Quadratic Funtions

7/23/2019 Quadratic Funtions

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Quadratic Functions



Introduction

This Chapter focuses on Quadratic Equations

We will be looking at Drawing and Sketchinggraphs of these

We are also going to be solving them usingvarious methods

s with Chapter !" some of this material willhave been covered at #CSE level



Quadratic $unctions

Plotting Graphs

%ou need to be able to accuratel& plot graphs ofQuadratic $unctions'

The general form of a Quadratic Equation is(

& ) a*+ , b* , c

Where a" b and c are constants and a - .'

This can sometimes be written as(

f/*0 ) a*+ , b* , c

f/*0 means 1the function of *2



Quadratic $unctions

Plotting Graphs

%ou need to be able to accuratel&plot graphs of Quadratic$unctions'

Example

a0 Draw the graph withequation & ) *+ 3 4* 3 5 forvalues of * from 6+ to ,7

b0 Write down the minimumvalue of & at this point

c0 8abel the line of s&mmetr&

9.65696965.9 &

!.5.6+6+.5!.*+

64*

!7!+:94.64694*

+7!9:5!.!5*+754+!.6!6+*

& ) *+ 3 4* 6 5

;E C<E$=8> Subtract what is in the14*2 bo*" from the 1*+2 bo*'

nd subtract 5 at the end?



The minimum value occurs at the *value halfwa& between 5 and 6!

Quadratic $unctions

Plotting Graphs

%ou need to be able to accuratel& plotgraphs of Quadratic $unctions'

-1

+

Example

a0 Draw the graph with equation & ) *+ 3 4* 3 5 for values of * from6+ to ,7

b0 Write down the minimum valueof &


& ) *+ 3 4* 6 5

* 6+ 6! . ! + 4 5 7

& 9 . 65 69 69 65 . 9

& ) *+ 3 4* 6 5

4

1.5

Substitute this value into theequation@

& ) *+ 3 4* 6 5

& ) !'7+ 3 /4 * !'70 6 5

& ) 69'+7



Quadratic $unctions

Plotting Graphs

%ou need to be able to accuratel& plotgraphs of Quadratic $unctions'

+

Example

a0 Draw the graph with equation & ) *+ 3 4* 3 5 for values of * from6+ to ,7

b0 Write down the minimum valueof &


& ) *+ 3 4* 6 5

* 6+ 6! . ! + 4 5 7

& 9 . 65 69 69 65 . 9

& ) *+ 3 4* 6 5x = 1.5

& ) 69'+7



Quadratic $unctions

Solving by Factorisation

%ou need to be able to solve QuadraticEquations b& factorising them'

Quadratic Equation will have ." ! or +solutions" known as 1roots2

If there is ! solution it is known as a1repeated root2

+;

Example

Solve the equation?

a0Subtract :*

$actorise

Either 1*2 or 1*6:2must be equal to

.

2 9 x x=2 9 0 x x− =

( 9) 0 x x − =

0 x = 9 0 x − =

9 x =



Quadratic $unctions



Quadratic Equation will have ." ! or +solutions" known as 1roots2


+;

Example

Solve the equation?

b0$actorise

2 2 15 0 x x− − =

( 3)( 5) 0 x x+ − =

3 0 x + = 5 0 x − =

3 x = − 5 x =



Quadratic $unctions



Quadratic Equation will have ." ! or +

solutions" known as 1roots2


+;

Example

Solve the equation?

c0$actorise

$actorising this is slightl& different'

There must be a 1+*2 at the start of abracket

The numbers in the brackets must stillmultipl& to give 1672

The number in the second bracket will bedoubled when the& are e*panded though"

so the numbers must add to give 16:2WAEB BE AS ;EEB D=;8ED

=sing 67 and ,!

