Quadratic functionsmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 02... · 2015-08-17 · 2...

2VCEVCEcocovverageerageAreas of studyUnit 1 • Functions and graphs

• Algebra

In thisIn this chachapterpter2A Expanding quadratic

expressions2B Factorising quadratic

expressions2C Factorising by completing

the square2D Solving quadratic equations

— Null Factor Law2E Solving quadratic equations

— completing the square2F The quadratic formula2G The discriminant2H Quadratic graphs —

turning point form2I Quadratic graphs —

intercepts method2J Using graphs to solve

quadratic equations2K Simultaneous quadratic

and linear equations

Quadraticfunctions

Ch 02 MM 1&2 YR 11 Page 55 Friday, June 29, 2001 11:07 AM

56

M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Expanding quadratic expressions

A quadratic expression is one that contains an

x

2

term, and possibly an

x

and a number(or constant) term. Examples of quadratic expressions include:

x

2

+

3

x

−

1

−

4

x

2

+

76

x

−

x

2

x

2

Quadratic expressions are produced when two linear expressions are multiplied.Consider the expansion (4

x

+

9)(2

x

−

3).When expanding brackets, ‘multiply everything by everything else’

as shown on the diagram at right. That is,First term

×

everything in the second brackets, thenSecond term

×

everything in the second brackets.The results of each part of the expansion are:

4

x

×

2

x

=

8

x

2

4

x

× −

3

= −

12

x

9

×

2

x

=

18

x

9

× −

3

= −

27

Parts and may be combined to give 6

x

, so we have:

(4

x

+

9)(2

x

−

3)

=

8

x

2

+

6

x

−

27

This method can be easily extended to deal with brackets containing more than twoterms.

The above method can be used on all types of binomial expansions, though a coupleof shortcuts for special cases are shown in the worked examples below.

14---

3 4

(4x + 9)(2x – 3)1

2

1 2 3 4

2 3

Expand the following.a (3x + 5)(6x − 7) b (2x − 9)2 c (5x − 3)(5x + 3) d −2(x − 9)(4 − x)

THINK WRITE

a Write the expression, and mark the required curved lines.

a

First term × everything in the second brackets gives 18x2 − 21x.Second term × everything in the second brackets gives 30x − 35.

= 18x2 − 21x + 30x − 35

Combine the middle (x) terms. = 18x2 + 9x − 35

b Write the expression. Recognise it as a ‘perfect square’.

b (2x − 9)2

Remember the shortcut for ( )2 type questions:Square the first term, double the product and square the last term.Square the first term to get 4x2.Double the product to get2 × (2x) × (−9) = −36x.Square the last term to get 81. = 4x2 − 36x + 81

1(3x + 5)(6x – 7)

2

3

4

1

2

34

5

1WORKEDExample

C h a p t e r 2 Q u a d r a t i c f u n c t i o n s 57

Expand (x − 7)(x + 2) − (2x − 1)(x + 4) and simplify.

THINK WRITE

Write the expression. (x − 7)(x + 2) − (2x − 1)(x + 4)Expand the first pair of brackets to get x2 − 5x − 14.Expand the second pair of brackets to get 2x2 + 7x − 4.Subtract the two expanded groups in the order given. Use new brackets for clarity as shown.

= (x2 − 5x − 14) − (2x2 + 7x − 4)

Apply the negative sign to the contents of the second brackets.

= x2 − 5x − 14 − 2x2 − 7x + 4

Collect like terms and simplify. = −x2 − 12x − 10

1

2

3

4

5

6

2WORKEDExample

THINK WRITE

c Write the expression. Recognise that the only difference between the two sets of brackets is the sign in the middle, so the answer is a ‘difference of squares’.

c (5x − 3)(5x + 3)

The shortcut for the expansion of a difference of squares is:(First term)2 − (second term)2.

= (5x)2 − 32

= 25x2 − 9

d Write the expression. d −2(x − 9)(4 − x)Rewrite the question so x is the first term in both brackets. This is not essential as long as all combinations of terms are multiplied in the next step.

= −2(x − 9)(−x + 4)

Expand the brackets first. = −2(−x2 + 13x − 36)Multiply the brackets’ contents by −2. = 2x2 − 26x + 72

1

2

12

34

remember1. To expand, multiply as follows.

(a) First term × everything in the second brackets, then(b) Second term × everything in the second brackets.

2. Perfect squares (ax + b)2 = a2x2 + 2abx + b2

3. Difference of squares (ax + b)(ax − b) = a2x2 − b2

4. Expand brackets first, then multiply if there is an ‘external’ factor in expressions like k(ax + b)(cx + d).

3 4

(4x + 9)(2x – 3)1

2

remember

58 M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Expanding quadratic expressions

Simplify your answers to questions in this exercise as fully as possible.1 Expand the following.

a (2x + 6)(x + 5) b (3x + 1)(4x − 3)c (5x − 7)(5x + 2) d (8x − 3)(6x − 1)e (x + 4)(7x − 9) f (x + 6)(x + 15)g (6x − 13)(2 − 3x) h (5 − x)(6 − x)i (x − 9)(9x − 1) j (4x + 21)(x − 3)

2 Expand using the shortcut for perfect squares.a (2x + 3)2 b (3x − 5)2

c (6x + 1)2 d (7x − 6)2

e (x − 8)2 f (x + 13)2

g (−2x + 9)2 h (2x + 9)2

i (4 − 3x)2 j (6 − x)2

3 Expand (remember the shortcut for difference of squares).a (2x − 6)(2x + 6) b (3x + 5)(3x − 5)c (6x + 1)(6x − 1) d (2x − 9)(2x + 9)e (11x + 3)(11x − 3) f (x − 12)(x + 12)g (x + 6)(x − 6) h (7 − 2x)(7 + 2x)i (1 − x)(1 + x) j (5x + 1)(5x − 1)

4 Expand:a 2(x + 6)(3x + 5) b 3(x − 4)(2x + 7)c −4(x + 2)(2x − 5) d 3(4x − 9)(2x − 1)e 8(3x − 1)(4x − 1) f −7(2x + 3)(5x − 10)g 4(4 − x)(7 − x) h −5(7x − 4)(2 − x)i a(2x + 9)(x − 6) j −b(8 − 2x)(x + 4)

5 Mixed expansions.a (6x − 13)(6x + 13) b (5x + 2)(12x − 5)c −3(7 + 2x)(x − 8) d (3x + 11)2

e (2x − 14)(2x + 14) f (6x − 5)2

g (x + 16)(x − 9) h −(x − 4)(x + 4)i −4(5x − 3)2 j 5(2x + 7)(2x − 7)

6 Expand and simplify.a (x − 9)(x + 2) + (x + 4)(x − 4) b (3x + 7)(2x − 1) + (4x − 3)(3x − 4)c 2(x + 5)2 + 5(2x + 7)(x − 3) d (x − 12)(x + 12) − 3(x + 1)(x + 5)e 4(x − 3)(3x + 5) − 2(2x + 1)2 f 6(2x − 5)(2x + 5) − (x − 6)(x + 6)

g hi (2x + 1)(4x + 7) j (5x + )(3x − 5)

7 Expand.a (2a + 4)(2a + b) b (x + 2y)(3x − 5y)c (6 − 7c)(2 − 7c) d (u + 4v)2

e (6r − s)(2r + 5s) f (3u − 2t)(3u + 2t)g 2(h − 8k)(h + 8k) h −3(m − 6n)2

2A

GCpro

gram

Expanding

Mathca

d

Expanding

WORKEDExample

1a

SkillSH

EET 2.1

WORKEDExample

1b

SkillSH

EET 2.2

WORKEDExample

1c

WORKEDExample

1d

WORKEDExample

2

x 3+( ) x 3–( ) 3x 2+( ) 4x 2–( )12--- 1

3---

C h a p t e r 2 Q u a d r a t i c f u n c t i o n s

59

Factorising quadratic expressions

‘Factorising’ means writing an expression as a product of two or more factors. Theeasiest type of factorisation is that involving a common factor.

For example, 4

x

2

+

8

x

−

12 can be factorised to 4(

x

2

+

2

x

+

3).Always look out for this type of factorisation first. The method of ‘sensible guesses’

is one that will be used in this section to factorise quadratic expressions or trinomials.This is sometimes referred to as ‘factorising by inspection’, or the ‘cross product’method.

Other methods rely on recognising perfect squares or difference of squares.

Perfect squares

a

2

x

2

+

2

abx

+

b

2

=

(

ax

+

b

)

2

A test for a perfect square is as follows: For example:1. Arrange the expression in order of decreasing powers of

x

.2. Does

=

middle term coefficient?(Coefficients are numbers or pronumerals in front of

x

2

and

x

terms.)3. If yes, you have a perfect square.

Difference of squares

ax

2

−

b

=

(

x

+

)(

x

−

)

where and may simplify to a whole number.

For example, 49

x

2

−

9

=

(7

x

+

3)(7

x

−

3).

4x2 + 20x + 25

Double

Multiply(2x x 5)

2x10x

5

first term coefficient last term 2××

a b a b a b

Factorise the following.a −6x2y + 15xy b x2 + 7x + 12 c 6x2 + 28x − 48 d 27x2 − 75 e 9x2 − 30x + 25

THINK WRITE

a Write the expression. a −6x2y + 15xyTake out a common factor of –3xy. Make the common factor negative so the leading term inside the brackets will be positive.

= −3xy(2x − 5)

b Write the expression. b x2 + 7x + 12Look for a common factor. (None)Write (x )(x ) and look for factors of +12 that add to give +7.+4 and +3 achieve this.

= (x + 4)(x + 3)

Note: The ‘. . . factors of _ which add to give _’ approach works only when there is a single x2 in the expression to be factorised.

c Write the expression. c 6x2 + 28x − 48Is there a common factor? Yes (2). = 2(3x2 + 14x − 24)

12

123

4

12

3WORKEDExample

Continued over page

Ch 02 MM 1&2 YR 11 9 Friday, June 29, 2001 11:07 AM


THINK WRITEAttempt to factorise by inspection. Write 2(3x )(x ) and try factors of −24.A few possibilities are:(3x + 12)(x − 2) = 3x2 + 6x − 24 NO(3x + 1)(x − 24) = 3x2 − 71x − 24 NO(3x + 4)(x − 6) = 3x2 − 14x − 24 NO(Nearly, just a wrong sign on the x term.)(3x − 4)(x + 6) = 3x2 + 14 − 24 YES = 2(3x − 4)(x + 6)

d Write the expression. d 27x2 − 75Look for a common factor. There is one (3). = 3(9x2 − 25)Recognise the difference of squares in the brackets.

= 3(3x + 5)(3x − 5)

e Write the expression. e 9x2 − 30x + 25There are no common factors.There seem to be a few square numbers in the expression, which looks suspiciously like a perfect square. The square root of the first term is 3x, and the square root of the last term is 5 or −5. Since we need a negative middle term, take –5. Double the product of these is 2 × 3x × −5 = −30x, which is the middle term, so we have a perfect square.

