$ in factored form.
2 marks
Write $y = x^{2} + 4 x + 1$ in vertex form.
2 marks
Write $y = - 2 {(x - 1)}^{2} + 8$ in expanded form.
2 marks
A parabola has roots 1 and 5 and passes through $(0, 10)$ . Find its rule.
3 marks
A parabola has vertex $(2, - 3)$ and passes through $(0, 5)$ . Find its rule.
3 marks

Read the worked solutions

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

Expand $y = (x - 2) (x + 5)$ .
2 marks
The middle terms are $5 x - 2 x = 3 x$ .
$y = x^{2} + 3 x - 10$
Write $y = x^{2} - 4 x - 5$ in factored form.
2 marks
The factor pair is $- 5, 1$ .
$y = (x - 5) (x + 1)$
Write $y = x^{2} + 4 x + 1$ in vertex form.
2 marks
Add and subtract $4$ .
$y = {(x + 2)}^{2} - 3$
Write $y = - 2 {(x - 1)}^{2} + 8$ in expanded form.
2 marks
Expand the square before multiplying by $- 2$ .
$y = - 2 x^{2} + 4 x + 6$
A parabola has roots 1 and 5 and passes through $(0, 10)$ . Find its rule.
3 marks
$y = a (x - 1) (x - 5)$ ; $10 = 5 a$ , so $a = 2$ .
$y = 2 (x - 1) (x - 5)$
A parabola has vertex $(2, - 3)$ and passes through $(0, 5)$ . Find its rule.
3 marks
Use $y = a {(x - 2)}^{2} - 3$ .
$5 = 4 a - 3$ , giving $a = 2$ .
$y = 2 {(x - 2)}^{2} - 3$

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

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