Quadratic functions - Year 9 Mathematics

What you will learn

recognise the shape and key features of a parabola,
identify the axis of symmetry, turning point and intercepts of $y = ax^2 + bx + c$ ,
sketch parabolas using a table of values and key features,
apply the null factor law: if $ab = 0$ then $a = 0$ or $b = 0$ ,
solve monic quadratic equations by factorising and applying the null factor law,
interpret solutions of a quadratic equation as $x$ -intercepts of the corresponding parabola.

Worked example 0 Real-world example: height of a ball

A ball is thrown upward from ground level. Its height (in metres) after $t$ seconds is modelled by $h = 20t - 5t^2$ . When does it hit the ground, and what is its maximum height?

Ground level means $h = 0$ : solve $20t - 5t^2 = 0$ .
Factorise: $5t(4 - t) = 0$ .
By the null factor law: $t = 0$ or $t = 4$ . The ball is at ground level at $t = 0$ s (launch) and $t = 4$ s (landing).
The axis of symmetry is midway: $t = \dfrac{0 + 4}{2} = 2$ s.
Maximum height: $h = 20(2) - 5(2)^2 = 40 - 20 = 20$ m.

Key idea: the $x$ -intercepts (here $t$ -intercepts) tell you when something happens; the turning point tells you the maximum or minimum value.

1. What is a quadratic function?

A quadratic function has the form:

Formula reference

y = ax^2 + bx + c \qquad (a \neq 0)

The highest power of $x$ is $2$ .
If $a > 0$ the parabola opens upward (concave up, “happy face”).
If $a < 0$ the parabola opens downward (concave down, “sad face”).
The constant $c$ gives the $y$ -intercept (set $x = 0$ ).

Worked example 1 Identifying a, b, c

For $y = 3x^2 - 5x + 2$ , state $a$ , $b$ and $c$ , and the $y$ -intercept.

$a = 3$ , $b = -5$ , $c = 2$ .
Since $a = 3 > 0$ , the parabola opens upward.
$y$ -intercept: $(0, 2)$ .

2. Key features of a parabola

Every parabola has:

Feature	How to find it
$y$ -intercept	Set $x = 0$ : the point is $(0, c)$ .
$x$ -intercepts	Set $y = 0$ and solve $ax^2 + bx + c = 0$ .
Axis of symmetry	$x = -\dfrac{b}{2a}$ , or the midpoint of the two $x$ -intercepts.
Turning point	Substitute the axis of symmetry value into $y = ax^2 + bx + c$ .

Formula reference

\text{Axis of symmetry: } x = -\frac{b}{2a}

Worked example 2 Finding key features

Find the turning point and intercepts of $y = x^2 - 6x + 8$ .

$y$ -intercept: $(0, 8)$ .
$x$ -intercepts: solve $x^2 - 6x + 8 = 0$ . Factorise: $(x - 2)(x - 4) = 0$ , so $x = 2$ or $x = 4$ .
Axis of symmetry: $x = \dfrac{2 + 4}{2} = 3$ (or $x = -\dfrac{-6}{2(1)} = 3$ ).
Turning point: $y = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1$ . The turning point is $(3, -1)$ .
Since $a = 1 > 0$ , the parabola opens upward, so $(3, -1)$ is a minimum.

3. The null factor law

Worked example 3 Applying the null factor law

Solve $(x - 5)(x + 2) = 0$ .

By the null factor law:

$x - 5 = 0 \Rightarrow x = 5$ , or
$x + 2 = 0 \Rightarrow x = -2$ .

Solutions: $x = 5$ or $x = -2$ .

4. Solving quadratics by factorisation

To solve $x^2 + bx + c = 0$ (monic, $a = 1$ ): find two numbers that multiply to give $c$ and add to give $b$ .

Worked example 4 Factorising a monic quadratic

Solve $x^2 + 5x + 6 = 0$ .

Find two numbers that multiply to $6$ and add to $5$ : the numbers are $2$ and $3$ .
Factorise: $(x + 2)(x + 3) = 0$ .
Null factor law: $x = -2$ or $x = -3$ .

Worked example 5 Quadratic with a negative constant

Solve $x^2 - 3x - 10 = 0$ .

Find two numbers that multiply to $-10$ and add to $-3$ : the numbers are $-5$ and $2$ .
Factorise: $(x - 5)(x + 2) = 0$ .
Solutions: $x = 5$ or $x = -2$ .

