Topic 1 Quadratic Function

Introduction

This Chapter focuses on Quadratic Equations

We will be looking at Drawing and Sketching
graphs of these

We are also going to be solving them using
various methods

As with Chapter 1, some of this material will
have been covered at GCSE level

Teachings for Exercise 2A

Quadratic Functions

Plotting Graphs

You need to be able to accurately plot graphs of

Quadratic Functions.

The general form of a Quadratic Equation is;

y = ax2 + bx + c

Where a, b and c are constants and a ≠ 0.

This can sometimes be written as;

f(x) = ax2 + bx + c

→ f(x) means ‘the function of x’

2A

Quadratic Functions

Plotting Graphs

You need to be able to accurately plot graphs

of Quadratic Functions.

2A

Example

a) Draw the graph with equation
y = x2 – 3x – 4 for values of x from
-2 to +5

b) Write down the minimum value
of y at this point

c) Label the line of symmetry

6

0

-4

-6

-6

-4

0

6

y

10

4

0

-2

-2

0

4

10
x2 -
3x

15

12

9

6

3

0

-3

-6

3x

25

16

9

4

1

0

1

4

x2

5

4

3

2

1

0

-1

-2

x

y = x2 – 3x - 4

BE CAREFUL! Subtract what is in the

‘3x’ box, from the ‘x2’ box.

And subtract 4 at the end…

The minimum value occurs at the x
value halfway between 4 and -1

Quadratic Functions

Plotting Graphs

You need to be able to accurately plot graphs

of Quadratic Functions.

-1

2A

Example

a) Draw the graph with equation
y = x2 – 3x – 4 for values of x from
-2 to +5

b) Write down the minimum value
of y

c) Label the line of symmetry

y = x2 – 3x - 4

x

-2

-1

0

1

2

3

4

5

y

6

0

-4

-6

-6

-4

0

6

y = x2 – 3x - 4

4

1.5

Substitute this value into the
equation:

y = x2 – 3x - 4

y = 1.52 – (3 x 1.5) - 4

y = -6.25

Quadratic Functions

Plotting Graphs

You need to be able to accurately plot graphs

of Quadratic Functions.

2A

Example

a) Draw the graph with equation
y = x2 – 3x – 4 for values of x from
-2 to +5

b) Write down the minimum value
of y

c) Label the line of symmetry

y = x2 – 3x - 4

x

-2

-1

0

1

2

3

4

5

y

6

0

-4

-6

-6

-4

0

6

y = x2 – 3x - 4
x = 1.5

y = -6.25

Teachings for Exercise 2B

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations

by factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated

root’

2B

Example

Solve the equation…

a)

29

x

x=

29

0

x

x−

=

(

9)

0

x x −

=

0x =

90

x −=

9x =

Subtract 9x

Factorise

Either ‘x’ or ‘x-9’
must be equal to

0

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations

by factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated

root’

2B

Example

Solve the equation…

b)

22

15

0

x

x−

−

=

Factorise

(

3)(

5)

0

x

x

+

−

=

30

x +=

50

x −=

3

x = −

5x =

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations

by factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated

root’

2B

Example

Solve the equation…

c)

22

9

50

x

x−

−=

Factorise

(2 )( )

0

x

x

=

Factorising this is slightly different.

→There must be a ‘2x’ at the start of a
bracket

→ The numbers in the brackets must still
multiply to give ‘-5’

→ The number in the second bracket will be
doubled when they are expanded though, so
the numbers must add to give ‘-9’ WHEN
ONE HAS BEEN DOUBLED

Using -5 and +1

→They multiply to give -5

→ If we double the -5, they add to give -9

→ So the -5 goes opposite the ‘2x’ term

(2

1)(

5)

0

x

x

+

−

=

1
2
x = −

5x =

or

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations

by factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated

root’

2B

Example

Solve the equation…

d)

26

13

50

x

x

+

−=
Factorise

Factorising this is even more difficult

→ The brackets could start with 6x and x,
or 2x and 3x (either of these would give the
6x2 needed)

→ So the numbers must multiply to give -5

→ And add to give 13 when either;

→One is made 6 times bigger

→One is made twice as big, and the
other 3 times bigger

Using 3x and 2x at the starts of the
brackets

And -1 and +5 inside the brackets…

→ They multiply to give -5

→ They will add to give 13 if the +5 is
tripled, and the -1 is doubled

→ So +5 goes opposite the 3x, and -1
opposite the 2x

(3 )(2 )

0

x

x

=

(3

1)(2

5)

0

x

x

−

+

=

3

10

x − =

2

50

x +=

13
x =
52
x = −

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations by

factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated root’

2B

Example

Solve the equation…

e)

25

18

23

x

x

x

−

+

=+
Subtract 2
Subtract 3x
28

16

0

x

x−

+

=

Factorise
(

4)(

4)

0

x

x

−

−

=

40

x −

=

4x =

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations by

factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated root’

2B

Example

Solve the equation…

f)

2

(2

3)

25

x −

=

Square root
both sides (2

possible
answers!)
2

3

5

x −= 

2

35

x −=

2

3

5

x −= −

2

8x =

4x =

2

2

x = −

1

x = −

Quadratic Functions

Solving by Factorisation

You need to be able to solve Quadratic Equations by

factorising them.