The& multipl& to give 67

If we double the 67" the& add to give6:

So the 67 goes opposite the 1+*2 term

22 9 5 0 x x− − =

(2 )( ) 0 x x =(2 1)( 5) 0 x x+ − =

1

2 x = − 5 x =or



Quadratic $unctions



Quadratic Equation will have ." ! or +

solutions" known as 1roots2


+;

Example

Solve the equation?

d0$actorise

$actorising this is even more difficult

The brackets could start with 9* and *"

or +* and 4* /either of these would givethe 9*+ needed0

So the numbers must multipl& to give 67

nd add to give !4 when either(

ne is made 9 times bigger

ne is made twice as big" and theother 4 times bigger

=sing 4* and +* at the starts of thebrackets

nd 6! and ,7 inside the brackets? The& multipl& to give 67

The& will add to give !4 if the ,7 istripled" and the 6! is doubled

So ,7 goes opposite the 4*" and 6!opposite the +*

26 13 5 0 x x+ − =(3 )(2 ) 0 x x =(3 1)(2 5) 0 x x− + =

3 1 0 x − = 2 5 0 x + =1

3 x = 5

2 x = −



Quadratic $unctions


%ou need to be able to solve Quadratic Equations b&factorising them'

Quadratic Equation will have ." ! or + solutions"known as 1roots2

If there is ! solution it is known as a 1repeated root2

+;

Example

Solve the equation?

e0 Subtract +Subtract 4*

$actorise

2 5 18 2 3 x x x− + = +2 8 16 0 x x− + =

( 4)( 4) 0 x x− − =

4 0 x − =4 x =



Quadratic $unctions





+;

Example

Solve the equation?

f0Square rootboth sides /+

possibleanswers>0

2(2 3) 25 x − =

2 3 5 x − = ±

2 3 5 x − = 2 3 5 x − = −

2 8 x =

4 x =

2 2 x = −

1 x = −



Quadratic $unctions





+;

Example

Solve the equation?

g0Square rootboth sides /+

possibleanswers>0

2( 3) 7 x − =

3 7 x − = ±

3 7 x − = 3 7 x − = −

3 7 x = −3 7 x = +



Quadratic $unctions

Completing the S!are

Quadratic Equations can be written inanother form b& 1Completing theSquare2

+C

Example

Complete the square for the followinge*pression?

a0

1So b+ is half of the

coefficient of *2

If we check b&e*panding our answer?

2 x bx+

2 2

2 2

b b x

+ − ÷ ÷

2 8 x x+

( )2 24 4 x + −

( )2 24 4 x + −

( ) 24 ( 4) 4 x x+ + −

2 24 4 16 4 x x x+ + + −

2

8 x x+



Quadratic $unctions



+C

Example


b0


coefficient of *2

2 x bx+

2 2

2 2

b b x

+ − ÷ ÷

2 12 x x+

( )2 26 6 x + −

( ) 2

6 36 x + −



Quadratic $unctions



+C

Example


c0


coefficient of *2

WithDecimals

With$ractions

2 x bx+

2 2

2 2

b b x

+ − ÷ ÷

2 3 x x+

( )2 21.5 1.5 x + −

( )2

1.5 2.25 x + −

2 2

3 3

2 2 x + − ÷ ÷

2

3 9

2 4 x

+ − ÷



Quadratic $unctions



+C

Example


d0


coefficient of *2

$actorisefirst

Complete thesquare insidethe bracket

%ou can work

out thesecondbracket

%ou can alsomultipl& it b&the + outside

2 x bx+

2 2

2 2

b b x

+ − ÷ ÷

22 10 x x+

22( 5 ) x x+

2 25 5

22 2

x + − ÷ ÷

2

5 252

2 4 x

+ − ÷

25 25

2

2 2

x + − ÷



Quadratic $unctions

"sing Completing the S!are

%ou can use the idea of completingthe square to solve quadraticequations'

This is vital as it needs minimalcalculations" and no calculator isneeded when using surds' /The Core! e*am is non6calculator0

+D

Example

Solve the following equation b& completingthe square?

a0Subtract !.

Completethe Square

dd !9

Square <oot

Subtract 5

2 8 10 0 x x+ + =

2

8 10 x x+ = −( )

2 24 (4) 10 x + − = −

( )2

4 10 16 x + = − +

( )2

4 6 x + =4 6 x + = ±

4 6 x = − ±



Quadratic $unctions

"sing Completing the S!are

%ou can use the idea of completingthe square to solve quadraticequations'

This is vital as it needs minimalcalculations" and no calculator isneeded when using surds' /The Core! e*am is non6calculator0

+D

ExampleSolve the following equation b& completingthe square?

b0Divide b& +

Subtract +

Completethe square

dd 5

Square <oot

dd +

22 8 7 0 x x− + =

2 74 0

2

x x− + =

2 74

2 x x− = −

( )2 2 7

2 ( 2)2

x − − − = −

( )2 1

22

x − =

12

2 x − = ± 1

22

x − = ±

1

2 2 x = ±



Quadratic $unctions

#he $!a%ratic Form!la

%ou will have used the Quadratic$ormula at #CSE level'