= (3x − 5)2

3

123

123

Factorise: a (x + 3)2 − 7 b (x − 6)2 + 5(x − 6) + 6THINK WRITEa Write the expression, and recognise a

difference of squares.a (x + 3)2 − 7

× = [(x + 3) + ][(x + 3) − ]= (x + 3 + )(x + 3 − )

b Write the expression, and notice the repeated portion, (x − 6).

b (x − 6)2 + 5(x − 6) + 6

Let X = (x − 6) and rewrite the expression. Let X = (x − 6)So X2 + 5X + 6

Factorise the new version of the expression. = [X + 2][X + 3]Replace X with x − 6. = [(x − 6) + 2][(x − 6) + 3]Simplify. = (x − 4)(x − 3)

1

2 ( first term second term)+

( first term second term)–

7 7

7 7

1

2

345

4WORKEDExample

remember1. Look for a common factor first.2. General quadratics: Write down (_x )(_x ) and try factors of the last term.3. Perfect squares: a2x2 + 2abx + b2 = (ax + b)2

4. Difference of squares: ax2 − b = ( + )( − )ax b ax b

remember


61

Factorising quadratic expressions

Factorise the following.

1

2

3

4

5

6

7

8

a

Which of the factors below could be multiplied by (

x

+

5) to get

x

2

+

14

x

+

45?

b

The factorised form of 6

x

2

– 67

x

– 60 is:

c

A factor of 6

x

2

– 54 is:

a

5

x

2

y

3

+

20

xy

2

b

9

x

2

−

36

x

c

−

8

ax

2

+

14

ax

d

6

x

−

24

x

2

e

7

p

2

q

−

21

p

+

7

f

22

r

4

s

3

+

11

r

2

s

g

9

m

5

−

24

m

2

h

g

4

h

2

+

4

g

3

h

5

i

5

x

2

+

10

x

+

75

j

−

3

x

2

−

18

x

+

6

a

x

2

+

16

x

+

63

b

x

2

−

4

x

−

21

c

x

2

−

17

x

+

66

d

2

x

2

+

22

x

+

48

e

−

3

x

2

−

6

x

+

24

f

4

x

2

−

16

x

−

20

g

−

x

2

−

4

x

+

96

h

6x2 − 11x − 35 i 6x2 − 37x + 45j −40x2 + 46x + 14

a 4x2 − 9 b 81x2 − 25 c x2 − 100d 2x2 − 32 e 3x2 − 108 f 18x2 − 162g (x + 1)2 − 4 h (3x − 4)2 − 1 i 2(x + 3)2 − 98j −3(5x − 8)2 + 27

a x2 + 14x + 49 b 16x2 − 24x + 9 c 25x2 − 10x + 1d 2x2 + 12x + 18 e 3x2 − 12x + 12 f −18x2 − 84x − 98g 72x2 − 24x + 2 h x2 + 2 + 3 i 4x2 − 4 + 5j 4x2 − 12 x + 18

a (x + 3)2 − 9 b (x − 3)2 − 16 c (2x + 7)2 − 36d (3x − 2)2 − 81 e 2(x + 1)2 − 8 f −3(x − 4)2 + 48g 4(2x − 3)2 − 36 h 50(3x + 2)2 − 98 i (6 − x)2 − 1j (2 − 3x)2 − 64

a (x + 7)2 + 9(x + 7) + 20 b (x − 2)2 + 7(x − 2) + 10c 6(x − 3)2 + 13(x − 3) + 5 d 4(x + 8)2 + 3(x + 8) − 10e (2x + 7)2 − 9(2x + 7) + 8 f 12(3x − 11)2 − 19(3x − 11) − 18

a 3x2 − 24x − 27 b 25x2 + 10x + 1 c (x − 13)2 − 2(x − 13) + 1d x2 − 28x + 196 e 196x2 − 49 f 60x2 + 40x + 5g 60x2 − 5x h 9 − 9x2y2 i 4(3x − 1)2 − (x + 2)2

j −12x2 + 70x + 98

A (x − 19) B (x + 5) C (x + 9) D (x + 14) E (x + 31)

A 2(3x + 5)(x + 6) B 2(3x − 5)(x − 6) C (3x + 6)(2x + 10)D (6x − 5)(x + 12) E (6x + 5)(x − 12)

A (x − 3) B (x – 6) C (x – 9) D (6x – 9) E 6x

2B

Mathcad

Factorising

WW

ORKEDORKEDE

Examplexample

3a

WW

ORKEDORKEDE

Examplexample

3b, c

WW

ORKEDORKEDEExamplexample

3d

WWORKEDORKEDEExamplexample3e

3x 5x2

WWORKEDORKEDEExamplexample4a

WWORKEDORKEDEExamplexample4b

mmultiple choiceultiple choice



Factorising by completing the squareFactorising by inspection (using ‘sensible guesses’) works best when whole numbersare involved — for example (2x + 3)(x − 7).

Sometimes it is clear (perhaps after several guesses fail) that it is impossible to findwhole numbers to complete a factorisation. For example, with x2 + 6x − 1, there is nopair of factors of 6 that add to −1! In such cases, completing the square may be the wayto go.

The method of completing the square involves ‘manufacturing’ a perfect square sothat an expression may be factorised as a difference of squares.

Because this method often produces surds (numbers like that cannot beexpressed as a ratio of whole numbers), factorising this way is sometimes referred to as‘factorising over R’, where R is the set of real numbers that includes surds.

3

Use the method of completing the square to factorise the following over R.a x2 + 6x − 1 b x2 − 7x + 8

THINK WRITE

a Write the expression. a x2 + 6x − 1Halve the x coefficient and jot down the result in a margin.

= 3

Square the value in the margin (to get 32 = 9 in this case) and add then subtract it as shown. Don’t forget the constant term (−1 in this case). Since the same value has been added and subtracted, the expression is equivalent to that in the question.

= x2 + 6x + 9 − 9 − 1

Combine the first three terms as a perfect square, as x2 + 6x + 9 = (x + 3)2. The number in brackets should be the one jotted in the margin earlier (3).

= (x + 3)2 − 10

Recognise a difference of squares. = [(x + 3) + ][(x + 3) − ]

Simplify. = (x + 3 + ) (x + 3 − )

b Write the expression. b x2 − 7x + 8Halve the x coefficient and jot down the result in a margin.Square the value in the margin (to get

= in this case) and add then subtract it as shown. Don’t forget the constant term (8 in this case).

= x2 − 7x + − + 8

1

262---

3

4

5 10 10

6 10 10

1

272---

372---

2 494------

494------ 49

4------

5WORKEDExample


The graphics calculator program CTSQUARE may be used to show the steps up to thestage of obtaining a difference of squares (step 5 in part b of the previous worked example).

The last line of the screen view may be factorised using a difference of squares asshown in steps 6 to 8 in part b of the previous worked example.

THINK WRITE

Combine the first three terms as a perfect square, since x2 − 7x + = (x − )2.The number in brackets should be the one jotted in the margin earlier (− ).

= (x − )2 − +

Combine the last two terms. = (x − )2 − +

= (x − )2 −

Recognise a difference of squares. = [(x − )2 + ][(x − )2 − ]Simplify the surd part if possible. = (x − + )(x − − )

Use a common denominator for the last two terms if desired.

or

4

494------ 7

2---

72---

72--- 49

4------ 8

1---

572--- 49

4------ 32

4------

72--- 17

4------

672--- 17

4------

72--- 17

4------

772--- 17

2---------- 7

2--- 17

2----------

8 x7 17–

2-------------------–

x7 17+

2-------------------–

GC program

Completingthe square

remember1. Use completing the square when whole number factors are not apparent.2. Halve and square the x coefficient, then add and subtract this new term.3. Form a perfect square from three of the terms.4. Continue to factorise using a difference of squares.

remember


Factorising by completing the square

1 Use the method of completing the square to factorise the following over R.a x2 + 4x − 3 b x2 + 10x + 20c x2 + 6x + 7 d x2 + 2x − 7e x2 + 8x + 13 f x2 − 4x − 1g x2 − 12x + 19 h x2 − 2x − 5i x2 − 8x + 10 j x2 − 6x − 4

2 Factorise:a x2 + 3x + 1 b x2 + 5x − 3c x2 − 7x + 2 d x2 − x − 1e x2 + 9x + 4 f x2 + 11x − 6g x2 − 3x + 5 h x2 + 5x + 2i x2 − 13x − 1 j x2 + x − 3

3

Which of the following could be added to x2 + 36x + 1 to assist factorising usingcompleting the square?

4

Which of the following could be added to x2 − 17x − 8 to assist factorising usingcompleting the square?

5

The expression x2 + 6x − 2 factorises to:

A (x − 3 + )(x − 3 − )B (x + 3 + )(x + 3 − )C (x + 3 + )(x + 3 − )D (x − 3 + )(x − 3 − )E (x + 3 + )(x + 3 − )

6

The expression x2 + 4x + 12:A can be factorised using whole numbersB can be factorised using the method of completing the squareC cannot be factorised using the methods covered in this topic so farD is a linear expression, and so cannot be factorisedE is already factorised.

7 Challenge: Factorise

A 6 B 18 C 72 D 144 E 324

A 4 B C 16 D E

a 3x2 − 18x − 3 b 2x2 + 10x + 4c −5x2 − 10x + 15 d −12x2 + 4x − 8

2CWORKEDExample

5a

GCpro

gram

Completing the square

WORKEDExample

5b



174------ 17

2------ 289

4---------


2 2

11 11

7 7

10 10

6 6


*


Solving quadratic equations —Null Factor Law

A quadratic equation is a quadratic expression that is put equal to zero, or which maybe put in such a form. Recall that a linear equation’s highest power of x (or anotherpronumeral) is 1, and linear equations have one solution. Quadratic equations have ahighest x (or another pronumeral) power of 2, and have at most two solutions.

Consider the quadratic equation:

x2 − 4 + 3 = 0On factorising, we have: (x − 1)(x − 3) = 0

or A × B = 0where A = (x – 1) and B = (x – 3).

For A × B to equal zero, either A or B or both must be zero.This is known as the ‘Null Factor Law’.That is, A = 0 or B = 0So (x − 1) = 0 or (x − 3) = 0Solving these two equations gives:

x − 1 = 0 or x − 3 = 0x = 1 or x = 3

Solve:a (x + 5)(7x − 11) = 0 b 12x2 − 11x − 15 = 0 c 4x2 + 25 = 20x d 6x2 = 54

Continued over page

THINK WRITE

a Write the equation. a (x + 5)(7x − 11) = 0Note that the equation is already factorised.Apply the Null Factor Law and solve two mini-equations.

x + 5 = 0 or 7x − 11 = 0x = −5 or 7x = 11x = 5 or x =

b Write the equation. b 12x2 − 11x − 15 = 0Factorise using sensible guesses. This may take several attempts. Just neatly cross out any failed attempts.

(4x + 3)(3x − 5) = 0

Solve two equations. 4x + 3 = 0 or 3x − 5 = 04x = −3 or 3x = 5

x = − or x =

c Write the equation. c 4x2 + 25 = 20xRearrange (transpose) so all terms are on the side on which the x2 term is positive. Write terms in order of decreasing powers of x.

4x2 − 20x + 25 = 0

12

3

117------

12

3

34--- 5

3---

12

6WORKEDExample


THINK WRITE

Factorise (in this case by recognising a perfect square).

(2x − 5)2 = 0

Solve (one equation here, or two identical equations if you like).

2x − 5 = 02x = 5

x =

d Write the equation. d 6x2 = 54Rearrange. 6x2 − 54 = 0Take out the common factor. 6(x2 − 9) = 0Notice a difference of squares. 6(x + 3)(x − 3) = 0Solve. x + 3 = 0 or x − 3 = 0

x = −3 or x = 3

3

4

52---

12345

The height of a triangle is 5 cm more than its base length. If the area of the triangle is 18 cm2, find the base length and height.

THINK WRITE

Define the length and height in terms of a pronumeral. Here, x is used.

Let base length = xThen height = x + 5

Draw a diagram.