A Quadratic Equation will have 0, 1 or 2 solutions,

known as ‘roots’

If there is 1 solution it is known as a ‘repeated root’

2B

Example

Solve the equation…

g)

2

(

3)

7

x −

=

Square root
both sides (2

possible
answers!)

3

7

x −= 

3

7

x −=

3

7

x −= −

3

7

x =

−

3

7

x =

+

Teachings for Exercise 2C

Quadratic Functions

Completing the Square

Quadratic Equations can be written in
another form by ‘Completing the Square’

2C

2x

bx+

2

2

2

2

b

b
x





+

−











Example

Complete the square for the following
expression…

a)

28

x

x+

‘So b/2 is half of the

coefficient of x’

(

)

2

2

4

4

x +

−

If we check by

expanding our answer…

(

)

2

2

4

4

x +

−

(

)

2

4 (

4)

4

x

x

+

+

−

2

2

4

4

16

4

x

x

x

+

+

+

−

28

x

x+

Quadratic Functions

Completing the Square

Quadratic Equations can be written in
another form by ‘Completing the Square’

2C

2x

bx+

2

2

2

2

b

b
x





+

−











Example

Complete the square for the following
expression…

b)

212

x

x

+

‘So b/2 is half of the

coefficient of x’

(

)

2

2

6

6

x +

−

(

)

26

36

x +

−

Quadratic Functions

Completing the Square

Quadratic Equations can be written in
another form by ‘Completing the Square’

2C

2x

bx+

2

2

2

2

b

b
x





+

−











Example

Complete the square for the following
expression…

c)

23

x

x+

‘So b/2 is half of the

coefficient of x’

(

)

2

2

1.5

1.5

x +

−

(

)

2

1.5

2.25

x +

−

2

2

3

3

2

2
x





+

−












23

9

2

4
x



+

−








With

Decimals

With

Fractions

Quadratic Functions

Completing the Square

Quadratic Equations can be written in
another form by ‘Completing the Square’

2C

2x

bx+

2

2

2

2

b

b
x





+

−











Example

Complete the square for the following
expression…

d)

22

10

x

x

+

‘So b/2 is half of the

coefficient of x’

22(

5 )

x

x+

2

2

5

5
2
2

2
x









+

−
























25

25
2
2

4
x







+

−




















25

25
2
2

2
x



+

−







Factorise

first

Complete the
square inside
the bracket

You can work

out the
second
bracket

You can also
multiply it by
the 2 outside

Teachings for Exercise 2D

Quadratic Functions

Using Completing the Square

You can use the idea of completing the
square to solve quadratic equations.

This is vital as it needs minimal
calculations, and no calculator is needed
when using surds. (The Core 1 exam is
non-calculator)

2D

Example

Solve the following equation by completing
the square…

a)

28

10

0

x

x+

+

=

28

10

x

x+

= −

(

)

2

2

4

(4)

10

x +

−

= −

(

)

24

10 16

x +

= −

+

(

)

24

6

x +

=

4

6

x +

= 

4

6

x = − 

Subtract 10

Complete
the Square

Add 16

Square Root

Subtract 4

Quadratic Functions

Using Completing the Square

You can use the idea of completing the
square to solve quadratic equations.

This is vital as it needs minimal
calculations, and no calculator is needed
when using surds. (The Core 1 exam is
non-calculator)

2D

Example

Solve the following equation by completing
the square…

b)

22

8

70

x

x−

+=

Divide by 2

2
7
4

0
2
x

x−

+

=

2
7
4
2
x

x−

= −

(

)

2

2
7
2

( 2)
2
x −

− −

= −

(

)

2
1
2
2
x −

=

1
2
2
x −

= 1
2

2

x −

= 

1
2

2

x =



Subtract

7/2

Complete
the square

Add 4

Square Root

Add 2

Teachings for Exercise 2E

Quadratic Functions

The Quadratic Formula

You will have used the Quadratic

Formula at GCSE level.

You can also use it at A-level for

Quadratics where it is more difficult
to complete the square.

We are going to see where this

formula comes from (you do not need
to know the proof!)

