Quadratic tables and graphs | Year 9

01 / LEARN

The idea, step by step

One input, one output

Choose several $x$ values and substitute each into the rule. Include negative inputs using brackets: if $x = - 2$ , then $x^{2} = {(- 2)}^{2} = 4$ . For $y = x^{2} - 1$ , the outputs at $- 2, - 1, 0, 1, 2$ are $3, 0, - 1, 0, 3$ .

Plot the ordered pairs with $x$ horizontal and $y$ vertical. A table gives sample points, not the whole graph. A quadratic relationship is continuous unless the context restricts inputs to separate values.

See the shape, not a zigzag

The graph is a parabola, with a smooth bend at its turning point. Use points on both sides of the turn. A straight join between sparse plotted points can hide the curve, so add more points near the turn if necessary.

Choose scales that display the important features. Label axes and the rule, and check symmetry when the table suggests it. A graphing tool can confirm a hand sketch, but it should not replace checking a suspicious table entry.

A common trap

At $x = - 2$ , replace $x^{2}$ with $- 4$ .

Instead: ${(- 2)}^{2} = 4$ because the whole negative input is squared.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Complete a table for $y = x^{2} - 1$ at $x = - 2, - 1, 0, 1, 2$ and describe the graph.

Square each input first, including the negative inputs.
Subtract $1$ from each square.
Plot the five pairs; the lowest point is $(0, - 1)$ .

y = 3, 0, - 1, 0, 3

; a smooth upward parabola.

Example 2

Worked example

Find the missing output for $y = x^{2} - 2 x$ when $x = - 3$ .

Substitute with brackets: ${(- 3)}^{2} - 2 (- 3)$ .
This is $9 + 6 = 15$ .
The point to plot is $(- 3, 15)$ , not $(15, - 3)$ .

y = 15

Check your understanding

For $y = x^{2} + 2$ , find $y$ when $x = - 3$ .

Show answer

$11$

${(- 3)}^{2} + 2 = 9 + 2 = 11$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $y = x^{2} + 1$ , find $y$ at $x = - 2, 0, 2$ .
2 marks
For $y = x^{2} - 3 x$ , find $y$ when $x = - 2$ .
2 marks
Complete the outputs for $y = x^{2} - 4$ at $x = - 3, - 2, - 1, 0, 1, 2, 3$ .
3 marks
Use the supplied graph of $y = x^{2} - 4$ to read its two points on the $x$ -axis.
2 marks
Sketch $y = x^{2} + 1$ using at least five points. State the lowest point.
4 marks
Why is joining only $(- 2, 4)$ and $(2, 4)$ with a straight line not a sketch of $y = x^{2}$ ?
3 marks

Read the worked solutions

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

For $y = x^{2} + 1$ , find $y$ at $x = - 2, 0, 2$ .
2 marks
Square the inputs to get $4, 0, 4$ , then add $1$ .
$5, 1, 5$
For $y = x^{2} - 3 x$ , find $y$ when $x = - 2$ .
2 marks
${(- 2)}^{2} - 3 (- 2) = 4 + 6 = 10$ .
$10$
Complete the outputs for $y = x^{2} - 4$ at $x = - 3, - 2, - 1, 0, 1, 2, 3$ .
3 marks
Square each input, then subtract $4$ .
$5, 0, - 3, - 4, - 3, 0, 5$
Use the supplied graph of $y = x^{2} - 4$ to read its two points on the $x$ -axis.
2 marks
On the $x$ -axis, $y = 0$ ; the curve crosses at $x = - 2, 2$ .
$(- 2, 0)$ and $(2, 0)$
Sketch $y = x^{2} + 1$ using at least five points. State the lowest point.
4 marks
A useful table is $(- 2, 5), (- 1, 2), (0, 1), (1, 2), (2, 5)$ .
Join with a smooth upward curve.
Lowest point $(0, 1)$ .
Why is joining only $(- 2, 4)$ and $(2, 4)$ with a straight line not a sketch of $y = x^{2}$ ?
3 marks
At $x = 0$ the rule gives $0$ , while the straight join gives $4$ .
More points are needed to show the turn.
It misses the curved shape and the point $(0, 0)$ .

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Roots, intercepts, symmetry and turning point

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

AC9M9A02, AC9M9A04–A06: monic quadratic expressions, integer-root equations, graphs and models.

Official curriculum source ↗

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