1.5.3 Quadratic functions

What you'll learn

Recognise quadratic functions and interpret their graphs.
Complete the square to find a quadratic's turning point and range.
Use the discriminant to determine the number of real roots.
Solve quadratic equations, including equations that are quadratic in another function of the unknown.

Starting Point: Quadratic Expressions

A quadratic function is a function whose highest power of the variable is two.

Definition

Quadratic function

A quadratic function has the form

f(x)=ax^2+bx+c

where $a$ , $b$ and $c$ are constants and $a≠0a\neq 0$ .

The condition $a≠0a\neq 0$ matters: if $a = 0$ , the $x^2$ term disappears and the function is linear rather than quadratic.

An equation such as $ax^2+bx+c=0$ is called a quadratic equation. A value of $x$ satisfying the equation is a root or solution.

Graphs of Quadratic Functions

The graph of a quadratic function is called a parabola. It has a single turning point and is symmetrical about a vertical line.

If $a > 0$ , the parabola opens upwards and has a minimum point.
If $a < 0$ , the parabola opens downwards and has a maximum point.
A larger value of $∣a∣\lvert a\rvert$ produces a narrower graph.

Definition

Vertex and axis of symmetry

The vertex is the turning point of a parabola. The vertical line passing through the vertex is its axis of symmetry.

The roots of $f (x) = 0$ are the points where the graph of $y = f (x)$ crosses or touches the $x$ -axis. The $y$ -intercept is found by setting $x = 0$ , giving the point $(0, c)$ .

Example

Sketching a quadratic graph

Sketch $y=x^2-5x+6$ .

Factorise the quadratic:
$x^2-5x+6=(x-2)(x-3)$
Therefore, the roots are $x = 2$ and $x = 3$ , giving the $x$ -intercepts $(2, 0)$ and $(3, 0)$ .
The axis of symmetry lies halfway between the roots:
$x=2+32=52x=\frac{2+3}{2}=\frac{5}{2}$
Substitute $x=52x=\frac{5}{2}$ to find the vertex:
$y=(52)2−5(52)+6=−14\begin{aligned} y&=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+6\\ &=-\frac{1}{4} \end{aligned}$
The vertex is therefore $(52,−14)\left(\frac{5}{2},-\frac{1}{4}\right)$ . Since the coefficient of $x^2$ is positive, the graph opens upwards.

Common Mistake

Forgetting the y-coordinate

The axis of symmetry gives only the $x$ -coordinate of the vertex. Substitute this value into the function to find the corresponding $y$ -coordinate.

Completing the Square

Completing the square rewrites a quadratic in a form that displays its vertex directly:

ax^2+bx+c=a(x-p)^2+q

This is called the completed-square form or vertex form. The vertex is $(p, q)$ and the axis of symmetry is $x = p$ .

Key Idea

Reading the vertex

In $f(x)=a(x-p)^2+q$ , the vertex is $(p, q)$ . Be careful with the sign: $x-3)^2$ gives $p = 3$ , while $x+3)^2$ gives $p = - 3$ .

To complete the square when the coefficient of $x^2$ is one, halve the coefficient of $x$ and use

x2+bx=(x+b2)2−(b2)2.x^2+bx=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2.

Example

Finding the vertex by completing the square

Write $x^2-8x+11$ in completed-square form and find its minimum value.

Halve the coefficient of $x$ , giving $- 4$ , and form the square:
$x-4)^2=x^2-8x+16.$
Adjust the constant because 16 has been added inside the square:
$x2−8x+11=(x−4)2−16+11=(x−4)2−5.\begin{aligned} x^2-8x+11 &=(x-4)^2-16+11\\ &=(x-4)^2-5. \end{aligned}$
Since $(x−4)2≥0(x-4)^2\geq 0$ , the smallest possible value occurs when $x = 4$ . The minimum value is therefore $- 5$ , and the vertex is $(4, - 5)$ .