Use the given information about area to form an equation involving the pronumeral.

Area = bh

18 = (x)(x + 5)Rearrange and form a quadratic equation. 36 = x(x + 5)

36 = x2 + 5xx2 + 5x − 36 = 0

Factorise. (x + 9)(x − 4) = 0Solve. x = −9 or x = 4Discard the negative answer, as length in this context must be positive.

x = 4 (as x > 0)

Calculate the height (x + 5). x + 5 = 4 + 5 = 9

Write the answer in words. Base length = 4 cm, height = 9 cm.

1x+5

x2

312---

12---

4

56

7

8

9

7WORKEDExample

rememberQuadratic equations:1. Factorise.2. Put each factor equal to zero.3. Solve two mini-equations.

remember


Solving quadratic equations — Null Factor Law

1 Solve:a (x + 6)(2x + 3) = 0 b (x + 2)(5x + 7) = 0c (x + 1)(6x − 1) = 0 d (x − 5)(x + 2) = 0e (3x − 5)(5x − 11) = 0 f (7x + 12)(9x + 2) = 0g (16x + 8)(2x − 6) = 0 h (7 + x)(23 − x) = 0i (6 − 7x)(x + 6) = 0 j (x + 0)(x + 4) = 0k x(x + 2) = 0 l −3x(x − 1) = 0

2 Factorise and solve:a x2 + 6x − 72 = 0 b x2 − 13x + 42 = 0c 2x2 − 7x − 4 = 0 d 4x2 + 13x + 10 = 0e 6x2 + 5x − 25 = 0 f 2x2 + 12x − 110 = 0g 3x2 − 27 = 0 h x2 + 16x + 64 = 0i 32x2 − 16x + 2 = 0 j 12 − 75x2 = 0k x2 − 6x = 0 l 15x − 3x2 = 0

3 Rearrange and solve:a x2 + 45 = 14x b 6x2 + 7x = 49c 1 + 8x = −16x2 d 36x2 = 1e −32x = 6x2 + 10 f 27 + 12x2 = 36xg 32x2 = 162 h 5x2 = 8x

i j

4Solutions to the equation (x – 3)(x + 7) = 0 are:

5Which of the following is a solution of 2x2 − 11x − 13 = 0?

6The solutions to a quadratic equation are x = −9 and x = . The equation could be:

7 The width of a rectangle is 3 cm less than its length. If the area of the rectangle is40 cm2, find the length and width.

A x = −3 and x = −7 B x = −3 and x = 7C x = 3 and x = 7 D x = 3 and x = −7E x = 4 and x = −21

A − B − C D 1 E

A (x − 9)(x − ) = 0 B 5(x − 9)(x − 2) = 0C 2(x + 9)(x − 5) = 0 D (x + 9)(5x − 2) = 0E (x + 9)(2x − 5) = 0

2D

GC program

Quadraticequations

EXCEL Spreadsheet

Quadraticequations

WORKEDExample

6a

WORKEDExample

6b

WORKEDExample

6c, d

x2

2----- 7x

2------=

x2

5----- 6

5---+ 7x

5------=



213------ 13

2------ 2

13------ 13

2------

mmultiple choiceultiple choice25---

25---

WORKEDExample

7


8 An engineer’s diagram of a rectangularcomponent is shown at right. If the totalarea of the surface shown is 63 cm2, findthe value of x.

9 The temperature, T, (in °C) in an office follows the equation T = t2 − 5t + 30, wheret is the time (in hours) the office’s climate control system has been operating. Howlong does it take the temperature to reach 20°C?

10 A number pattern follows the rule n2 + 3n + 2; that is, the number pattern is 6, 12,20, . . . (found by substituting n = 1, n = 2, n = 3 into the rule).

a Find the next value by substituting n = 4 into the rule.b What positive n value gives the number 210 when substituted into the rule?

11 The diagram below demonstrates the idea of ‘rectangular’ numbers.

The formula N = w(w + 1) gives the value of a rectangular number based on a shapewith a width of w dots.

a Which rectangular number has a width of 6?b What is the width of the rectangular number 272?

12 The amount of bending, B mm, of a particular wooden beam under a load is given byB = 0.2m2 + 0.5m + 2.5, where m kg is the mass (or load) on the end of the beam.What mass will produce a bend of 8.8 mm?

13 A window washer drops a squeegeefrom a scaffold 100 m off the ground.The relationship between the heightof the squeegee (h), in metres, andthe length of time it has beenfalling (t), in seconds, is given byh = 100 − 5t2.

a When does the squeegee passa window 30 m off the ground?

b How long does it take for thesqueegee to hit the ground?

x

x6 cm

4 cm

58---

w = 1

N = 1

w = 2

N = 6

w = 3

N = 12

WorkS

HEET 2.1


Fixed point iterationFixed point (or simple) iteration is a way of solving equations numerically rather than algebraically. To use this method, the equation to be solved must be manipulated into the form x = g(x).

For example, if the equation to be solved is f(x) = x2 – 7x + 1 = 0, a possible manipulation is

x2 − 7x + 1 = 0x2 + 1 = 7x

= x

In this case, we have x = g(x), where g(x) =

Fixed point iteration now works as follows:1 Make an initial guess, say x = 1.

2 Substitute the guess into g(x). g(x) = = = = 0.2857

3 Substitute the improved guess, 0.2857. g(x) = = = 0.1545



Each step above is called an ‘iteration’. If accuracy to 3 decimal places is acceptable, we may halt iterations here, as the last two answers equal 0.146 to this degree of accuracy. The process is said to have ‘converged’ to the fixed point 0.146.

Checking this solution we have: x2 − 7x + 1 = 0.1462 − 7 × 0.146 + 1 = 0 as required.

Exercise1 The following equations are already in the form x = g(x). Find a solution using

fixed point iteration with an initial guess of x = 1.

a x = b x =

2 Find a solution to each of the following (correct to 3 decimal places) using fixed point iteration. First manipulate each equation into the form x = g(x) in a similar way to that shown above.a x2 − 9x + 2 = 0 b x2 + 5x − 7 = 0c x2 − 4x + 1 = 0 d 2x2 − 11x + 8 = 0

3 Show that an alternative manipulation of x2 − 7x + 1 = 0 into the form x = g(x)

is x = . Does this form converge using fixed point iteration? If so, state the solution found.

4 Find a manipulation of 4x2 + x − 3 = 0 that converges using fixed point iteration. State the solution.

5 Investigate when equations of the form x2 − bx + 1 are likely to converge to a

solution using fixed point iteration with a manipulation of the form x = .

GC program

Fixedpoint

iteration

x2 1+7

--------------

x2 1+7

--------------

x2 1+7

-------------- 12 1+7

-------------- 27---

x2 1+7

-------------- 0.28572 1+7

----------------------------

x2 1+7

-------------- 0.15452 1+7

----------------------------

x2 1+7

-------------- 0.14632 1+7

----------------------------

x2 6+5

-------------- x2 2–3

--------------

7x 1–

x2 1+b

--------------


Solving quadratic equations — completing the square

Recall that when you cannot factorise quadratics by the method of sensibly guessingwhole numbers, the method of completing the square may be used. Completing thesquare may also be used to solve quadratic equations that don’t appear to easily factorise.

Solve, giving answers in exact (surd) form.a x2 + 10x + 2 = 0 b x2 − 7x + 1 = 0 c x2 − 8 = 0

THINK WRITE

a Write the equation. Notice that there are no factors of 2 that add to 10, so this equation is a candidate for completing the square.

a x2 + 10x + 2 = 0

Halve the x coefficient and write it in the margin. Here, = 5.

= 5

Square the margin number and add and subtract it as shown. Here we add and subtract 52 = 25.

x2 + 10x + 25 − 25 + 2 = 0

Partially simplify (the last two values) as shown.

x2 + 10x + 25 − 23 = 0

Form a perfect square with the first three terms. Check that the number in the margin matches the one in brackets.

(x + 5)2 − 23 = 0

Take the constant term (in this case the −23) to the other side of the equation, remembering to swap the sign.

(x + 5)2 = 23

Take the square root of both sides. (x + 5) = Solve for x. x = −5

b Write the equation. b x2 − 7x + 1 = 0Decide that completing the square is appropriate.Halve the x coefficient (to get − ). Note it in the margin.Square the margin value (to get ), and add and subtract it.

x2 − 7x + − + 1 = 0

Partially simplify. x2 − 7x + − + = 0

x2 − 7x + − = 0

Form a perfect square. Check that the number in the margin matches the one in brackets.

(x − )2 − = 0

1

2102------

102------

3

4

5

6

7 23±8 23±

12

372--- 7

2---–

4494------ 49

4------ 49

4------

5494------ 49

4------ 4

4---

494------ 45

4------

672--- 45

4------

8WORKEDExample


Solving x2+ bx+c=0

The general quadratic equation x2 + bx + c = 0 can be solved using a method like that in the previous worked example. Show that the solution to the general quadratic equation

x2 + bx + c = 0 is x = .

(The solution to equations of the form ax2 + bx + c = 0 is discussed in the next section.)

THINK WRITE

Solve for x. Note that the surd part has been simplified where possible.

(x − )2 =

x − = ±

= ±

= ±

x = ±

=

c The equation may be solved for x immediately, without completing the square.

c x2 − 8 = 0

Rearrange the equation. x2 = 8Take the square root of both sides and simplify.

x = ±

= ±

= ± 2

772--- 45

4------

72--- 45

4------

9 5×4

------------

3 52

----------

72--- 3 5

2----------

7 3 5±2

-------------------

1

23 8

4 2×

2

–b b2 4c–±

2----------------------------------

rememberWhen completing the square to solve:1. Halve the x-coefficient (note in margin).2. Square the margin value, and add and subtract it.3. Form a perfect square and solve for x.

remember


Solving quadratic equations — completing the square

1 Solve, giving answer in exact (surd) form.a x2 + 8x + 1 = 0 b x2 + 12x + 3 = 0c x2 + 4x − 2 = 0 d x2 − 6x + 4 = 0e x2 − 10x + 18 = 0 f x2 − 6x + 6 = 0g x2 + 2x − 9 = 0 h x2 − 8x − 1 = 0

2 Solve:a x2 + 3x − 1 = 0 b x2 + 5x + 2 = 0c x2 − 7x + 5 = 0 d x2 − 9x − 2 = 0e x2 + 11x + 4 = 0 f x2 − x − 6 = 0g x2 − 5x + 1 = 0 h x2 + 7x − 3 = 0

3 Rearrange and solve:a x2 = 4x + 1 b x2 + 2 = 6xc 9x − 2 = x2 d 4 − x2 = 7xe 2(3x + 5) = x2 f x2 − 3(5x − 2) = 0

g 14x − x2 = −1 h

4 Solve:a x2 − 14 = 0 b 6 − x2 = 0c 3x2 = 36 d −2x2 + 18 = 0

5

When completing the square to solve x2 + 16x + 1 = 0, the perfect square part of theexpression will be:

6

When completing the square to solve x2 − 20x − 4 = 0, the value to be simultaneouslyadded and subtracted is:

7

An expression which is being solved using completing the square is at the stage shownbelow.

(x − 9)2 − 15 = 0

The solution is:

A 8 B 16 C 64 D (x + 8) E (x + 8)2

A 4 B 10 C 16 D 100 E 400

A (x − 9)2 = 15 B x = ±C x = −9 ± D x = 9 ±

E x = 3 ±

2EWORKEDExample

8aSkillSH

EET 2.3

WORKEDExample

8bSkillSH

EET 2.4

x2 3x+4

----------------- 2–=

WORKEDExample

8c




24

15 15

15


The quadratic formulaAn alternative to the method of factorising by inspection or completing the square is touse the quadratic formula. This method isn’t always as quick, but it is reliable! Thederivation of the formula follows, and is based on the completing-the-square method,but all you have to remember is the formula in the last step.