Worked example 6 Factorising with a common factor first

Solve $2x^2 + 10x + 12 = 0$ .

Take out the common factor of $2$ : $2(x^2 + 5x + 6) = 0$ .
Factorise the monic quadratic: $2(x + 2)(x + 3) = 0$ .
Null factor law: $x = -2$ or $x = -3$ .

5. Connecting graphs and solutions

The solutions of $ax^2 + bx + c = 0$ are the $x$ -intercepts of $y = ax^2 + bx + c$ . This means:

A parabola that crosses the $x$ -axis at two points has two solutions.
A parabola that just touches the $x$ -axis has one repeated solution.
A parabola that does not cross the $x$ -axis has no real solutions.

Worked example 7 Reading solutions from a graph

A parabola crosses the $x$ -axis at $x = -1$ and $x = 4$ . Its equation is $y = (x + 1)(x - 4) = x^2 - 3x - 4$ .

The axis of symmetry is $x = \dfrac{-1 + 4}{2} = 1.5$ .
The turning point: $y = (1.5)^2 - 3(1.5) - 4 = 2.25 - 4.5 - 4 = -6.25$ . So the turning point is $(1.5,\;-6.25)$ .

Practice

Fluency

Tier 1: identify and factorise

For $y = 2x^2 - 4x + 1$ , state $a$ , $b$ , $c$ and whether the parabola opens up or down.
Find the $y$ -intercept of $y = -x^2 + 3x - 5$ .
Factorise $x^2 + 7x + 12$ .
Factorise $x^2 - 9x + 20$ .
Factorise $x^2 + 2x - 15$ .
Solve $(x - 3)(x + 7) = 0$ .
Solve $x^2 - 5x + 6 = 0$ .
Solve $x^2 + x - 12 = 0$ .
Solve $x^2 - 16 = 0$ .
Solve $3x^2 - 12x = 0$ .

Reasoning

Tier 2: features and graphs

Find the axis of symmetry and turning point of $y = x^2 - 4x + 3$ .
Find the $x$ -intercepts, turning point and $y$ -intercept of $y = x^2 - 2x - 8$ . Sketch the parabola.
A parabola has $x$ -intercepts at $x = -3$ and $x = 1$ . Find the axis of symmetry and the turning point if the equation is $y = x^2 + 2x - 3$ .
Solve $x^2 = 7x - 10$ by first rearranging to standard form.
Solve $2x^2 - 14x + 20 = 0$ by first taking out a common factor.
Explain why $x^2 + 4 = 0$ has no real solutions. What does this mean for the graph?
A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm. Its area is $21$ cm $^2$ . Find $x$ .
The product of two consecutive integers is $72$ . Find the integers.

Reasoning

Tier 3: explain and apply

Explain the connection between the factorised form $y = (x - p)(x - q)$ and the $x$ -intercepts. How do you find the axis of symmetry from $p$ and $q$ ?
A ball is launched upward with height $h = 24t - 4t^2$ metres after $t$ seconds. Find when it hits the ground and its maximum height.
Two numbers add to $12$ and their product is $32$ . Set up and solve a quadratic equation to find them.
The parabola $y = x^2 + bx + 9$ has only one $x$ -intercept. Find the two possible values of $b$ .

Challenge

Reasoning

Harder reasoning

A farmer has $60$ m of fencing to enclose a rectangular paddock against a wall (only three sides need fencing). If the width is $x$ m, show that the area is $A = 60x - 2x^2$ and find the dimensions that maximise the area.
The parabola $y = x^2 - 6x + k$ passes through the point $(1, 2)$ . Find $k$ , then find the turning point and $x$ -intercepts.
Show that if a monic quadratic $x^2 + bx + c = 0$ has solutions $p$ and $q$ , then $p + q = -b$ and $pq = c$ .
The sum of the squares of two consecutive positive odd numbers is $130$ . Find the numbers.

Interactive

Try it yourself: match the parabola

Work through 3 examples on one graph. Drag the vertex and slide the coefficient a.

Example 1 (easy). Match the dashed target y = x². Drag the vertex to the origin and set a = 1.

Coefficient a:1

Vertex (h, k): (0, 0)
a: 1
Your curve: y = x²

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 9 core - answers

Fluency

Tier 1: identify and factorise

$a = 2$ , $b = -4$ , $c = 1$ . Since $a = 2 > 0$ , the parabola opens upward.
$y$ -intercept: $(0, -5)$ .
$(x + 3)(x + 4)$ .
$(x - 4)(x - 5)$ .
$(x + 5)(x - 3)$ .
$x = 3$ or $x = -7$ .
$(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
$(x + 4)(x - 3) = 0$ : $x = -4$ or $x = 3$ .
$(x - 4)(x + 4) = 0$ : $x = 4$ or $x = -4$ .
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .

Reasoning

Tier 2: features and graphs

$x = -\dfrac{-4}{2(1)} = 2$ . Turning point: $y = 4 - 8 + 3 = -1$ . Turning point is $(2, -1)$ .
$x$ -intercepts: $x^2 - 2x - 8 = (x - 4)(x + 2) = 0$ , so $x = 4$ or $x = -2$ . Axis of symmetry: $x = 1$ . Turning point: $y = 1 - 2 - 8 = -9$ , so $(1, -9)$ . $y$ -intercept: $(0, -8)$ . Parabola opens upward.
Axis of symmetry: $x = \dfrac{-3 + 1}{2} = -1$ . Turning point: $y = 1 - 2 - 3 = -4$ , so $(-1, -4)$ .
$x^2 - 7x + 10 = 0$ . $(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$2(x^2 - 7x + 10) = 0$ . $2(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$x^2 = -4$ has no real solution because a square is never negative. The parabola $y = x^2 + 4$ has its turning point at $(0, 4)$ , entirely above the $x$ -axis, so it never crosses it.
$(x + 3)(x - 1) = 21$ . Expand: $x^2 + 2x - 3 = 21$ , so $x^2 + 2x - 24 = 0$ . $(x + 6)(x - 4) = 0$ : $x = -6$ or $x = 4$ . Since $x - 1 > 0$ , we need $x > 1$ , so $x = 4$ . The rectangle is $7$ cm by $3$ cm.
Let the integers be $n$ and $n + 1$ . $n(n + 1) = 72$ , so $n^2 + n - 72 = 0$ . $(n + 9)(n - 8) = 0$ : $n = 8$ or $n = -9$ . The consecutive integers are $8$ and $9$ (or $-9$ and $-8$ ).

Reasoning

Tier 3: explain and apply

In $y = (x - p)(x - q)$ , the $x$ -intercepts are $x = p$ and $x = q$ (by the null factor law). The axis of symmetry is $x = \dfrac{p + q}{2}$ (midpoint of the intercepts).
Ground: $24t - 4t^2 = 0$ , $4t(6 - t) = 0$ , so $t = 0$ or $t = 6$ s. Axis of symmetry: $t = 3$ . Max height: $h = 24(3) - 4(9) = 72 - 36 = 36$ m.
Let the numbers be $x$ and $12 - x$ . Product: $x(12 - x) = 32$ , so $12x - x^2 = 32$ , $x^2 - 12x + 32 = 0$ . $(x - 4)(x - 8) = 0$ : $x = 4$ or $x = 8$ . The numbers are $4$ and $8$ .
One $x$ -intercept means the discriminant is zero: $b^2 - 4(1)(9) = 0$ , so $b^2 = 36$ , $b = 6$ or $b = -6$ .

Reasoning

Challenge

Width $= x$ , length $= 60 - 2x$ . Area $= x(60 - 2x) = 60x - 2x^2$ . This is a downward parabola. Axis of symmetry: $x = -\dfrac{60}{2(-2)} = 15$ . Maximum area: $A = 60(15) - 2(225) = 900 - 450 = 450$ m $^2$ . Dimensions: $15$ m wide, $30$ m long.
Substitute $(1, 2)$ : $2 = 1 - 6 + k$ , so $k = 7$ . Equation: $y = x^2 - 6x + 7$ . Turning point: $x = 3$ , $y = 9 - 18 + 7 = -2$ , so $(3, -2)$ . $x$ -intercepts: $x^2 - 6x + 7 = 0$ . Using the quadratic formula (or completing the square): $x = \dfrac{6 \pm \sqrt{36 - 28}}{2} = \dfrac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$ .
$(x - p)(x - q) = x^2 - (p+q)x + pq$ . Comparing with $x^2 + bx + c$ : $-(p+q) = b$ so $p + q = -b$ , and $pq = c$ .
Let the numbers be $n$ and $n + 2$ . $n^2 + (n+2)^2 = 130$ . $2n^2 + 4n + 4 = 130$ . $n^2 + 2n - 63 = 0$ . $(n + 9)(n - 7) = 0$ : $n = 7$ (positive). The numbers are $7$ and $9$ .

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