If the coefficient of $x^2$ is not one, first take it outside the relevant terms.

Example

Completing the square with a leading coefficient

Write $3x^2+12x-7$ in completed-square form.

Factor 3 from the terms containing $x$ :
$3x^2+12x-7=3(x^2+4x)-7.$
Complete the square inside the brackets:
$x^2+4x=(x+2)^2-4.$
Substitute and simplify:
$3(x2+4x)−7=3((x+2)2−4)−7=3(x+2)2−19.\begin{aligned} 3(x^2+4x)-7 &=3\left((x+2)^2-4\right)-7\\ &=3(x+2)^2-19. \end{aligned}$
The vertex is $(- 2, - 19)$ .

Common Mistake

Missing the outside multiplier

In $3((x+2)2−4)3\left((x+2)^2-4\right)$ , both terms inside the brackets must be multiplied by 3. The constant contribution is $- 12$ , not $- 4$ .

Solving Quadratic Equations

To solve a quadratic equation, first rearrange it into the form

ax^2+bx+c=0.

You can then use factorisation, completing the square or the quadratic formula.

Definition

Quadratic formula

The solutions of $ax^2+bx+c=0$ , where $a≠0a\neq 0$ , are

x=−b±b2−4ac2a.x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

Example

Using the quadratic formula

Solve $2x^2+3x-7=0$ , giving exact answers.

Identify $a = 2$ , $b = 3$ and $c = - 7$ . Substitute these carefully:
$x=−3±32−4(2)(−7)2(2).x=\frac{-3\pm\sqrt{3^2-4(2)(-7)}}{2(2)}.$
Simplify the expression under the square root:
$3^2-4(2)(-7)=9+56=65.$
Therefore,
$x=−3±654.x=\frac{-3\pm\sqrt{65}}{4}.$

Tip

Choosing a method

Try factorisation first when integer factors are easy to see. Otherwise, the quadratic formula is usually the most reliable method for finding roots.

The Discriminant

The expression inside the square root in the quadratic formula is called the discriminant.

Definition

Discriminant

For $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac.\Delta=b^2-4ac.

Its sign determines the number of real roots.

If $Δ>0\Delta>0$ , there are two distinct real roots.
If $Δ=0\Delta=0$ , there is one repeated real root.
If $Δ<0\Delta<0$ , there are no real roots.

Graphically, these correspond to a parabola crossing the $x$ -axis twice, touching it once, or not meeting it.

Three quadratic graphs showing two real roots, one repeated root and no real roots according to the discriminant

Example

Finding a condition for repeated roots

Find the values of $k$ for which $x^2+kx+9=0$ has a repeated root.

A repeated root occurs when the discriminant equals zero:
$k^2-4(1)(9)=0.$
Simplify and solve:
$k^2-36=0.$
Factorise:
$(k - 6) (k + 6) = 0,$
so $k = 6$ or $k = - 6$ .

Common Mistake

Dropping a negative sign

When calculating $b^2-4ac$ , put negative values of $b$ or $c$ in brackets. For example, if $c = - 3$ , then $- 4 a c$ becomes positive when $a > 0$ .

Quadratics in a Function of the Unknown

Some equations are not quadratics in $x$ itself but become quadratics after replacing a repeated expression with a new variable.

Definition

Quadratic in a function

An equation is quadratic in a function of the unknown if it can be written as

a[g(x)]^2+b[g(x)]+c=0,

where $g (x)$ is an expression involving $x$ .

Use a substitution such as $u = g (x)$ , solve the resulting quadratic in $u$ , and then return to the original variable.

Example

Solving a disguised quadratic

Solve $x^4-5x^2+4=0$ .

