Expanded, factored and vertex forms | Year 10

01 / LEARN

Understand the idea.

One function, three useful views

Each quadratic form highlights a different feature. Expanded form gives the vertical intercept $(0, c)$ ; real factored form gives the roots; vertex form gives the turning point $(h, k)$ . The coefficient $a \neq 0$ controls opening and vertical scale.

These are alternative descriptions of the same function, so corresponding inputs must give the same output. Real root factors exist only when real roots exist; a repeated root supplies the same factor twice.

\begin{aligned} y & = a x^{2} + b x + c \\ = a (x - r_{1}) (x - r_{2}) \\ = a {(x - h)}^{2} + k \end{aligned}

Choose the form before doing extra work

Choose a starting form that uses the given information. Two roots give $y = a (x - r_{1}) (x - r_{2})$ ; a vertex gives $y = a {(x - h)}^{2} + k$ . Substitute another suitable point to determine the remaining scale factor.

Convert by expanding, factorising or completing the square when another feature is needed. Keep a useful form visible rather than expanding automatically: a correct rule can often answer several questions directly.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Connect all three forms

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

Graph of y=2x squared-8x+6 on x from -1 to 5 and y from -4 to 12. Grid lines are one unit apart. Only the expanded equation is labelled, without adding roots or vertex coordinates.

Worked solution

1
Remove the common factor.
$y = 2 (x^{2} - 4 x + 3)$
Factoring out the leading coefficient leaves a monic quadratic.
2
Factor the monic quadratic.
$y = 2 (x - 1) (x - 3)$
The numbers $- 1$ and $- 3$ have product $3$ and sum $- 4$ .
3
Complete the square from the expanded form.
$\begin{aligned} y & = 2 (x^{2} - 4 x) + 6 \\ = 2 [{(x - 2)}^{2} - 4] + 6 \\ = 2 {(x - 2)}^{2} - 2 \end{aligned}$
The square adds $4$ inside the brackets, so the compensation is multiplied by $2$ .
4
Read and connect the features.
$\begin{array}{c} x = 1 or x = 3 when y = 0 \\ vertex (2, - 2) \end{array}$
The roots are symmetric about the vertex’s horizontal coordinate, consistent with both forms.

Answer

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

EXAMPLE 02

Build a model from roots

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

Worked solution

1
Build factors from the roots.
$y = a (x + 2) (x - 4), a \neq 0$
The factors must vanish at $x = - 2$ and $x = 4$ .
2
Substitute the additional point.
$- 16 = a (0 + 2) (0 - 4) = - 8 a$
The point fixes the scale that the roots alone cannot determine.
3
Solve for the leading coefficient.
$a = \frac{- 16}{- 8} = 2$
Dividing by the nonzero product gives a unique scale factor.
4
State and verify the rule.
$\begin{array}{c} y = 2 (x + 2) (x - 4) \\ y (0) = 2 (2) (- 4) = - 16 \end{array}$
The rule has both required zero factors and passes through the supplied point.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The vertex form ${(x - 3)}^{2} - 4$ displays the vertex $(3, - 4)$ because the square is zero at $x = 3$ . Expanding gives $x^{2} - 6 x + 5$ , confirming the forms agree.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

One function, three useful views

Choose the form before doing extra work

See the method in action.

Connect all three forms

Remove the common factor.

Factor the monic quadratic.

Complete the square from the expanded form.

Read and connect the features.

Build a model from roots

Build factors from the roots.

Substitute the additional point.

Solve for the leading coefficient.

State and verify the rule.

Try it yourself.

Check your understanding

Keep practising