Roots, intercepts, symmetry and turning points | Year 10

01 / LEARN

Understand the idea.

Intersections and symmetry

Roots are inputs for which $y = 0$ , so their intercept points are $(r, 0)$ . The vertical intercept instead comes from $x = 0$ . Keeping inputs and coordinates separate makes graph labels precise.

A parabola mirrors across a vertical axis through its vertex. With two real roots, this axis lies halfway between them. Without roots, vertex form or $x = - \frac{b}{2 a}$ still locates the axis.

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

A turning point is not a root

After finding the axis’s horizontal coordinate, substitute it into the rule to find the vertex’s height. Positive $a$ gives a minimum; negative $a$ gives a maximum. A vertex need not be an intercept.

For a model, apply the meaningful input range after finding its algebraic features. An unrestricted graph may have a root before launch or a vertex outside an allowed interval, so context determines which features are usable.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read a downward-opening graph

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

Graph of y=-(x-2) squared+9. The x window is -2 to 6 and the y window is -4 to 10. Fine grid lines are one unit apart; numbered labels are two units apart.

Worked solution

1
Read the vertex and opening.
$\begin{array}{c} y = - {(x - 2)}^{2} + 9 \\ vertex (2, 9) \\ axis x = 2 \end{array}$
The negative square term makes $9$ the maximum, attained when $x - 2 = 0$ .
2
Set the height equal to zero.
$\begin{aligned} 0 & = - {(x - 2)}^{2} + 9 \\ {(x - 2)}^{2} & = 9 \end{aligned}$
Roots occur on the horizontal axis, where the output is zero.
3
Solve both branches.
$\begin{aligned} x - 2 & = \pm 3 \\ x & = - 1 or 5 \end{aligned}$
The roots lie three units to either side of the symmetry axis $x = 2$ .
4
Find the vertical intercept.
$y (0) = - {(0 - 2)}^{2} + 9 = - 4 + 9 = 5$
The vertical intercept is $(0, 5)$ because its input is zero.

Answer

Roots $- 1, 5$ ; y-intercept $(0, 5)$ ; maximum at $(2, 9)$ .

EXAMPLE 02

Use the midpoint of roots

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

Worked solution

1
Find the two roots.
$x + 3 = 0 or x - 1 = 0 ⟹ x = - 3 or 1$
Each root makes one of the displayed factors zero.
2
Take their midpoint.
$x_{axis} = \frac{- 3 + 1}{2} = - 1$
Symmetry places the vertex halfway between the two roots.
3
Calculate the vertex height.
$y (- 1) = (- 1 + 3) (- 1 - 1) = (2) (- 2) = - 4$
Substituting the axis coordinate supplies the missing vertical coordinate.
4
State the vertex and axis.
$vertex (- 1, - 4), axis x = - 1$
The vertex is a point, while the symmetry axis is a vertical line.

Answer

Vertex $(- 1, - 4)$ ; axis $x = - 1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The square ${(x + 2)}^{2}$ is zero at $x = - 2$ . Hence the vertex is $(- 2, - 7)$ and its symmetry axis is $x = - 2$ ; the positive coefficient makes this a minimum.

Keep practising

6 worksheet questions · 19 marks

Open worksheet 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 $x = - \frac{- 2}{2} = 1$ ; 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 ↗

In this lesson

Understand the idea.

Intersections and symmetry

A turning point is not a root

See the method in action.

Read a downward-opening graph

Read the vertex and opening.

Set the height equal to zero.

Solve both branches.

Find the vertical intercept.

Use the midpoint of roots

Find the two roots.

Take their midpoint.

Calculate the vertex height.

State the vertex and axis.

Try it yourself.

Check your understanding

Keep practising