Since $x^4=(x^2)^2$ , let $u=x^2$ . The equation becomes
$u^2-5u+4=0.$
Factorise and solve for $u$ :
$(u - 1) (u - 4) = 0,$
so $u = 1$ or $u = 4$ .
Replace $u$ with $x^2$ and solve both equations:
$x2=1⇒x=±1,x2=4⇒x=±2.\begin{aligned} x^2&=1 &&\Rightarrow x=\pm1,\\ x^2&=4 &&\Rightarrow x=\pm2. \end{aligned}$
Therefore, the solutions are $x = - 2, - 1, 1, 2$ .

Common Mistake

Return to the original variable

Solving the substituted quadratic is not the end. If $u=x^2$ , each positive value of $u$ may produce two values of $x$ , while a negative value produces no real values of $x$ .

Exam technique

In the exam

Rearrange every quadratic equation so that one side is zero before choosing a method.
Use the graph, completed-square form and discriminant as connected information: roots are $x$ -intercepts, while the vertex controls the maximum or minimum.
Write discriminant conditions precisely: use $> 0$ for two distinct roots, $= 0$ for a repeated root and $≥0\geq0$ when at least one real root is required.
Keep exact roots as fractions or surds unless the question specifically requests decimal answers.

Self review

Check yourself

Can you complete the square for $2x^2-12x+5$ and identify its vertex?
What condition on $k$ makes $x^2+4x+k=0$ have two distinct real roots?
How would you solve $x^6-7x^3+6=0$ using a substitution?

What you'll learn

Recognise quadratic functions and interpret their graphs.
Complete the square to find a quadratic's turning point and range.
Use the discriminant to determine the number of real roots.
Solve quadratic equations, including equations that are quadratic in another function of the unknown.

Starting Point: Quadratic Expressions

A quadratic function is a function whose highest power of the variable is two.

Definition

Quadratic function

A quadratic function has the form

f(x)=ax^2+bx+c

where $a$ , $b$ and $c$ are constants and $a≠0a\neq 0$ .

The condition $a≠0a\neq 0$ matters: if $a = 0$ , the $x^2$ term disappears and the function is linear rather than quadratic.

An equation such as $ax^2+bx+c=0$ is called a quadratic equation. A value of $x$ satisfying the equation is a root or solution.

Graphs of Quadratic Functions

The graph of a quadratic function is called a parabola. It has a single turning point and is symmetrical about a vertical line.

If $a > 0$ , the parabola opens upwards and has a minimum point.
If $a < 0$ , the parabola opens downwards and has a maximum point.
A larger value of $∣a∣\lvert a\rvert$ produces a narrower graph.

Definition

Vertex and axis of symmetry

The vertex is the turning point of a parabola. The vertical line passing through the vertex is its axis of symmetry.

The roots of $f (x) = 0$ are the points where the graph of $y = f (x)$ crosses or touches the $x$ -axis. The $y$ -intercept is found by setting $x = 0$ , giving the point $(0, c)$ .

Example

Sketching a quadratic graph

Sketch $y=x^2-5x+6$ .

Factorise the quadratic:
$x^2-5x+6=(x-2)(x-3)$
Therefore, the roots are $x = 2$ and $x = 3$ , giving the $x$ -intercepts $(2, 0)$ and $(3, 0)$ .
The axis of symmetry lies halfway between the roots:
$x=2+32=52x=\frac{2+3}{2}=\frac{5}{2}$
Substitute $x=52x=\frac{5}{2}$ to find the vertex:
$y=(52)2−5(52)+6=−14\begin{aligned} y&=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+6\\ &=-\frac{1}{4} \end{aligned}$
The vertex is therefore $(52,−14)\left(\frac{5}{2},-\frac{1}{4}\right)$ . Since the coefficient of $x^2$ is positive, the graph opens upwards.

Common Mistake

Forgetting the y-coordinate

The axis of symmetry gives only the $x$ -coordinate of the vertex. Substitute this value into the function to find the corresponding $y$ -coordinate.

