Roots, intercepts, symmetry and turning point | Year 9

01 / LEARN

The idea, step by step

Intercepts answer different questions

An $x$ -intercept occurs where $y = 0$ . Its $x$ coordinate is a root of the quadratic equation. A $y$ -intercept occurs where $x = 0$ . Substitute the relevant zero rather than confusing the two axes.

In $y = (x - 1) (x - 5)$ , the roots are $1$ and $5$ , giving intercepts $(1, 0)$ and $(5, 0)$ . The $y$ -intercept is $(0, 5)$ because $(- 1) (- 5) = 5$ .

Symmetry finds the middle

The vertical line halfway between the roots is the axis of symmetry. Here it is $x = (1 + 5) / 2 = 3$ . Substitute $x = 3$ into the rule to find the turning point: $y = (2) (- 2) = - 4$ , so the point is $(3, - 4)$ .

An upward-opening parabola has a minimum turning point. A downward-opening one has a maximum. Some parabolas never meet the $x$ -axis; they still have a turning point and an axis of symmetry.

A common trap

The turning point is always an $x$ -intercept.

Instead: The turning point is where the curve changes direction; it only lies on the $x$ -axis when the root is repeated.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Find the roots, axis and turning point of $y = (x - 1) (x - 5)$ .

Set each factor equal to zero.
Average the roots: $(1 + 5) / 2 = 3$ .
Substitute $3$ : $(3 - 1) (3 - 5) = - 4$ .

Roots

1, 5

; axis

x = 3

; turning point

(3, - 4)

Example 2

Worked example

For $y = - x^{2} + 4$ , find the intercepts and say whether the turning point is a maximum.

$0 = - x^{2} + 4$ gives $x^{2} = 4$ , so $x = \pm 2$ .
At $x = 0$ , $y = 4$ .
The negative coefficient makes the parabola open downwards.

x

-intercepts

(- 2, 0), (2, 0)

;

y

-intercept/maximum

(0, 4)

Check your understanding

A parabola has roots $- 2$ and $6$ . What is its axis?

Show answer

$x = 2$

Average the roots: $(- 2 + 6) / 2 = 2$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

For $y = (x + 2) (x - 4)$ , state both $x$ -intercepts.
2 marks
Find the $y$ -intercept of $y = x^{2} - 6 x + 8$ .
2 marks
Find the axis of a parabola with roots $3$ and $9$ .
2 marks
Find the turning point of $y = (x - 2) (x - 6)$ .
3 marks
Find the turning point of $y = - x^{2} + 9$ and state its type.
2 marks
Use the graph to give the minimum value and the number of real roots of $y = {(x - 2)}^{2} + 1$ .
3 marks

Read the worked solutions

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

For $y = (x + 2) (x - 4)$ , state both $x$ -intercepts.
2 marks
$x + 2 = 0$ or $x - 4 = 0$ .
$(- 2, 0), (4, 0)$
Find the $y$ -intercept of $y = x^{2} - 6 x + 8$ .
2 marks
Substitute $x = 0$ .
$(0, 8)$
Find the axis of a parabola with roots $3$ and $9$ .
2 marks
$(3 + 9) / 2 = 6$ .
$x = 6$
Find the turning point of $y = (x - 2) (x - 6)$ .
3 marks
The axis is $x = 4$ .
At $4$ , $y = 2 (- 2) = - 4$ .
$(4, - 4)$
Find the turning point of $y = - x^{2} + 9$ and state its type.
2 marks
The axis is $x = 0$ and $y = 9$ there.
The parabola opens downwards.
$(0, 9)$ , maximum.
Use the graph to give the minimum value and the number of real roots of $y = {(x - 2)}^{2} + 1$ .
3 marks
The lowest point is $(2, 1)$ , above the $x$ -axis.
The curve never reaches $y = 0$ .
Minimum $1$ ; no real roots.

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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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