Roots, intercepts, symmetry and turning points | Year 10

01 / LEARN

The idea, step by step

Intersections and symmetry

Roots are x-values where y is zero. The y-intercept has x=0. A parabola is symmetric about the vertical line through its turning point.

When two roots are known, the axis lies halfway between them. Substitute that x-value into the function to find the vertex height.

x_{axis} = \frac{r_{1} + r_{2}}{2} = - \frac{b}{2 a}

A turning point is not a root

The vertex is a maximum if a is negative and a minimum if a is positive. It need not lie on either axis.

In a practical model, check which part of the graph is meaningful. For example, a negative time before launch is not a later landing time, even if it is an algebraic root.

A common trap

Treating the y-intercept as a root.

Instead: An x-intercept has y=0; a y-intercept has x=0. They are different substitutions.

02 / WORKED EXAMPLES

See the thinking

Example 1

Read a downward-opening graph

Find the main features of $y = - {(x - 2)}^{2} + 9$ .

The vertex is $(2, 9)$ , axis $x = 2$ , and the maximum is $9$ .
For roots, ${(x - 2)}^{2} = 9$ , so $x = - 1$ or $5$ .
At $x = 0$ , $y = 5$ ; the y-intercept is $(0, 5)$ .

Roots

- 1, 5

; y-intercept

(0, 5)

; maximum at

(2, 9)

Example 2

Use the midpoint of roots

Find the vertex of $y = (x + 3) (x - 1)$ .

The roots are $- 3$ and $1$ .
Their midpoint is $(- 3 + 1) / 2 = - 1$ .
At $x = - 1$ , $y = (2) (- 2) = - 4$ .

Vertex

(- 1, - 4)

; axis

x = - 1

Check your understanding

What are the vertex and axis of $y = 3 {(x + 2)}^{2} - 7$ ?

Show answer

Vertex $(- 2, - 7)$ ; axis $x = - 2$ .

The square is zero when $x = - 2$ .

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

State the vertex, axis and minimum of $y = {(x - 4)}^{2} + 2$ .
2 marks
Find the roots and y-intercept of $y = (x + 2) (x - 6)$ .
3 marks
Find the vertex of $y = x^{2} - 2 x - 8$ .
2 marks
Find all intercepts and the vertex of $y = 2 {(x - 1)}^{2} - 8$ .
4 marks
The graph shown is $y = - x^{2} + 4 x + 5$ . State its roots, axis and maximum.
4 marks
A ball has height $h = - t^{2} + 2 t + 3$ metres for times from launch until it first reaches the ground. Find the greatest height and the landing time.
4 marks

Read the worked solutions

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

State the vertex, axis and minimum of $y = {(x - 4)}^{2} + 2$ .
2 marks
The squared term is zero at $x = 4$ .
Vertex $(4, 2)$ ; axis $x = 4$ ; minimum $2$ .
Find the roots and y-intercept of $y = (x + 2) (x - 6)$ .
3 marks
Set each factor to zero; then set $x = 0$ .
Roots $- 2, 6$ ; y-intercept $(0, - 12)$ .
Find the vertex of $y = x^{2} - 2 x - 8$ .
2 marks
The axis is $x = - (- 2) / (2) = 1$ ; substitute to get $1 - 2 - 8 = - 9$ .
$(1, - 9)$
Find all intercepts and the vertex of $y = 2 {(x - 1)}^{2} - 8$ .
4 marks
For roots, ${(x - 1)}^{2} = 4$ .
For the y-intercept, use $x = 0$ .
Roots $- 1, 3$ ; y-intercept $(0, - 6)$ ; vertex $(1, - 8)$ .
The graph shown is $y = - x^{2} + 4 x + 5$ . State its roots, axis and maximum.
4 marks
$- x^{2} + 4 x + 5 = - (x - 5) (x + 1)$ .
The midpoint of the roots is $2$ and $y (2) = 9$ .
Roots $- 1, 5$ ; axis $x = 2$ ; maximum $9$ .
A ball has height $h = - t^{2} + 2 t + 3$ metres for times from launch until it first reaches the ground. Find the greatest height and the landing time.
4 marks
$h = - {(t - 1)}^{2} + 4$ , so the maximum is $4$ .
$- t^{2} + 2 t + 3 = 0$ has roots $- 1, 3$ ; use the non-negative landing time.
Greatest height $4$ m at $t = 1$ s; landing at $t = 3$ s.

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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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