Completing the Square

Completing the square rewrites a quadratic in a form that displays its vertex directly:

ax^2+bx+c=a(x-p)^2+q

This is called the completed-square form or vertex form. The vertex is $(p, q)$ and the axis of symmetry is $x = p$ .

Key Idea

Reading the vertex

In $f(x)=a(x-p)^2+q$ , the vertex is $(p, q)$ . Be careful with the sign: $x-3)^2$ gives $p = 3$ , while $x+3)^2$ gives $p = - 3$ .

To complete the square when the coefficient of $x^2$ is one, halve the coefficient of $x$ and use

x2+bx=(x+b2)2−(b2)2.x^2+bx=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2.

Example

Finding the vertex by completing the square

Write $x^2-8x+11$ in completed-square form and find its minimum value.

Halve the coefficient of $x$ , giving $- 4$ , and form the square:
$x-4)^2=x^2-8x+16.$
Adjust the constant because 16 has been added inside the square:
$x2−8x+11=(x−4)2−16+11=(x−4)2−5.\begin{aligned} x^2-8x+11 &=(x-4)^2-16+11\\ &=(x-4)^2-5. \end{aligned}$
Since $(x−4)2≥0(x-4)^2\geq 0$ , the smallest possible value occurs when $x = 4$ . The minimum value is therefore $- 5$ , and the vertex is $(4, - 5)$ .

If the coefficient of $x^2$ is not one, first take it outside the relevant terms.

Example

Completing the square with a leading coefficient

Write $3x^2+12x-7$ in completed-square form.

Factor 3 from the terms containing $x$ :
$3x^2+12x-7=3(x^2+4x)-7.$
Complete the square inside the brackets:
$x^2+4x=(x+2)^2-4.$
Substitute and simplify:
$3(x2+4x)−7=3((x+2)2−4)−7=3(x+2)2−19.\begin{aligned} 3(x^2+4x)-7 &=3\left((x+2)^2-4\right)-7\\ &=3(x+2)^2-19. \end{aligned}$
The vertex is $(- 2, - 19)$ .

Common Mistake

Missing the outside multiplier

In $3((x+2)2−4)3\left((x+2)^2-4\right)$ , both terms inside the brackets must be multiplied by 3. The constant contribution is $- 12$ , not $- 4$ .

Solving Quadratic Equations

To solve a quadratic equation, first rearrange it into the form

ax^2+bx+c=0.

You can then use factorisation, completing the square or the quadratic formula.

Definition

Quadratic formula

The solutions of $ax^2+bx+c=0$ , where $a≠0a\neq 0$ , are

x=−b±b2−4ac2a.x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

Example

Using the quadratic formula

Solve $2x^2+3x-7=0$ , giving exact answers.

Identify $a = 2$ , $b = 3$ and $c = - 7$ . Substitute these carefully:
$x=−3±32−4(2)(−7)2(2).x=\frac{-3\pm\sqrt{3^2-4(2)(-7)}}{2(2)}.$
Simplify the expression under the square root:
$3^2-4(2)(-7)=9+56=65.$
Therefore,
$x=−3±654.x=\frac{-3\pm\sqrt{65}}{4}.$

Tip

Choosing a method

Try factorisation first when integer factors are easy to see. Otherwise, the quadratic formula is usually the most reliable method for finding roots.

The Discriminant

The expression inside the square root in the quadratic formula is called the discriminant.

Definition

Discriminant

For $ax^2+bx+c=0$ , the discriminant is

Δ=b2−4ac.\Delta=b^2-4ac.

Its sign determines the number of real roots.

If $Δ>0\Delta>0$ , there are two distinct real roots.
If $Δ=0\Delta=0$ , there is one repeated real root.
If $Δ<0\Delta<0$ , there are no real roots.

Graphically, these correspond to a parabola crossing the $x$ -axis twice, touching it once, or not meeting it.