%ou can also use it at 6level forQuadratics where it is moredifficult to complete the square'

We are going to see where thisformula comes from /&ou do notneed to know the proof>0

+E

2 4

2

b b ac

a

− ± −



Quadratic $unctions


+E

Divide all b& a

Subtract ca

Complete the Square/Aalf of ba is

b+a0

Square the+nd bracket

dd b+5a+

Top andbottom of +nd

fractionmultiplied b&

5a

Combine the<ight side

Square <oot

Square <oottopbottomseparatel&

Subtractb+a

Combine the<ight side

2 0ax bx c+ + =

2 0b c

x xa a

+ + =

2 b c x x

a a+ = −

2 2

2 2

b b c x

a a a

+ − = − ÷ ÷

2 2

22 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

−+ =

2 4

2 2

b b ac

x a a

± −

+ =2 4

2 2

b b ac x

a a

−= − ±

2 4

2

b b ac

x a

− ± −

=



Quadratic $unctions


%ou need to be able to recognisewhen the formula is better touse'

E*amples would be when thecoefficient of *+ is larger" orwhen the 4 parts cannot easil& bedivided b& the same number'

+E

Example

Solve 5*+ 3 4* 3 + ) . b& using the formula'

a ) 5 b ) 64 c ) 6+

2 4

2

b b ac xa

− ± −=

23 3 (4 4 2)

2 4 x

± − − × × −=

×

3 9 328

x ± − −=

3 41

8 x

±=

3 41

8 x

+=

3 41

8 x

−=



Quadratic $unctions

S&etching Graphs

%ou need to be able tosketch a Quadratic b&working out ke& co6

ordinates" and knowingwhat shape it should be'

+$

&

*

&

*

&

*

&

*

&

*

&

*b+ 3 5ac is known as the

1discriminant2

Its value determineshow man& solutions the

equation has

2 4

2

b b ac x

a

− ± −=

2 0ax bx c+ + =

24 0b ac

− >0a >

24 0b ac

− =0a >

24 0b ac

− <0a >

24 0b ac− >

0a <

24 0b ac− =

0a <

24 0b ac− <

0a <



Quadratic $unctions

S&etching Graphs

To sketch a graph" &ou need towork out(

!0 Where it crosses the &6a*is

+0 Where /if an&where0 it crossesthe *6a*is

Then confirm its shape b& lookingat the value of a" as well as thediscriminant /b+ 3 5ac0

+$

ExampleSketch the graph of the equation(

& ) *+ 3 7* , 5

'here it crosses the y-axis

The graph will cross the &6a*is where *)."

so sub this into the original equation'

Co6ordinate /."50

'here it crosses the x-axis

The graph will cross the *6a*is where &)."so sub this into the original equation'

Co6ordinates /!".0and /5".0

/."50

/!".0 /5".0

2 5 4 y x x= − +4 y =

2 5 4 y x x= − +20 5 4 x x= − +

0 ( 4)( 1) x x= − −

1 or 4 x x= =



Quadratic $unctions

S&etching Graphs

To sketch a graph" &ou need to workout(

!0 Where it crosses the &6a*is

+0 Where /if an&where0 it crossesthe *6a*is

Then confirm its shape b& looking atthe value of a" as well as thediscriminant /b+ 3 5ac0

& ) *+ 3 7* , 5

+$

/."50

/!".0 /5".0

Confirmation a F . so a 1=2 shape

b+ 3 5ac

67+ 3 /5*!*50

:

#reater than . so + solutions

&

*



Quadratic $unctions

S&etching Graphs

%ou can also use the informationon the discriminant to calculateunknown values'

%ou need to remember(

1real roots2 b+ 6 5ac F .

1equal roots2 b+ 3 5ac ) .

1no real roots2 b+ 3 5ac G .

+$

Example$ind the values of k for which(

*+ , k* , : ) .

has equal roots'

Sub in a" b and c fromthe equation /b ) k>0

Work out the bracket

dd 49

Square <oot

•

2 4 0b ac− =2 (4 1 9) 0k − × × =

2 36 0k − =2

36k =



Summar&

We have recapped solving a Quadratic Equation

We have learnt how to use 1completing the square2

We have also solved questions on sketching graphsand using the 1discriminant2

Documents

Quadratic Funtions