The following examples show how the quadratic formula may be used.

Consider the general quadratic equation ax2 + bx + c = 0 (where a ≠ 0)

Divide every term by a. x2 + + = 0

Use the completing-the-square method.

Halve and square to get

and add and subtract it to the equation. =

Rearrange. =

The left side is now a perfect square. Get a common denominator on the right side. =

Take the square root of both sides. =

Simplify the RHS denominator. =

Subtract from both sides. = –

Write using the one denominator (2a). =

ba---x

ca---

ba--- b

2a------

2 b2

4a2--------=

x2 ba---x

ca--- b2

4a2-------- b2

4a2--------–+ + + 0

x2 ba---x

b2

4a2--------+ + b2

4a2-------- c

a---–

xb

2a------+

2 b2 4ac–4a2

--------------------

xb

2a------+ b2 4ac–

4a2--------------------±

xb

2a------+ b2 4ac–±

2a----------------------------

b2a------ x

b2a------

b2 4ac–2a

------------------------±

xb– b2 4ac–±

2a--------------------------------------

Use the quadratic formula to solve the following.a 3x2 − 8x − 9 = 0 b x2 + 5x + 6 = 0Give answers in exact (surd) and decimal form.

Continued over page

THINK WRITE

a Write the equation, and match up a, b and c. a a b c3x2 – 8x – 9 = 0

Write the quadratic formula.

Substitute the matched values for a, b and c. x =

1

2 xb– b2 4ac–±

2a--------------------------------------=

3 – 8–( ) 8–( )2 4 3 9–××–±2 3×

---------------------------------------------------------------------

9WORKEDExample


Notice how the quadratic formula still works for cases which could have beenfactorised by inspection.

x2 + 5x + 6 = 0 could have been expressed as (x + 2)(x + 3) = 0,and the solutions or ‘roots’ x = −2 or x = −3 obtained by inspection.

THINK WRITE

Simplify. x =

In particular, note the splitting of 172 into 4 × 43 inside the square root.

x =

x =

Write the two solutions separately. Use a calculator to obtain decimal answers if required. (Note that the surd forms of the answers are exact, but answers produced using a calculator may not be.)

x =

x =

x =

x = or x =

x = 3.519 or x = −0.852

b Write the equation, and match up a, b and c. b a b c1x2 + 5x + 6

Write the quadratic formula. x =

Substitute the matched values for a, b and c. x =

Simplify. x =

x =

x =

x = or x =

x = or x =

x = −2 or x = −3

48 64 108+±+

6---------------------------------------------

5

8 172±

6----------------------

8 4 43×±6

----------------------------

6

8 4 43×( )±6

-------------------------------------

8 2 43±6

----------------------

4 43±3

-------------------

4 43+3

------------------- 4 43–3

-------------------

1

2b– b2 4ac–±

2a--------------------------------------

35– 52 4 1 6××–±

2 1×---------------------------------------------------

45– 25 24–±

2------------------------------------

5– 1±2

---------------------

5– 1±2

----------------

5– 1+2

---------------- 5– 1–2

----------------

4–2

------ 6–2

------


The quadratic formula

1 Identify a, b and c (the coefficients of x2, x and the constant respectively) in each ofthe following quadratic equations.a x2 + 4x − 3 = 0 b x2 − 7x + 9 = 0c 9 − 4x + x2 = 0 d 7 − 3x − 6x2 = 0e x2 − 7x + 5 = 0 f (x + 1)2 = 0

g (2x − 3)2 = 0 h 5 − 2(x2 + 2) = 0

i 3x2 − 10x + 4 + 4x2 − 11x = 0 j

2 Use the quadratic formula to solve the following (even though the equations may befactorised by inspection).

3 Find exact (surd) answers to each of the following.a x2 + 5x + 3 = 0 b x2 + 8x + 5 = 0c x2 + 3x + 1 = 0 d x2 + 10x + 12 = 0e x2 − 6x + 2 = 0 f x2 − 7x + 6 = 0g x2 − 4x − 2 = 0 h x2 − 9x − 8 = 0i −2x2 + 3x + 1 = 0 j −4x2 + 12x − 1 = 0k −6x2 − 5x + 2 = 0 l −x2 + 14x − 5 = 0

4 Find decimal answers (3 decimal places) to each of the following.a x2 − 6x − 2 = 0 b x2 + 3x − 9 = 0c −2x2 + 7x + 1 = 0 d 9x2 − 2x − 2 = 0e −x2 − 8x + 1 = 0 f 3x2 + x − 9 = 0g x2 + 8x + 13 = 0 h x2 − 10x + 30 = 0i −2x2 + 3x − 2 = 0 j 5x2 − 3x + 7 = 0

5 Find decimal answers to the following.a x2 + 6x = 11 b 2x2 = 7 − 4xc 10x + 2 = −5x2 d x2 = 8x – 6e 5 = 9x − 2x2 f x2 − 2 = 7x + 4g 5x2 + 6x + 2 = 0 h −x2 + 4x = 8

6 The population of a colony of rare African person-eating ants is given by the equationN = x2 + 2x + 300, where N is the number of ants, and x is the height of the anthill incentimetres. How high could the anthill be when there are 850 ants in the colony?

a x2 + 9x + 20 = 0 b x2 − 10x + 16 = 0c 2x2 − 13x − 24 = 0 d −4x2 + 13x − 3 = 0

rememberIf ax2 + bx + c = 0, then solutions may be found using .x

b– b2 4ac–±2a

--------------------------------------=

remember

2F

12---

6x2 4+–2

--------------------- 0=

WWORKEDORKEDEExamplexample

9

Mathcad

The quadraticformula

Mathcad

Quadraticroots


7 The profit $P made when a particular make of car is manufactured in t hours is givenby the equation P = 40t2 − 7t − 5000. In order to just ‘break even’, how long must themanufacturing process take?

8 The position x metres (from the start of an assembly line) of an item is given byx = −4t2 + 20t + 5 where t is the time in minutes elapsed since the item started on theline. When is the item at position x = 0 metres on the assembly line?

9 A golf ball is hit from the bottom of abunker as shown below. The height hmetres of the ball above the ground isgiven by h = 60t − 5t2 − 1, where t secondsis the time the ball has been in flight.

a How deep is the bunker?b When is the ball first level with the top of the bunker?c When is the ball at a height of 4 metres?

10 Debbi plans to pave sections of her backyard as shown at right. If she has a total of 12 square metres of pavers, find the value of x.

11 The surface area (in m2) of cement transport containers made by a certain company is given by 4πr2 + 24πr, where r is the radius of the container. If the surface area of a particular container is 60 m2, determine its radius.

12 The Gateway arch in St. Louis, Missouri, was designed by Eero Saarinen in 1948 butnot completed until 1964. The line of the arch can be represented by the quadraticequation y = − (x − 95)2 + 190 where y is the height of thearch at a horizontal distance x along thebase (ground level) from one side of thearch. Both x and y are in metres.a Expand and simplify the quadratic

equation.b How wide is the arch at its base?c This arch is the tallest in the

world. How high is theGateway arch?

h

x m

x m

4 m

SloppoCement Co.

r

295------


The formula that ‘doesn’t work’!1 Among the following quadratic equations in the form ax2 + bx + c = 0 are 3 that

cannot be solved to obtain decimal answers using the quadratic formula.Try to determine which ones they are, and copy and complete the table.

2 What does the last column tell you? How?

The expression b2 − 4ac is called ‘the discriminant’, and is studied further in the next section. If you wish to check whether or not a quadratic will factorise in terms of whole numbers, or whether it will factorise at all, the discriminant is a useful tool.

3 Make up five quadratic equations of your own.

4 A Mathcad file that calculates the discriminant for you is provided on the Maths Quest CD-ROM. Try it with the equations you listed in 3.The screen dump shown below is for Equation A in question 1 above.

5 Decide whether each of these quadratic equations will factorise.

Equation a b c

Number of solutions using the quadratic

formula b2 − 4ac

A x2 + 6x + 9 = 0

B x2 + 6x + 5 = 0

C x2 + 6x + 10 = 0

D x2 − 2x + 4 = 0

E x2 − 2x + 1 = 0

F 2x2 − 4x − 5 = 0

G x2 + 5x + 6 = 0

H x2 − 9x + 14 = 0

Mathcad

Calculatingthe

discriminant


The discriminantYou may have found on occasions that no solutions (or ‘roots’) can be found for aquadratic equation. If you were using the quadratic formula at the time, you would havefound the trouble started when you tried to evaluate the square root part of the formula.The expression under the square root sign is called ‘the discriminant’.

The discriminant is used to determine how many roots of an equation exist and isdenoted by the upper case Greek letter ‘delta’ (∆).

The quadratic formula The discriminant

If ∆ < 0 (that is, negative), then no real solutions exist, as we cannot yet evaluate thesquare root of a negative value. The word ‘real’ is used to describe numbers we candeal with at present. The set of real numbers includes positives, negatives, fractions,decimals, surds, rationals (numbers that may be expressed as a ratio — for example )and irrationals.

(In further studies of Maths, a way of dealing with square roots of negative numbershas been devised using what are known as ‘imaginary numbers’.)If ∆ > 0, then the discriminant can be evaluated, and its square root added andsubtracted in the quadratic formula (see above) to produce two real solutions.If ∆ = 0, then there is nothing to add or subtract in the quadratic formula, and so thereis only one real solution to evaluate.

xb– b2 4ac–±

2a--------------------------------------= ∆ b2 4ac–=

49---

Find the value of the discriminant for the equation 3x2 − 5x + 2 = 0.

THINK WRITE

Write the expression.Note the value of a, b and c.

3x2 − 5x + 2 = 0a = 3, b = −5, c = 2

Write the formula for ∆. ∆ = b2 − 4acSubstitute for a, b and c. ∆ = (−5)2 − 4(3)(2)Evaluate. ∆ = 25 − 24

∆ = 1

1

234

10WORKEDExample

How many real solutions are there to the equation − 7x2 + 3x − 1 = 0?

THINK WRITE


−7x2 + 3x − 1 = 0a = −7, b = 3, c = −1

Write the formula for ∆. ∆ = b2 − 4acSubstitute for a, b and c. ∆ = (3)2 − 4(−7)(−1)Simplify. ∆ = 9 − 28

= −19Comment on the number of solutions. Since ∆ < 0, there are no real solutions.

1

234

5

11WORKEDExample


For what values of k does −x2 + 2kx − 9 = 0 have:i two distinct solutions? ii one solution? iii no solutions?

THINK WRITE


−x2 + 2kx − 9 = 0a = −1, b = 2k, c = −9

Write the formula for ∆. ∆ = b2 − 4acSubstitute for a, b and c. ∆ = (2k)2 − 4(−1)(−9)Simplify. ∆ = 4k2 − 36 Consider case i, two distinct solutions.Note: |k| means the ‘size’ or ‘absolute value’ of k without regard to + or −. Since k2 is always positive, we require k > 3 (values such as 3.1, 4, 5 etc.)or k < −3 (values such as −3.1, −4, −5 etc.) if k2 is to be > 9.

i Two distinct solutions, require ∆ > 0.So 4k2 − 36 > 0

4k2 > 36k2 > 9|k| > 3k > 3 or k < −3

Consider case ii, one solution.The ‘size’ of k must be = 3.(That is, 3 or −3 are our only choices.)

ii One solution, require ∆ = 0.So 4k2 − 36 = 0

4k2 = 36k2 = 9|k| = 3k = 3 or k = −3

(k = ±3)Consider case iii, no solutions.The ‘size’ of k must be < 3.(values like –2.9, –2, –1, 0, 1, 2, 2.9 etc.)An alternative method is shown in the next example.

iii No solutions, require ∆ < 0.So 4k2 − 36 < 0

4k2 < 36k2 < 9|k| < 3−3 < k < 3

1

2345

6

7

12WORKEDExample

For which values of k does 2x2 + 2kx + (k + 4) = 0 have:i two solutions? ii one solution? iii no solutions?