2E

2

4

2

b

b

ac

a

− 

−

Quadratic Functions

The Quadratic Formula

2E

2

0

ax

bx

c

++=

2

0
b

c
x

x
a

a
+

+=

2
b

c
x

x
a

a
+

= −

2

2

2

2

b

b

c
x
a

a

a






+

−

= −












2

2

2

2

4

b

b

c
x
a

a

a




+

−

= −








2

2

2

2

4

b

b

c
x
a

a

a




+

=

−








2

2

2

2

4

2

4

4

b

b

ac
x
a

a

a




+

=

−








2

2

2
4

2

4

b

b

ac
x
a

a
−




+

=








2

2
4

2

4

b

b

ac
x
a

a
−
+

=

24

2

2

b

b

ac
x
a

a



−
+

=

24

2

2

b

b

ac
x
a

a
−
= −



24

2

b

b

ac
x
a

− 

−
=

Divide all by a

Subtract c/a

Complete the Square

(Half of b/a is b/2a)

Square the
2nd bracket

Add b2/4a2

Top and

bottom of 2nd

fraction

multiplied by

4a

Combine the

Right side

Square Root

Square Root
top/bottom
separately

Subtract

b/2a

Combine the

Right side

2

2

2

2

4

b

b

c
x
a

a

a




+

=

−








Quadratic Functions

The Quadratic Formula

You need to be able to recognise when
the formula is better to use.

Examples would be when the
coefficient of x2 is larger, or when
the 3 parts cannot easily be divided
by the same number.

2E

Example

Solve 4x2 – 3x – 2 = 0 by using the formula.

a = 4 b = -3 c = -2

24

2

b

b

ac
x
a

− 

−
=

2

3

3

(4 4

2)

2 4
x
−−

 −
=


3

9

32

8
x


− −
=

3

41

8
x

=

3

41

8
x
+
=
3

41

8
x
−
=

Teachings for Exercise 2F

Quadratic Functions

Sketching Graphs

You need to be able to sketch
a Quadratic by working out
key co-ordinates, and knowing
what shape it should be.

2F

y

x

y

x

y

x

y

x

y

x

y

x

24

2

b

b

ac
x
a

− 

−
=

b2 – 4ac is known as the

‘discriminant’

→ Its value determines
how many solutions the

equation has

2

0

ax

bx

c

+

+=

24

0

b

ac−



0a 

24

0

b

ac−

=

0a 

24

0

b

ac−



0a 

24

0

b

ac−



0a 

24

0

b

ac−

=

0a 

24

0

b

ac−



0a 

Quadratic Functions

Sketching Graphs

To sketch a graph, you need to work
out;

1) Where it crosses the y-axis

2) Where (if anywhere) it crosses the
x-axis

Then confirm its shape by looking at
the value of a, as well as the
discriminant (b2 – 4ac)

2F

Example

Sketch the graph of the equation;

y = x2 – 5x + 4

Where it crosses the y-axis

The graph will cross the y-axis where
x=0, so sub this into the original equation.

25

4

y

x

x

=

−

+

4y =
Co-ordinate (0,4)

Where it crosses the x-axis

The graph will cross the x-axis where y=0,
so sub this into the original equation.

25

4

y

x

x

=

−

+

2

0

5

4

x

x

=

−

+

0(

4)(

1)

x

x

=

−

−

1 or

4

x

x

=

=

Co-ordinates (1,0)

and (4,0)

(0,4)

(1,0)

(4,0)

Quadratic Functions

Sketching Graphs

To sketch a graph, you need to work
out;

1) Where it crosses the y-axis

2) Where (if anywhere) it crosses the
x-axis

Then confirm its shape by looking at
the value of a, as well as the
discriminant (b2 – 4ac)

y = x2 – 5x + 4

2F

(0,4)

(1,0)

(4,0)

Confirmation → a > 0 so a ‘U’ shape

→ b2 – 4ac

→ -52 – (4x1x4)

→ 9

→ Greater than 0 so 2 solutions

y

x

Quadratic Functions

Sketching Graphs

You can also use the information on
the discriminant to calculate
unknown values.

You need to remember;

‘real roots’ → b2 - 4ac > 0

‘equal roots’ → b2 – 4ac = 0

‘no real roots’ → b2 – 4ac < 0

2F

Example

Find the values of k for which;

x2 + kx + 9 = 0

has equal roots.

24

0

b

ac−

=

2(4 1 9)

0

k −

 

=

236

0

k −

=

2

36

k =

Sub in a, b and c from
the equation (b = k!)

Work out the bracket

Add 36

Square Root

Summary

We have recapped solving a Quadratic Equation

We have learnt how to use ‘completing the square’

We have also solved questions on sketching graphs and using
the ‘discriminant’