Three quadratic graphs showing two real roots, one repeated root and no real roots according to the discriminant

Example

Finding a condition for repeated roots

Find the values of $k$ for which $x^2+kx+9=0$ has a repeated root.

A repeated root occurs when the discriminant equals zero:
$k^2-4(1)(9)=0.$
Simplify and solve:
$k^2-36=0.$
Factorise:
$(k - 6) (k + 6) = 0,$
so $k = 6$ or $k = - 6$ .

Common Mistake

Dropping a negative sign

When calculating $b^2-4ac$ , put negative values of $b$ or $c$ in brackets. For example, if $c = - 3$ , then $- 4 a c$ becomes positive when $a > 0$ .

Quadratics in a Function of the Unknown

Some equations are not quadratics in $x$ itself but become quadratics after replacing a repeated expression with a new variable.

Definition

Quadratic in a function

An equation is quadratic in a function of the unknown if it can be written as

a[g(x)]^2+b[g(x)]+c=0,

where $g (x)$ is an expression involving $x$ .

Use a substitution such as $u = g (x)$ , solve the resulting quadratic in $u$ , and then return to the original variable.

Example

Solving a disguised quadratic

Solve $x^4-5x^2+4=0$ .

Since $x^4=(x^2)^2$ , let $u=x^2$ . The equation becomes
$u^2-5u+4=0.$
Factorise and solve for $u$ :
$(u - 1) (u - 4) = 0,$
so $u = 1$ or $u = 4$ .
Replace $u$ with $x^2$ and solve both equations:
$x2=1⇒x=±1,x2=4⇒x=±2.\begin{aligned} x^2&=1 &&\Rightarrow x=\pm1,\\ x^2&=4 &&\Rightarrow x=\pm2. \end{aligned}$
Therefore, the solutions are $x = - 2, - 1, 1, 2$ .

Common Mistake

Return to the original variable

Solving the substituted quadratic is not the end. If $u=x^2$ , each positive value of $u$ may produce two values of $x$ , while a negative value produces no real values of $x$ .

Exam technique

In the exam

Rearrange every quadratic equation so that one side is zero before choosing a method.
Use the graph, completed-square form and discriminant as connected information: roots are $x$ -intercepts, while the vertex controls the maximum or minimum.
Write discriminant conditions precisely: use $> 0$ for two distinct roots, $= 0$ for a repeated root and $≥0\geq0$ when at least one real root is required.
Keep exact roots as fractions or surds unless the question specifically requests decimal answers.

Self review

Check yourself

Can you complete the square for $2x^2-12x+5$ and identify its vertex?
What condition on $k$ makes $x^2+4x+k=0$ have two distinct real roots?
How would you solve $x^6-7x^3+6=0$ using a substitution?

What you'll learn

Starting Point: Quadratic Expressions

Graphs of Quadratic Functions

Completing the Square

Solving Quadratic Equations

The Discriminant

Quadratics in a Function of the Unknown

Review 1.5.3 Quadratic functions by teaching Genie

1.5.3 Quadratic functions

What you'll learn

Starting Point: Quadratic Expressions

Graphs of Quadratic Functions

Completing the Square

Solving Quadratic Equations

The Discriminant

Quadratics in a Function of the Unknown

Review 1.5.3 Quadratic functions by teaching Genie

1.5.3 Quadratic functions

What you'll learn

Starting Point: Quadratic Expressions

Graphs of Quadratic Functions

Completing the Square

Solving Quadratic Equations

The Discriminant

Quadratics in a Function of the Unknown

Review 1.5.3 Quadratic functions by teaching Genie

1.5.3 Quadratic functions Revision Guide

1.5.3 Quadratic functions

What you'll learn

Starting Point: Quadratic Expressions

Graphs of Quadratic Functions

Completing the Square

Solving Quadratic Equations

The Discriminant

Quadratics in a Function of the Unknown

Review 1.5.3 Quadratic functions by teaching Genie

1.5.3 Quadratic functions Revision Guide