Continued over page

THINK WRITE


2x2 + 2kx + (k + 4) = 0a = 2, b = 2k, c = k + 4

Write the formula for ∆. ∆ = b2 − 4acSubstitute for a, b and c. ∆ = (2k)2 − 4(2)(k + 4)Simplify and factorise. ∆ = 4k2 − 8(k + 4)

= 4k2 − 8k − 32 = 4(k2 − 2k − 8) = 4(k + 2)(k − 4)

1

234

13WORKEDExample


The steps below show how to repeat discriminant calculations using the formula∆ = b2 − 4ac. To find the discriminant for 3x2 + 7x − 5 = 0:1. Type 3→A:7→B:–5→C:B2 − 4*A*C on the

home screen. (Use ALPHA for A, B, C: andSTO for →.)

2. Press to calculate.3. Press ENTRY to recall the previous

calculation instructions.4. Use the arrow keys to move over old

values of A, B and C, and type newvalues. (Note: To insert a negative sign (–)or extra digit, press [INS] before typing the character/s to be inserted to the leftof the cursor. To delete a character, place the cursor over it and press .)

5. Press to calculate the expression for the new set of values.

THINK WRITE

Since ∆ is a more complicated expression than those in the previous example, a graph of ∆ versus k (∆ on vertical axis, k on horizontal axis) is useful. (Recall how you sketched quadratic graphs in previous work, or see the next section.)For case i, Require ∆ > 0From graph, ∆ > 0 when k < −2 or k > 4.

i Two solutions, ∆ > 0.So k < −2 or k > 4

For case ii, Require ∆ = 0 From graph, ∆ > 0 when k = −2 or k = 4.

ii One solution, ∆ = 0.So k = −2 or k = 4

For case iii, Require ∆ < 0From graph, ∆ < 0 when −2 < k < 4.Note: This method involving sketching a graph of ∆ may be used as an alternative to the method shown in the previous example.

iii No solutions, ∆ < 0.So −2 < k < 4

5��0 ��0

��0

��0

–2

–32

4 k

�

6

7

8

Graphics CalculatorGraphics Calculator tip!tip! Repeated calculation of the discriminant

ENTER2nd

2ndDEL

ENTER

remember1. If ax2 + bx + c = 0, then ∆ = b2 − 4ac.2. If ∆ < 0, no real solutions exist.3. If ∆ > 0, there are two real solutions.4. If ∆ = 0, there is only one real solution.

remember


81

The discriminant

1

Find the value of the discriminant in each case.

2

How many real solutions are there to the following equations?

Do not

actually work out any solutions.

3

Find an expression for the discriminant of each of the following.

4

Which values of

m

below both result in the expression

x

2

−

mx

+

5

=

0 having two distinct solutions?

5

For what values of

k

does each equation have:

i

two distinct solutions?

ii

one solution?

iii

no solutions?

a

x

2

+

9

x

+

2

=

0

b

x

2

−

4

x

−

1

=

0

c

5

x

2

+

6

x

−

7

=

0

d

2

x

2

−

3

x

+

10

=

0

e

−

3

x

2

+

x

+

3

=

0

f

−

x

2

−

2

x

−

6

=

0

g

x

2

+

15

x

=

1

h

9

−

7

x

=

4

x

2

i

−

3

x

2

=

5

a

5

x

2

+

x

+

2

=

0

b

−

x

2

+

4

x

+

4

=

0

c

3

x

2

−

3

x

+

1

=

0

d

3

x

2

+

6

x

+

3

=

0

e

−

2

x

2

−

8

x

−

8

=

0

f 9 − x2 + x = 0g 5x2 = 2 − x h 6 − 6x = x2 i 12x = 9x2 + 4

a x2 + ax + 1 = 0 b ax2 + 2x + 3 = 0c x2 + 6x + a = 0 d ax2 + bx + 1 = 0e mx2 + 2mx + 1 = 0 f x2 + (m + 1)x + 3 = 0g x2 − mx − (m + 4) = 0 h (k − 1)x2 − kx + 2 = 0

A −20 and 3 B 0 and 20 C 4 and 5 D −6 and −5 E 0 and 5

a x2 + kx + 4 = 0 b x2 − 4x + k = 0c x2 + 4kx + 4 = 0 d kx2 − 18x + 20 = 0e x2 − 4x + (k + 1) = 0 f 6x2 + 4kx + (k + 3) = 0g 4kx2 + 12kx + 9k = 0 h (k + 4)x2 + 10x + 5 = 0i (k − 1)x2 − (k + 1)x + 2 = 0

Quadratic graphs — turning point formUse one of the Maths Quest CD-ROM files below to investigate the effect of a, b and c on the graph of y = a(x − b)2 + c.

2GWW

ORKEDORKEDE

Examplexample10

EXCEL Spreadsheet

Calculating thediscriminant

Mathcad

Calculating thediscriminant

WW

ORKEDORKEDE

Examplexample

11


WWORKEDORKEDEExamplexample12

WWORKEDORKEDEExamplexample13

Mathca

d

EXCEL

Spreadsheet

Quadratic graphs – turning point form


82

M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Quadratic graphs — turning point form

If you completed the previous investigation, you may have discovered the followingconnections between a quadratic equation in turning point (TP) form, and its graph.

y dilation(bigger* = thinner,negative = upside down)

*ignoring any − signs.

x-coordinate of turning point (horizontal shift).

y-coordinate of turning point (vertical shift). Minimum value of y for ∪ shape graphs. Maximum value of y for ∩ shape graphs.

y = a(x − b)2 + c

x

y a positive, increasing

a negative, increasingly negativex

y

(b, c)

For the graph of y = −3(x + 2)2 − 1:i State the turning point coordinates.ii Describe the width of the graph as S, the Same width as y = x2; T, Thinner than y = x2;

or W, Wider than y = x2.iii State whether the graph is a minimum (∪) or maximum (∩) type, and state the

maximum or minimum value of y.iv Find the y-intercept.v Sketch the graph. (x-intercepts are not required.)

THINK WRITE

Write the equation. Comparey = −3(x + 2)2 − 1 with y = a(x − b)2 + ca = −3, b = −2, c = 1

iii y = −3(x + 2)2 − 1

x TP coordinate = b = −2y TP coordinate = c = −1

iii TP (–2, –1)

y dilation factor = −3 (magnitude greater than 1 means it is thinner)

iii T

a is negative, so the graph is ‘upside down’, or a maximum shape.

iii Graph is a maximum (∩) type.

Maximum y-value is c, which equals 1. iii Maximum value of y is −1.

The y-intercept can be found by putting x = 0 in the original equation.

iv y-intercept: y = −3(0 + 2)2 − 1= −3(2)2 − 1= −13

1

2

3

4

5

6

14WORKEDExample



A quadratic equation is not always in turning point form, however, and must bemanipulated in order to answer questions like those posed in the previous example. Todo this, we use the method of completing the square as demonstrated in the followingexample.

THINK WRITE

Use the information above to sketch the graph. v7

x

y

(–2, –1)

–13

Convert the equation y = x2 − 6x + 1 to turning point form, state the coordinates of the turning point, and the maximum or minimum value of y.

THINK WRITE

Write the equation. Ensure the x2 coefficient is 1. It is.(If not, divide the equation by whatever will change the x2 coefficient to 1. Such problems are beyond the scope of this course, however.)

y = x2 − 6x + 1

Halve the x coefficient and square it.− = −3, −32 = 9

Add and subtract this value after the x term.

y = x2 − 6x + 9 − 9 + 1

Group terms to make a perfect square. y = (x2 − 6x + 9) − 8Factorise the perfect square part. y = (x − 3)2 − 8Compare with y = a(x − b)2 + ca = 1, b = 3, c = −8TP = (b, c) = (−3, −8) TP (3, −8)a is positive, so there is a minimum value which equals c = −8.

Minimum value of y is −8

1

262---

3

456

78

15WORKEDExample

remember1. Turning point form: y = a(x − b)2 + c.2. TP is at (b, c), dilation factor a.3. Positive (greater than zero) a means ∪ (minimum) shape.4. Negative a means ∩ (maximum) shape.5. To convert to TP form, complete the square.

remember


Quadratic graphs — turning point form

1 Which of the following are quadratics in turning point form?a y = x2 + 4x + 2 b y = x2 − 3c y = (x − 1)2 + 1 d y = 3(x + 6)2

e y = (x2 − 2x + 3)2 f y = −(x + 2)3 − 5g y = −(x − 3)2 + 7 h y = −2(x + 1)2 + 5

2 State the coordinates of the turning point for each of the following.a y = (x − 5)2 b y = (x + 7)2 + 3c y = (x − 2)2 − 7 d y = (x − 1)2 + 8e y = 2(x + 3)2 − 4 f y = 6(x − 2)2 + 2g y = −3(x − 4)2 − 9 h y = −4(x + 6)2 − 1i y = −(x − 7)2 − 3 j y = − (x − 4)2 + 8

3 For each of the following:i State the turning point coordinates.ii Describe the width of the graph as S − the Same width as y = x2, T − Thinner than

y = x2 or W − Wider than y = x2.iii State whether the graph is a minimum (∪) or maximum (∩) type, and state the

maximum or minimum value of y.iv Find the y-intercept.v Sketch the graph. (x-intercepts are not required.)

a y = (x − 4)2 + 2 b y = (x − 2)2 − 3c y = (x + 5)2 − 8 d y = (x − 1)2 − 1e y = 2(x + 5)2 − 9 f y = (x − 4)2 + 4g y = (x + 1)2 − 12 h y = −7(x − 3)2 + 3i y = −(x − 8)2 − 9 *j y = (1 − x)2 + 20

4 Sketch graphs having the following properties, but do not show intercepts.a Turning point (2, 5), y dilation of 3b Turning point (−1, 3), y dilation of 1c Turning point (0, −4), y dilation of 2d Turning point (6, 0), y dilation of −1e Turning point (7, −7), y dilation of −4f Turning point (−10, −1), y dilation of g Turning point (−2, −5), y dilation of −h Turning point (0, 2), y dilation of

5 Determine the equations of the following graphs, given that they are all of the formy = (x − b)2 + c (that is, dilation factors all equal to 1).

a b c

2H

WORKEDExample

14i

12---

WORKEDExample

14

13---

65---

14---

13---

12---

x

y

(2, 4)8

x

y(–1, 5)

4

x

y

(2, –5)

–1


6 Find the y-intercepts of graphs with the following properties.a Turning point (1, 4), y dilation of 2b Turning point (−2, −3), y dilation of −1

7 Convert each of the following to turning point form, state the coordinates of theturning point and the maximum or minimum value of y.

8 A section of perspex for a parabolic window is drawn on acoordinate grid as shown at right. What equation should beprogrammed into the automatic glass cutting machine, usingthe grid system on the diagram?

