Higher Maths 2.1.2 - Quadratic Functions

1Higher Maths 2 1 2 Quadratic Functions

Any function containing an term is called a Quadratic

Function.

The Graph of a Quadratic Function


The graph of a Quadratic Function is

a type of symmetrical curve called a

parabola.

x

2

f ( x ) = ax 2 + bx + cGeneral Equation

of a Quadratic Function

a > 0

a < 0

Minimum turning

point Maximum turning

point

with a ≠ 0

turning point

Sketching Quadratic Functions


Before it is possible to sketch the graph of a

quadratic function, the following information must be

identified:• the nature of the turning point

i.e.

• the coordinates of the y-

intercept• the zeroes or ‘roots’ of the function

i.e. the x-intercept(s), if any

minimum or maximum

• the location of the axis of

symmetry

and coordinates of the turning

point

a > 0 or a < 0

substitute x

= 0solve f ( x ) =

0 :ax 2 + bx + c = 0

evaluate f ( x ) at

axis of symmetry

Perfect Squares


x 2 + 6 x +

2x

2 + 6 x

+ 2( )2

+ 9 – 9( ) x+ 3 – 7

A perfect square is an

expression that can be written

in the formExample

Complete the square

for

x 2 + 6 x +

2

= =

( ... )2

x 2 + 4 x +

4( )2x+ 2= x

2 + 5 x + 9= ( )2x+ ?

Step One:Separate number

term

Step Two:Try to form a perfect

square from remaining terms

Step Three:Remember to balance the extra number, then write

out result

Completing the Square


y = x 2 – 8 x +

19

= x 2 – 8 x

+ 19 = ( )2

+ 16 – 16( )

x – 4 + 3

It is impossible for a

square number to be

negative.

The minimum possible

value of is

zero.

( x – 4 )2

The minimum possible value of

y is 3 .x = 4 .This happens

when

The minimum

turning point is at( 4 , 3 )

A turning point can often be

found by completing the

square.Example

y-coordinatex-coordinate

Find the minimum turning

point of

Quadratic Equations can be

solved in several different

ways:

• using a graph to identify

roots

• factorising

• completing the square

• using the quadratic formula


Solving Quadratic Equations

Example

f ( x ) = 0Use the

graph to

solvex = -2 -2 5x =

5

or

Example 2

Solve6 x 2 + x – 15 =

0

6 x 2 + x – 15 =

0(2 x – 3)(3 x + 5)

= 0

The trinomial can

be factorised...

or

2 x – 3 = 0

3 x + 5 = 0

2 x = 3

3 x = -5x =3

2 x = -53

If a Quadratic Equation cannot be factorised, it is

sometimes possible to solve by completing the

square.


Solving Quadratic Equations by Completing the Square

Example

Solvex 2 – 4 x – 1 =

0The trinomial cannot be

factorised so complete the

square...x

2 – 4 x – 1 = 0

– 1 = 0( )

( )2

+ 4 – 4x 2 – 4 x

– 5 = 0x – 2

( )2 = 5x – 2

x – 2 = 5±

x = 2 5±

Now solve for

x ...

x ≈ 4.24 -0.24orNot

accurate!

most accurate answer

Quadratic Inequations


A Quadratic Inequation can be solved by using a

sketch to identify where the function is positive

or negative.Exampl

eFind the values of x for

which12 – 5 x – 2 x 2

> 0Factorise 12 – 5 x – 2 x

2

= 0 (4 + x)(3 – 2 x) = 0

The graph has

roots

x = -4 and

and a y-intercept at

12

-432

12

Now sketch the graph:

The graph is

positive for-4 < x <and negative

forx < -4

32 x >and

32

32

The Quadratic Formula


x = -b b 2 – (4

ac )

±

2 a

with a ≠ 0

f ( x ) = ax 2 + bx + c

If a quadratic function has roots, it is

possible to find them using a formula.

This is very useful if the roots cannot be found

algebraically, i.e. by factorising or completing the

square.The roots

of

are given

by

root

If roots cannot be found

using the quadratic

formula, they are

impossible to find.

no roots

Real and Imaginary Numbers

10

Higher Maths 2 1 2 Quadratic Functions

36= ±6

-36 =

It is impossible to

find the square

root of a negative

number.

In Mathematics, the square

root

of a negative number still

exists

and is called an imaginary

number.

It is possible for a quadratic

equation to have roots which

are not real.

1 real

root

no real

roots

2 real

roots

• If

• If

The Discriminant

11


x = -b b 2 – (4

ac )

±

2 a= b

2 – (4 ac )

The part of the quadratic formula inside the square root is

known

as the Discriminant and can be used to find the nature of

the roots.

The Discriminant

• If b 2 – (4 ac ) > 0 there are two real

roots.b 2 – (4 ac ) = 0 there is only one real

root.b

2 – (4 ac ) < 0 the roots cannot be calculated

and are imaginary or non-real.

(‘real and

unequal’)(‘real and

equal’)

Using the Discriminant to Find Unknown Coefficients

12


Example

The quadratic equation2 x 2 + 4 x + p

= 0Find all possible values of

p.

has real

roots.

b2 – (4 ac ) 0a = 2 b = 4 c = p

For real roots,

16 – 8 p 0

16 8 p

8 p 16p 2

The equation has

real roots for

.

p 2

(for the roots

are

imaginary or non-

real)

p >

2

Finding Unknown Coefficients using Quadratic Inequations

13


Exampl

eFind q given that x

2 + (q – 3) x + q = 0 has non-

real roots.

a = 1 b = (q – 3) c = q

b2 – (4 ac ) < 0

For non-real

roots,

(q – 3)

2 – 4 q < 0

q2 – 6q + 9 – 4 q < 0

q2 – 10q + 9 < 0

(q – 9) (q – 1) < 0

Sketch graph of the inequation:

1 9

q2 – 10q + 9 < 0for 1 < q < 9

The roots of the original equationare non-real when1 < < 9q

Straight Lines and Quadratic Functions

14


When finding points of intersection between a line and a

parabola:

• equate the functions

• rearrange into a quadratic equation equal to zero

• solve for

• substitute back to find

x

Exampl

eFind the points of

intersection ofy = x 2 + 3 x

+ 2

x 2 + 2 x =

0

y = x + 2

and

x 2 + 3 x + 2 = x

+ 2

x ( x + 2 ) = 0

x = 0 x = -2or

y = 2 y = 0or

y

Points of intersection are (-2,0)

and (0,2).

Tangents to Quadratic Functions

15


The discriminant can be used to find the number of

points of intersection between a parabola and a

straight line. • equate the functions and rearrange into an equation equal to zero

• evaluate the discriminant of the new quadratic equation

b 2 – (4 ac ) > 0 Two points of

intersection

b 2 – (4 ac ) = 0 One point of

intersection

b 2 – (4 ac ) < 0 No points of

intersection

the line is a tangent

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