quadratic funtions
TRANSCRIPT
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Quadratic Functions
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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
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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
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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?
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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 &
c0 8abel the line of s&mmetr&
& ) *+ 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
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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 &
c0 8abel the line of s&mmetr&
& ) *+ 3 4* 6 5
* 6+ 6! . ! + 4 5 7
& 9 . 65 69 69 65 . 9
& ) *+ 3 4* 6 5x = 1.5
& ) 69'+7
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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 =
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Quadratic $unctions
Solving by Factorisation
%ou need to be able to solve QuadraticEquations b& factorising them'
Quadratic Equation will have ." ! or +solutions" known as 1roots2
If there is ! solution it is known as a1repeated root2
+;
Example
Solve the equation?
b0$actorise
2 2 15 0 x x− − =
( 3)( 5) 0 x x+ − =
3 0 x + = 5 0 x − =
3 x = − 5 x =
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Quadratic $unctions
Solving by Factorisation
%ou need to be able to solve QuadraticEquations b& factorising them'
Quadratic Equation will have ." ! or +
solutions" known as 1roots2
If there is ! solution it is known as a1repeated root2
+;
Example
Solve the equation?
c0$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
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Quadratic $unctions
Solving by Factorisation
%ou need to be able to solve QuadraticEquations b& factorising them'
Quadratic Equation will have ." ! or +
solutions" known as 1roots2
If there is ! solution it is known as a1repeated root2
+;
Example
Solve the equation?
d0$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 = −
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Quadratic $unctions
Solving by Factorisation
%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 =
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Quadratic $unctions
Solving by Factorisation
%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?
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 = −
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Quadratic $unctions
Solving by Factorisation
%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?
g0Square rootboth sides /+
possibleanswers>0
2( 3) 7 x − =
3 7 x − = ±
3 7 x − = 3 7 x − = −
3 7 x = −3 7 x = +
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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+
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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?
b0
1So b+ is half of the
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 + −
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Quadratic $unctions
Completing the S!are
Quadratic Equations can be written inanother form b& 1Completing theSquare2
+C
Example
Complete the square for the followinge*pression?
c0
1So b+ is half of the
coefficient of *2
WithDecimals
With$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
+ − ÷
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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?
d0
1So b+ is half of the
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 + − ÷
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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 = − ±
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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 = ±
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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
− ± −
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Quadratic $unctions
#he $!a%ratic Form!la
+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
− ± −
=
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Quadratic $unctions
#he $!a%ratic Form!la
%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
−=
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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 <
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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= =
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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
&
*
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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 =
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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