9

What is the effect on the graph of increasing k in the equation y = k(x − 2)2 + 1?The graph is:

10

What is the effect on the graph of increasing k in the equation y = (x + k)2?The graph is:

11

What is the effect on the graph of increasing k in the equation y = (x + 1)2 − k?The graph is:

12 For the graph y = M(x + h)2 + n, state:a the x-coordinate of the turning pointb the y-coordinate of the turning pointc the y dilation factord the coefficient of x2 in the expanded versione the coefficient of x in the expanded versionf the constant term in the expanded version.

d e f

a y = x2 − 4x + 9 b y = x2 − 6x + 17 c y = x2 − 12x + 37d y = x2 + 8x + 13 e y = x2 − 4x − 5 f y = x2 + 7g y = x2 + 18x + 0 h y = 2x2 − 12x + 22 i y = 3x2 + 12x + 15

A raised B lowered C thinnedD widened E shifted left F shifted right



x

y(–3, 1)

–8

x

y(6, 6)

–30x

y

(–3, –2)

7

WORKEDExample

15

GC program

Completingthe square

* *

0

2

4

6

8

2 4 6 8mmultiple choiceultiple choice



WorkS

HEET 2.2


Quadratic graphs — intercepts methodIn this section we will consider graphs of quadratic functions of the form y = ax2 + bx + c.

When we talk about sketching a graph, we mean drawing a diagram showing themain features — not a true scale graph showing every point plotted accurately using acomputer package or other means.

To sketch a quadratic graph, the following features should generally be apparent orlabelled.

1 The y-interceptThe y-intercept is found by substituting x = 0 into theequation, and solving for the corresponding y-value.That is, it is the c in the general form y = ax2 + bx + c.

2 Any x-intercepts if they existAny x-intercepts are found by substituting y = 0 into the equation and solving for the corresponding y-value. Let’s say the equation factorises to y = (x − d)(x − e). Then substituting y = 0 into the equation gives 0 = (x − d)(x − e).

For this to be true, we must have x = d or x = e.

3 The turning point coordinates

Consider the general quadratic func-tion, and complete the square as follows:

y = ax2 + bx + c

=

=

=

=

=

You may recognise this form as turning point form. In this case, the coordinates ofthe turning point are

x

y

y-intercept

x

y

x-intercepts

Axis of symmetry

x-interceptsy-intercepts

Turning point(– , c– )b—

2ab2—4a

x

y

a x2 b

a---x

ca---+ +

a x2 b

a---x

b2a------

2 b2a------

2 ca---+–+ +

a xb

2a------+

2 b2

4a2-------- c

a---+–

a xb

2a------+

2 b2

4a------– c+

a xb

2a------+

2c

b2

4a------–

+

b–2a------ c

b2

4a------–,

C h a p t e r 2 Q u a d r a t i c f u n c t i o n s 87Using xt and yt for the coordinates of the turning point, we have

and

If a quadratic function has two x-intercepts, If there is only one x-intercept, then itthe x-coordinate of the turning point may is the x-coordinate of the turningbe found by averaging them. point.

The y-coordinate of the turning point may be found using the general turning pointform above, by completing the square from scratch, or by substituting the x-coordinateinto the original equation.

Recall also the two main types of parabola:

Minimum parabola Maximum parabolay = ax2 . . . (a positive, not zero) y = ax2 . . . (a negative)

xtb–

2a------= yt c

b2

4a------–=

x

y

d+e——2

d e

x

y

d

x

y

x

y

Sketch graphs of the following, showing all intercepts and the turning point in each case.a y = x2 − 4x − 32 b y = x2 + 10x + 25 c y = x2 − 3x − 10 d y = −2x2 + 11x − 15

Continued over page

THINK WRITE

a Write the equation. a y = x2 − 4x − 32

Find the y-intercept (when x = 0). If x = 0, y = −32

Factorise before finding x-intercepts. y = (x + 4)(x − 8)

Find x-intercepts (when y = 0). If y = 0, 0 = (x + 4)(x − 8)x = −4 or x = 8

Find the x-coordinate of the turning point.

TP xt = , b = −4, a = 1

=

= 2

1

2

3

4

5b–

2a------

4–( )–2 1( )

--------------

16WORKEDExample


THINK WRITE

Substitute into the original equation to find the y-coordinate of the TP.

Alternatively, use

yt = 22 − 4(2) − 32 = 4 − 8 − 32 = −36

Write the turning point coordinates. TP (2, −36)Combine the above information and sketch.Note: The method of completing the square may also be used to find the turning point.

b Write the equation. b y = x2 + 10x + 25Find the y-intercept (when x = 0). If x = 0, y = 25Factorise before finding x-intercepts. y = (x + 5)(x + 5)

= (x + 5)2

Find the x-intercept (when y = 0). If y = 0, 0 = (x + 5)2

x = −5

Find the x-coordinate of the turning point.Alternatively, since there is only one x-intercept (−5), it must be the turning point x-coordinate.

TP xt = , b = 10, a = 1

=

= −5

Substitute into the original equation to find the y-coordinate of the turning point.

yt = (–5)2 + 10(–5) + 25 = 25 – 50 + 25 = 0

Write the turning point coordinates. TP (−5, 0)Combine the above information and sketch.

c Write the equation. c y = x2 − 3x − 10Find the y-intercept (when x = 0). If x = 0, y = −10Factorise before finding x-intercepts. y = (x − 5)(x + 2)Find x-intercepts (when y = 0). If y = 0, 0 = (x − 5)(x + 2)

x = 5 or x = −2

Find the x-coordinate of the turning point.Alternatively, average the two x-intercepts:

=

TP xt = , b = −3, a = 1

=

= (or x = 1.5)

6

yt cb2

4a------–=

7

8

–4

(2, –36)

8 x

y

1

2

3

4

5b–

2a------

10–2 1( )-----------

6

7

8

(–5, 0)

25

x

y

1

2

3

4

5

xt5( ) 2–( )+

2-------------------------= 3

2---

b–2a------

3–( )–2 1( )

--------------

32---


Continued over page

THINK WRITE

Substitute into the original equation to find the y-coordinate of the turning point.Use a calculator if you wish to avoid working with fractions.

yt = 2 − 3 − 10

= − − 10

= − −

= −

= −12 (or y = −12.25)

Write the turning point coordinates.Note: The method of completing the square may also be used to find the turning point.

TP ( , − )

Sketch.

d Write the equation. d y = −2x2 + 11x − 15Find the y-intercept (when x = 0). If x = 0, y = −15Factorise before finding x-intercepts. y = –(2x2 − 11x + 15)

= –(2x − 5)(x − 3)Find x-intercepts (when y = 0). If y = 0, 0 = −(2x − 5)(x − 3)

2x − 5 = 0 or x − 3 = 02x = 5 or x = 3

x = or = 3


TP xt = , b = 11, a = −2

=

= (or 2.75)


Use a calculator if you wish to avoid working with fractions.

yt = −2 2 + 11 − 15

= + − 15

= + −

=

= (or 0.125)

632---

32---

94--- 9

2---

94--- 18

4------ 40

4------

494------

14---

732--- 49

4------

8

–2

–10

5 x

y

( , – )3–2

49—4

1

2

3

4

52---

5b–

2a------

11–2 2–( )--------------

114------

6114------

114------

242–16

------------ 1214

---------

242–16

------------ 48416--------- 240

16---------

216------

18---


Don’t be put off if asked to sketch a quadratic graph whose equation doesn’t have 3 terms. Such cases are easier to sketch, as the following example shows.

THINK WRITE

Write the turning point coordinates. TP ( , )

Sketch.

7114------ 1

8---

8

–15

3x

y ( , )11—4

1–8

5–2

Sketch:a y = x2 − 16 b y = −2x2 + 14x

THINK WRITE

a Write the equation. a y = x2 − 16

Find the y-intercept (when x = 0). If x = 0, y = 02 − 16 = −16

Factorise before finding x-intercepts. In this case, recognise a difference of squares.

y = (x + 4)(x − 4)

Find x-intercepts (when y = 0). If y = 0, 0 = (x + 4)(x − 4)So x = −4 or x = 4

Find the x-coordinate of the turning point. xt = , b = 0, a = 1

xt =

= 0


yt = 02 − 16 = −16

Write the turning point coordinates. TP (0, −16)

Sketch.

1

2

3

4

5b–

2a------

02 1( )-----------

6

7

8

–4 4 x

y

(0, –16)

17WORKEDExample


A graphics calculator may be used to check sketch graphsquickly by entering the equation in the Y= menu, thenpressing and .

Hint: Ensure the WINDOW settings for Xmin and Xmaxdiffer by 9.4 (for example Xmin 4.7, Xmax 4.7, or Xmin1.7 and Xmax 7.7 as on the right) so tracing moves inconvenient steps of 0.1.

THINK WRITE

b Write the equation. b y = –2x2 + 14x

Find the y-intercept (when x = 0). If x = 0, y = −2(0)2 + 14(0) = 0

Factorise before finding x-intercepts. In this case, use a common factor of −2x.

y = –2x(x − 7)

Find x-intercepts (when y = 0). If y = 0, 0 = −2x(x − 7)So x = 0 or x = 7


xt = , b = 14, a = −2

xt =

=

= (or 3.5)


We could also use

Use a calculator if you wish to avoid working with fractions.

yt = (2 )2 + 14( )

= −2 + 49

= − + 49

= − +

= (or 24.5)

Write the turning point coordinates. TP ( , )

Sketch.

1

2

3

4

5b–

2a------

142 2–( )--------------

144------

72---

6

yt cb2

4a------–=

72--- 7

2---

494------

492------

492------ 98

2------

492------

772--- 49

2------

8

0 7 x

y

( , )7–2

49—2

GRAPH TRACE


If factorisation ‘by inspection’ (as in the previous examples) is not possible, thequadratic formula may be used.

Sketch the graph of the following.a y = 3x2 + 6x – 1 b y = −2x2 + x – 7

THINK WRITE

a Write the equation. a y = 3x2 + 6x − 1

Find the y-intercept (when x = 0). If x = 0, y = −1

Factorising before finding x-intercepts (when y = 0) is not easily done, so use the quadratic formula.

If y = 0, 0 = 3x2 + 6x − 1

a = 3, b = 6, c = −1

x = 0.155 or x = −2.155


TP xt = , b = 6, a = 3

xt =

= −1


yt = 3(−1)2 + 6(−1)−1 = 3 − 6 − 1 = −4

TP (−1, −4)

Sketch.

1

2

3

xb– b2 4ac–±

2a--------------------------------------=

x6– 62 4 3( ) 1–( )–±

2 3( )----------------------------------------------------=

6– 36 12+±

6------------------------------------=

6– 48±

6------------------------=

6– 16 3×±

6---------------------------------=

6– 4 3±

6------------------------=

3– 2 3±

3------------------------=

4b–

2a------

6–2 3( )-----------

5

6

–2.155 0.155 x

y

(–1, –4)–1

18WORKEDExample


THINK WRITE

b Write the equation. b y = −2x2 + x − 7

Find the y-intercept (when x = 0). If x = 0, y = −2(0) + (0) −7If x = 0, y = −7

Factorising before finding x-intercepts (when y = 0) is not easily done, so try the quadratic formula.

If y = 0, 0 = −2x2 + x − 7

x =

a = −2, b = 1, c = −7

x =

=

=

cannot be evaluated so there are no x-intercepts.

No x-intercepts


xt =

=

=

= (or 0.25)


yt = −2( )2 + − 7

= −2( ) + − 7

= − + − 7

= − + −

= − (or −6.875)

Sketch.

