Roots, intercepts, symmetry and turning point | Year 9

01 / LEARN

Understand the idea.

Intercepts answer different questions

A root is an input making the quadratic output zero. Roots give the $x$ -coordinates of horizontal-axis intercepts, so write each crossing as $(x, 0)$ . For the vertical-axis intercept, substitute $x = 0$ instead.

For $y = (x - 1) (x - 5)$ , the zero-product principle gives roots $1$ and $5$ . At $x = 0$ , both factors are negative, so $y = 5$ . The intercepts answer different questions and should be labelled separately.

Symmetry finds the middle

A parabola is symmetric about a vertical line through its turning point. When two roots are known, their average gives this axis. Substitute that middle input into the original rule to find the turning point's height.

A positive squared-term coefficient gives an upward opening and a minimum; a negative one gives a downward opening and a maximum. A parabola can miss the horizontal axis entirely and still have a turning point.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

LaTeX graph of y equals (x minus one) times (x minus five), with x from minus four to six and y from minus six to twelve. Horizontal grid spacing is one unit and vertical grid spacing is two units. Roots, axis and turning point remain unmarked.

Worked solution

1
Find the zero-output inputs.
$x - 1 = 0 or x - 5 = 0$
The factored quadratic is zero when either bracket is zero.
2
State the roots and their midpoint.
$x = 1, 5, \frac{1 + 5}{2} = 3$
Symmetry puts the vertical axis halfway between the roots.
3
Substitute the axis input.
$y = (3 - 1) (3 - 5) = 2 (- 2) = - 4$
The turning point lies on the curve at x equal to 3.
4
State the features.
$axis: x = 3, turning point: (3, - 4)$
The two coordinates specify the turn; its positive leading coefficient makes it a minimum.

Answer

Roots $1, 5$ ; axis $x = 3$ ; turning point $(3, - 4)$ .

EXAMPLE 02

Worked example

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

Worked solution

1
Set the output to zero.
$\begin{aligned} 0 & = - x^{2} + 4 \\ x^{2} & = 4 \\ x & = - 2 or x = 2 \end{aligned}$
Both positive and negative square roots solve the equation.
2
Write the horizontal intercepts.
$(- 2, 0), (2, 0)$
Each root pairs with output zero.
3
Find the vertical intercept.
$x = 0 \Rightarrow y = 4, (0, 4)$
The y-axis crossing occurs at input zero.
4
Verify that the turn is a maximum.
$x^{2} \geq 0 \Rightarrow - x^{2} + 4 \leq 4$
The height 4 is attained at x equal to zero and no larger output is possible.

Answer

$x$ -intercepts $(- 2, 0), (2, 0)$ ; $y$ -intercept/maximum $(0, 4)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = 2$

The axis lies halfway between the roots: $\frac{- 2 + 6}{2} = 2$ . Write the vertical line $x = 2$ , rather than a point or a horizontal line.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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
$\frac{3 + 9}{2} = 6$ .
$x = 6$
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.

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Intercepts answer different questions

Symmetry finds the middle

See the method in action.

Worked example

Find the zero-output inputs.

State the roots and their midpoint.

Substitute the axis input.

State the features.

Worked example

Set the output to zero.

Write the horizontal intercepts.

Find the vertical intercept.

Verify that the turn is a maximum.

Try it yourself.

Check your understanding

Keep practising