Building quadratic models from information | Year 11

01 / LEARN

Understand the idea.

Choose the form first

Choose a quadratic form that incorporates the given information. Known roots suggest $y = a (x - r) (x - s)$ ; a known vertex suggests $y = a {(x - h)}^{2} + k$ . Substitute a remaining point to determine $a$ , then check all supplied conditions.

Define each quantity and its units before building the rule. An equation may have negative or otherwise impossible roots that cannot represent the situation. State the allowed input interval alongside a model whenever the context supplies one.

Interpret and check

A model’s practical meaning depends on its assumptions and domain. An arch rule between supports need not describe the structure beyond them. Similarly, a simple projectile model can neglect air resistance without asserting that real motion is exactly quadratic.

For an optimisation, find the vertex and verify that its input belongs to the allowed domain. Explain its input and output in context. If the vertex is excluded, the practical optimum must instead be determined from the permitted inputs and boundaries.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A symmetric arch

An arch is at ground level at x=0 m and x=8 m. At x=4 m its height is 6 m. Find its rule.

Worked solution

1
Use the two ground-level positions as roots.
$h = a x (x - 8)$
Height is zero at input zero and input eight, so both factors belong to the model.
2
Use the known height to find the coefficient.
$\begin{aligned} 6 & = a (4) (4 - 8) \\ 6 & = - 16 a \\ a & = - \frac{3}{8} \end{aligned}$
The point at horizontal position four metres has height six metres.
3
State the rule over the arch’s span.
$h = - \frac{3}{8} x (x - 8), 0 \leq x \leq 8$
The model applies between the two ground-level supports, with both position and height measured in metres.
4
Check the given conditions.
$h (0) = 0, h (8) = 0, h (4) = 6$
Both supports and the specified interior height are reproduced by the rule.

Answer

$h = - \frac{3}{8} x (x - 8)$ , $0 \leq x \leq 8$ .

EXAMPLE 02

Greatest rectangular area

A rectangle has perimeter 28 cm. Find its maximum area.

Worked solution

1
Express the second dimension using the perimeter.
$\begin{aligned} 2 x + 2 ℓ & = 28 \\ ℓ & = 14 - x, 0 < x < 14 \end{aligned}$
Let x be the width in centimetres; both width and length must be positive.
2
Write the area model.
$A = x (14 - x) = - x^{2} + 14 x$
A rectangle’s area is the product of its width and length, measured in square centimetres.
3
Complete the square.
$\begin{aligned} A & = - (x^{2} - 14 x) \\ = - [{(x - 7)}^{2} - 49] \\ = 49 - {(x - 7)}^{2} \end{aligned}$
This form exposes the greatest area because the subtracted square is non-negative.
4
Locate and interpret the feasible maximum.
$x = 7, ℓ = 14 - 7 = 7, A_{max} = 49 {cm}^{2}$
Seven lies inside the permitted interval and makes the square zero, so the maximum occurs for a seven-centimetre square.

Answer

49 square centimetres, at dimensions 7 cm by 7 cm.

03 / YOUR TURN

Try it yourself.

Check your understanding

A parabola has vertex (2,5) and passes through (0,1). Find its rule.

Show answer & reasoning

$y = - {(x - 2)}^{2} + 5$

Use the given vertex to write $y = a {(x - 2)}^{2} + 5$ . The point $(0, 1)$ gives $1 = 4 a + 5$ , hence $a = - 1$ and $y = - {(x - 2)}^{2} + 5$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Roots are -2,4 and the graph passes through $(0, - 16)$ . Find its rule.
3 marks
Vertex is $(3, - 2)$ and the graph passes through $(1, 6)$ . Find its rule.
3 marks
A rectangle has perimeter 40 cm. Find maximum area and dimensions.
3 marks
A ball has height $h = - 5 t^{2} + 20 t + 1$ m. Find maximum height/time.
3 marks
A rectangle’s length is 3 cm more than its width; area is 40 square centimetres. Find dimensions.
3 marks
An arch has supports $x = - 3$ m, $x = 3$ m and maximum height 9 m. Give its rule/domain.
3 marks

Read the worked solutions

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

Roots are -2,4 and the graph passes through $(0, - 16)$ . Find its rule.
3 marks
Use factored form.
At $x = 0$ , $- 16 = - 8 a$ .
$y = 2 (x + 2) (x - 4)$
Vertex is $(3, - 2)$ and the graph passes through $(1, 6)$ . Find its rule.
3 marks
Use vertex form.
Substitute $(1, 6)$ : $6 = 4 a - 2$ .
$y = 2 {(x - 3)}^{2} - 2$
A rectangle has perimeter 40 cm. Find maximum area and dimensions.
3 marks
Area is $x (20 - x) = - {(x - 10)}^{2} + 100$ .
The vertex is within $0 < x < 20$ .
100 square centimetres; 10 cm by 10 cm.
A ball has height $h = - 5 t^{2} + 20 t + 1$ m. Find maximum height/time.
3 marks
Complete the square: $h = - 5 {(t - 2)}^{2} + 21$ .
Interpret the vertex.
21 m at $t = 2$ s.
A rectangle’s length is 3 cm more than its width; area is 40 square centimetres. Find dimensions.
3 marks
Use $x (x + 3) = 40$ .
Factor $(x + 8) (x - 5) = 0$ and reject -8.
5 cm by 8 cm.
An arch has supports $x = - 3$ m, $x = 3$ m and maximum height 9 m. Give its rule/domain.
3 marks
Use $h = a (x + 3) (x - 3)$ .
At $x = 0$ , $9 = - 9 a$ .
$h = 9 - x^{2}$ , $- 3 \leq x \leq 3$ .

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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.

Choose the form first

Interpret and check

See the method in action.

A symmetric arch

Use the two ground-level positions as roots.

Use the known height to find the coefficient.

State the rule over the arch’s span.

Check the given conditions.

Greatest rectangular area

Express the second dimension using the perimeter.

Write the area model.

Complete the square.

Locate and interpret the feasible maximum.

Try it yourself.

Check your understanding

Keep practising