1

2

3−b b2 4ac–±

2a--------------------------------------

−1 12 4 2–( ) 7–( )–±2 2–( )

------------------------------------------------------

−1 12 56–±4–

----------------------------------

−1 55–±4–

--------------------------

4 55–

5b–

2a------

1–2 2–( )--------------

1–4–

------

14---

6 14--- 1

4---

116------ 1

4---

18--- 1

4---

18--- 2

8--- 56

8------

558------

7

–7

x

y

( , – )1–4

55—8


An alterative, if graphics calculators are permitted, is to enter the equation in the Y= menu, and calculate important points as follows. The equation y = x2 − 5x + 6 is used as an example. Be sure to use ZOOM and choose WINDOW settings that provide a clear picture of key points.

The y-intercept The x-intercepts The turning point

Press [CALC], Press [CALC] Press [CALC] andselect 1:value and at X=, and select 2:zero. select 3:minimum to find the

enter 0. Press . TP of a minimum parabola.

or [CALC] and select4:maximum to find the TP ofa maximum parabola.

Using the discriminantSince finding x-intercepts for a quadratic graph involves solving a quadratic equation,we can use the discriminant to decide the number of x-intercepts such a graph has.

∆ > 0 ∆ = 0 ∆ < 0

2nd 2nd 2nd

ENTER

2nd

x

y

x

y

x

y

rememberTo sketch a quadratic graph:1. Find the y-intercept (when x = 0).2. Factorise if possible, and find the x-intercepts (when y = 0).

It may be easier to use the quadratic formula in some cases.If ∆ < 0, there are no x-intercepts. If ∆ = 0, one intercept only. If ∆ > 0, two intercepts.

3. Find the x-coordinate of the TP using or completing the square.

4. Find the y-coordinate of the TP by substituting xt into the equation for y, or

using or by completing the square.

5. Combine all the information and sketch. Functions with a positive x2 coefficient are ∪ shaped, and those with negative x2 coefficients are ∩ shaped. If the information you have gathered doesn’t seem to fit, check for calculation errors.

xtb–

2a------=

yt cb2

4a------–=

remember


Quadratic graphs — intercepts method

1 Find the y-intercepts for each of the following.

2 State the x-intercepts for each of the following.

3 State the x-intercepts for each of the following.

4 Use the quadratic formula to find exact values (if possible) for the x-intercepts of:

5 Sketch graphs of the following, showing all intercepts and the turning point in eachcase.(Hint: Factorise first.)

6 Sketch the following graphs, showing all intercepts and the turning point in each case.

7 Sketch:

8 Sketch:

9 Sketch the graph of the following. Verify your sketches using a graphics calculator ifone is available.

a y = x2 + 9x + 2 b y = −3x2 + 6x − 4 c y = 4x2 + 2xd y = 6 − x2 e y = 5x + 2 − 4x2 f y = 1 − x − x2

a y = (x − 1)(x − 6) b y = (x − 3)(x + 2) c y = (x + 5)(x + 1)d y = (x − 4)(x − 5) e y = (x + 7)2 f y = x(x − 2)g y = (3 − x)(4 − x) h y = (x + 5)(5 − x) i y = −x(x + 8)j y = −(x + 9)2 k y = 2(x + 6)(x − 6) l y = −3(2 − x)(x + 10)

a y = (2x − 5)(x + 1) b y = (3x + 8)(x − 6) c y = (7x + 2)(2x − 1)d y = (−3x + 1)(x + 2) e y = (x − 9)(4x − 9) f y = −(x + 1)(6 − 5x)g y = (9x − 1)(3x − 1) h y = (2x + 3)(4x + 1) i y = 3x(x + 4)j y = (Ax + a)(Bx + b)

a y = 4x2 − 2x + 3 b y = −x2 + 7x − 7c y = −2x2 − 9x − 1 d y = 10 − 3x + 3x2

a y = x2 − 4x + 3 b y = x2 + 2x + 1 c y = x2 + 6x + 8d y = x2 + 12x + 35 e y = x2 − 8x + 12 f y = x2 + 2x − 63g y = x2 + 3x + 2 h y = x2 − 5x + 6 i y = x2 −11x − 12j y = x2 + 14x + 49 k y = x2 − 16x + 64 l y = x2 + 8x − 153

a y = 3x2 + 2x − 8 b y = 5x2 + 18x − 8 c y = 3x2 − 4x − 15d y = 4x2 − 8x + 3 e y = 8x2 − 10x + 3 f y = 7x2 + 18x − 9g y = 15x2 + 48x + 9 h y = 9x2 − 2x − 7 i y = 2x2 + x − 28j y = 3x2 + 5x + 2 k y = 2x2 − 3x − 9

a y = −x2 − 8x + 33 b y = −x2 + 2x + 3 c y = −x2 − 18x − 45d y = −x2 + 18x − 81 e y = −4x2 + 12x − 5 f y = −8x2 − 6x + 5

a y = x2 − 25 b y = x2 − 121 c y = −x2 + 1d y = 3 − x2 e y = 2x2 − 18 f y = −3x2 + 12g y = x2 + 5x h y = x2 − 8x i y = 5x2 − 10x j y = −4x2 − 24x k y = 21x − 3x2

a y = x2 + 2x − 7 b y = x2 + 4x + 1 c y = x2 + 8x + 3d y = −4x2 + 2x + 3 e y = −x2 − 18x − 1 f y = x2 − 3x + 1g y = x2 − 7x − 2 h y = 2x2 + 7x + 4 i y = 3x2 − 9x − 5j y = −3x2 + 17 k y = 4x2 − 5x

2I

Mathcad

Quadraticgraphs –factored

form

EXCEL Spreadsheet

Quadraticgraphs –factored

form

WORKEDExample

16a, b ,c

Mathcad

Quadraticgraphs –general

form

EXCEL Spreadsheet

Quadraticgraphs –general

form

WORKEDExample

16d

WORKEDExample

17

WORKEDExample

18


10

If a and b are positive numbers, which of the following graphs could be that of y = (ax + 1)(x + b)?

11

If k and p are positive numbers, which of the following could be the graph ofy = x2 − 2kx + p?

12 Without sketching, determine how many x-intercepts each of the following graphshave.

13 Consider the value of the discriminant in deciding which graph matches whichequation below.

a y = 5x2 + x + 1 b y = 6x2 − 2x − 1 c y = −7x2 − x + 2 d y = −4x2 + 8x − 4

A B C D

A B C

D E

A B C

D E

a y = x2 + 37x + 208 b y = −3x2 − 4x + 8c y = −9x2 + 78x − 169 d y = 4x2 + x + 9e y = −x2 − 12x − 35 f y = 4x2 + 72x + 324


x

y

x

y

x

y

x

y

x

y


x

y

x

y

x

y

x

y

x

y

x

yx

y

x

y

x

y


Using graphs to solve quadratic equations

With the advent of graphic calculators, using graphs to solve equations is now mucheasier than it was when graphs were hand-drawn from tables of values.Consider the quadratic equation:

2.7x2 + 3.1x − 4.519 = 0We may write:

Let y = 2.7x2 + 3.1x − 4.519and then say:

‘If the left side of the original equation equals zero, then so must y’.One way to find solutions (also known as ‘roots’ or ‘zeros’ or ‘x-intercepts’) using a

graphics calculator is to enter the equation in the Y= menu of a graphics calculator, and

find x-intercepts by zooming in — or better still, using the [CALC] 2:zerofacility of the calculator.

As an example, use a graphics calculator to find all solutions of 2.7x2 + 3.1x − 4.519 = 0.To find the solution:

1. Enter 2.7x2 + 3.1x − 4.519 as Y1 in the Y= menu.

2. Press to view an initial graph screen.

3. ZOOM and/or alter WINDOW settings until a clear graph is produced, with intercepts in view.

4. Use [CALC] 2:zero to locate all roots or solu-tions. (You’ll need to scroll and press at threex-values — one below, one above and one anywhereclose to the intercept in each case.)

2nd

GRAPH

2ndENTER

remember1. Enter the equation in the Y= menu.2. Find the x-intercepts of the graph using [CALC] 2:zero.2nd

remember


5 mh

d

h = –0.5d2 + 2d + 5

Using graphs to solve quadratic equations

1 Use a graphics calculator graph to find all solutions of each of the following.a x2 + 4x + 1 = 0 b x2 − 11x − 2 = 0 c −x2 + 4x − 2 = 0d x2 − 8x + 8 = 0 e x2 + 12x + 9 = 0 f −3x2 + 5x − 1 = 0

2 Find all roots of the quadratics below using a graphics calculator graph.

3 The distance, d, of a cometfrom one of the moons ofJupiter is given by the equationd = 47.9t2 + 0.03t − 908.7where t is the number of hourssince the comet was first dis-covered on June 28, 2001. Atwhat value of t will the cometreach this moon?

4 The number of marine organisms, N, in a marine research organisation’s testing tank is foundto follow the equation (or model) N = −0.0751h2 + 0.69h + 200, where h is the number ofhours since the tank was supplied with nutrient and stocked with 200 organisms. How longafter being fed could the colony survive without further food before none were left?

5 A diver follows a parabolic path from the diving board to the water below. What is thehorizontal distance travelled by the diver from leaving the diving board to entering thewater? (h represents the height of the diver above the water for a distance, d, from thediving board in the equation shown. Both h and d are in metres.)

a 2.3x2 + 0.7x − 0.59 = 0 b −0.811x2 − 5.2x − 3.1 = 0c −3.97x2 + 17x + 8.05 = 0 d 5.18x2 − 2.66x − 9 = 0e 0.006x2 + 0.923x + 0.361 = 0 f x2 + 500x − 47 = 0

2J


Simultaneous quadratic and linear equations

In previous studies you have dealt with pairs of simultaneous linear equations andsolved these using algebra. You may recall that the solution could also be representedby the intersection of graphs of the equations and the equations solved by finding thecoordinates of the point of intersection. The same is true when we have one linear andone quadratic equation as a pair of simultaneous equations.

Consider the following pair of simultaneous equations:y = x2 + x − 2 and y = 3x + 1

If x = 3 is substituted into the first equation, y = 32 + 3 − 2 = 10 is obtained.If x = 3 is substituted into the second equation, y = 3 × 3 + 1 = 10 is obtained.That is, the coordinate pair (3, 10) ‘fits’ both equations, so is a solution.If x = −1 is substituted into the first equation, y = (−1)2 + (−1) − 2 = −2 is obtained.If x = −1 is substituted into the second equtation, y = 3(−1) + 1 = −2 is obtained.That is, the coordinate pair (−1, −2) is also a solution.The following graphs illustrate how simultaneous quadratic and linear equations may

have 0, 1 or 2 solutions.

When solving a linear and quadratic equation simultaneously, a new quadraticequation is formed, as you will see in the following examples.

The number of solutions relates to the discriminant (positive = 2 solutions; negative= no solutions; zero = one solution) of the ‘new’ quadratic.

No solution

x

y

Onesolution

x

y

Twosolutions

x

y

a Solve the system of equations y = x2 + x − 2 and y = 3x + 1.b Illustrate the solution using a sketch graph. The turning point of the quadratic graph

is not required.

Continued over page

THINK WRITE

a Write the original equations and label them. a y = x2 + x − 2 [1]y = 3x + 1 [2]

Equate [1] and [2]. Put RHS [1] = RHS [2]. x2 + x − 2 = 3x + 1Collect terms on the side that makes the x2 term positive.

x2 + x − 2 − 3x − 1 = 0

Combine like terms. x2 − 2x − 3 = 0Factorise if possible. (x − 3)(x + 1) = 0

Solve for x. x = 3 or x = −1

1

23

4

5

6

19WORKEDExample


THINK WRITE

Substitute x-values into the linear equation [2] to find the corresponding y-values.

If x = 3, y = 3(3) + 1= 10

If x = −1, y = 3(−1) + 1= −2

Write the solution coordinates. Solutions: (3, 10) and (−1, −2)

b Find intercepts for a sketch of the linear graph.

b For y = 3x + 1,if x = 0, y = 1 (0, 1)if y = 0, 0 = 3x + 1

−1 = 3x

x = − (− , 0)

Find intercepts for a sketch of the quadratic graph. (Turning point not required here.)

For y = x2 + x − 2,if x = 0, y = −2 (0, −2)if y = 0, 0 = x2 + x − 2so 0 = (x + 2)(x − 1)and x = −2 or x = 1 (−2, 0)

(1, 0)Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

7

8

1

13--- 1

3---

2

3

x

y

1–2 1

(–1, –2)

(3, 10)

–2

1–3–

a Solve the system of equations y = x2 − 8x + 12 and y = −6x + 11.b Illustrate the solution using a sketch graph. The turning point of the quadratic graph

is not required.

THINK WRITE

a Write the original equations and label them.

a y = x2 − 8x + 12 [1]y = −6x + 11 [2]

Put RHS [1] = RHS [2]. x2 − 8x + 12 = −6x + 11Collect terms on the side that makes the x2 term positive.

x2 − 8x + 12 + 6x − 11 = 0

Combine like terms. x2 − 2x + 1 = 0Factorise if possible. (x − 1)(x − 1) = 0Solve for x. x = 1 onlySubstitute x-values into the linear equation [2] to find the corresponding y-value.

If x = 1, y = −6x + 11y = −6(1) + 11y = −6 + 11y = 5

Write the solution coordinates. Solution: (1, 5) only

1

23

4

5

6

7

8

20WORKEDExample


THINK WRITE

b Find intercepts for a sketch of the linear graph.

b For y = −6x + 11,if x = 0, y = 11if y = 0, 0 = −6x + 11

6x = 11

x = or 1


For y = x2 − 8x + 12,if x = 0, y = 12if y = 0, 0 = x2 − 8x + 12so 0 = (x − 6)(x − 2)and x = 6 or x = 2

Sketch the linear and quadratic graphs on the same axes, and include the points of intersection found in part a.

1

116------ 5

6---

2

31211

x

y

6211—6

(1, 5)

Tangentpoint

a Solve the system of equations y = −x2 + 3x + 18 and y = 4x + 22.b Illustrate the solution (or lack of solution) using a sketch graph. The turning point of

the quadratic graph is not required.

Continued over page

THINK WRITE

a Write the original equations and label them.

a y = –x2 + 3x + 18 [1]y = 4x + 22 [2]

Put RHS [1] = RHS [2]. −x2 + 3x + 18 = 4x + 22Collect terms on the side that makes the x2 term positive.

0 = x2 − 3x − 18 + 4x + 22

Combine like terms. x2 + x + 4 = 0Try to factorise. No solution using the Null Factor Law is apparent. Check the value of the discriminant ∆ of the quadratic in step 4.

(x )(x ) = 0??∆ = b2 − 4ac

= (1)2 − 4(1)(4)= 1 − 16= −15

Since ∆ < 0, there is no solution. ∆ < 0, ∴ no solution.b Find intercepts for a sketch of the

linear graph.b For y = 4x + 22,

if x = 0, y = 22 (0, 22)if y = 0, 0 = 4x + 22

−22 = 4x

x = −

x = − or −5 (− , 0)

1

23

4

5

6

1

224------

112------ 1

2--- 11

2------

21WORKEDExample


THINK WRITE


For y = −x2 + 3x + 18,if x = 0, y = 18 (0, 18)if y = 0, 0 = −x2 + 3x + 18

x2 − 3x − 18 = 0so (x − 6)(x + 3) = 0and x = 6 or x = −3 (6, 0) (−3, 0)

Sketch the linear and quadratic graphs on the same axes. Note that the graphs do not intersect, indicating no solution.

2

3

22

18

x

y

6–3– 11—

2

The graphs with equations y = x2 + 4x + 33 and y = mx + 24 intersect once only. Find thepossible values of m.

THINK WRITE

Write and label the equations. y = x2 + 4x + 33 [1]y = mx + 24 [2]

Put equation [1] = equation [2] and form a new quadratic equation [3].

Put [1] = [2]x2 + 4x + 33 = mx + 24

x2 + 4x − mx + 9 = 0x2 + (4 − m)x + 9 = 0 [3]

For one solution only, we require ∆ = 0. ∆ = (4 − m)2 − 4(1)(9)= 16 − 8m + m2 − 36= m2 − 8m − 20= 0 for one solution only.

Factorise and solve for m. (m − 10)(m + 2) = 0m = 10 or m = −2

1

2

3

4

22WORKEDExample

rememberIf given equations of the form y = ax2 + bx + c and y = mx + k:1. put y = ax2 + bx + c = mx + k2. rearrange to form a new quadratic equation Ax2 + Bx + C3. solve to find any x-coordinates of intersection4. substitute any x-coordinates into the linear equation to find the corresponding

y-coordinates5. state the solutions (or state that there are none).

remember


Simultaneous quadratic and linear equations

1 For each of the following systems of equations:i solve to find any solution coordinatesii illustrate the solution (or lack of solution) using a sketch graph. The turning point ofii the quadratic graph is not required.Your teacher may permit the use of a graphics calculator to find or verify solutions. (Ifyou do use a graphics calculator, be sure to zoom in enough to determine whether thegraphs intersect once, twice or not at all, as looks can be deceptive in some WINDOWsettings.)

2 State how many points of intersection exist with each of the following pairs of simultaneous equations.

3 The graphs of equations y = x2 − 14x + 49 and y = mx + 48 intersect once only, at x = 1.Find the value of m.

4 The system of equations y = x2 − 4x + c and y = −7x + 8 has two solutions, one at x = 1,and another at x = −4. Find the value of c.

5 The graphs of y = x2 + bx − 14 and y = −9x + c intersect at (−1, 8) and (−3, 10). Findthe values of b and c.

6 Find the intersection points (if any exist) of each of the following graphs.

7 An engineer’s plans for a proposed road througha mountain are shown at right. At what heightsabove sea level will the entrance and exit to thetunnel be, given the equations of the mountainprofile and road path as shown on the plan?

8 A graphic designer draws a logo involving a parabola‘sitting’ in a V shape on a set of axes as shown at right.

Find the equation of the parabola, given it is of the formy = kx2 and the points of intersection of the V with theparabola.

a y = x2 + 6x + 5 and y = 11x − 1 b y = x2 + 5x − 6 and y = 8x − 8c y = x2 + 9x + 14 and y = 3x + 5 d y = x2 − 7x + 10 and y = −11x + 6e y = x2 − 2x − 3 and y = x − 6 f y = x2 + 11x + 28 and y = 10x + 40g y = x2 + 5x − 36 and y = 15x − 61 h y = x2 − 6x − 16 and y = −4x − 17i y = x2 − 2x − 24 and y = 4x + 3 j y = x2 − 7x + 10 and y = −4x + 6k y = −x2 + 4x + 21 and y = x + 11 l y = −x2 + 14x − 48 and y = 13x − 54m y = −x2 + 4x + 12 and y = 9x + 16 n y = x2 + 7x + 12 and y = 20o y = −x2 − 4x + 5 and y = −4x + 9 p y = x2 − 4x + 4 and y = 8x − 32

a y = x2 + x − 6 and y = −9x − 31 b y = x2 − 16 and y = 6x + 11c y = −x2 + 3x + 4 and y = −7x + 25 d y = x2 − 6x + 5 and y = 2x − 12

a y = −x2+ 6x − 5 and y = 10x − 1 b y = −x2 + x and y = −x + 1c y = 2x2 − 5x + 3 and y = −2x + 23 d y = 9x2 + 21x + 10 and y = 27x + 9

2KWORKEDExample

19, 20, 21

Mathcad

Simultaneousquadraticand linearequations

EXCEL Spreadsheet

Simultaneousquadraticand linearequations

WORKEDExample

22

x

y

y = – +5x2–2

y = +2x–4

Proposed roadEntranceof tunnel

Sea level

x

yy = kx2

–1 1

–2


Expanding quadratic expressions• To expand:

1. first term × everything in the second brackets, then2. second term × everything in the second brackets.

• Perfect squares (ax + b)2 = a2x2 + 2abx + b2

• Difference of squares (ax + b)(ax − b) = a2x2 − b2

• Expand brackets first, then multiply if there is an ‘external’ factor in expressions like k(ax + b)(cx + d).

Factorising quadratic trinomials• Look for a common factor first.• General quadratics: Write down (_x )(_x ) and try factors of the constant term.• Perfect squares: a2x2 + 2abx + b2 = (ax + b)2

• Difference of squares: ax2 − b = ( x + )( x − )

Factorising by completing the square• Use when whole number factors are not apparent.• Halve and square the x-coefficient, then add and subtract this new term.• Form a perfect square from three of the terms.• Continue to factorise using a difference of squares.

Solving quadratic equations — Null Factor Law• Factorise.• Put each factor equal to zero.• Solve two mini-equations.

Solving quadratic equations — completing the square• Halve the x-coefficient (note in margin).• Square the margin value, and add and subtract it.• Form a perfect square and solve for x.

The quadratic formula• If ax2 + bx + c = 0, then solutions may be found using .

The discriminant• ∆ = b2 − 4ac.• If ∆ < 0, no real solutions exist.• If ∆ > 0, there are two real solutions.• If ∆ = 0, there is only one real solution.

summary

3 4

(4x + 9)(2x – 3)1

2

a b a b

x−b b2 4ac–±

2a--------------------------------------=

C h a p t e r 2 Q u a d r a t i c f u n c t i o n s 105Quadratic graphs — turning point form• Turning point form: y = a(x − b)2 + c.• TP is at (b, c), dilation factor a.• To convert to TP form, complete the square.

Quadratic graphs — intercepts method• To sketch a quadratic graph of the form y = ax2 + bx + c:

1. Find the y-intercept (when x = 0)2. Factorise if possible, and find the x-intercepts (when y = 0)It may be easier to use the quadratic formula in some cases.If ∆ < 0, there are no x-intercepts. If ∆ = 0, one intercept only. If ∆ > 0, two intercepts.3. Find the x-coordinate of the

TP using or

completing the square.4. Find the y-coordinate of the TP by substituting xt into the equation for y, or

using or by completing the square.

5. Combine all the information and sketch it. Functions with a positive x2 coefficient are ∪ shaped, and those with negative x2 coefficients are ∩ shaped. If the information you have gathered doesn’t seem to fit, check for calculation errors.

Using graphs to solve quadratic equations• Enter the equation in the menu.

• Find the x-intercepts of the graph using [CALC] 2:zero.

Simultaneous quadratic and linear equations• If given equations of the form y = ax2 + bx + c and y = mx + k,

1. Put ax2 + bx + c = mx + k.2. Rearrange to form a new quadratic equation Ax2 + Bx + C.3. Solve to find any x-coordinates of intersection.4. Substitute any x-coordinates into the linear equation to find the

corresponding y-coordinates.

x

y

(b, c)

Axis of symmetry

x-interceptsy-intercepts

Turning point(– , c– )b—

2ab2—4a

x

y

xtb–

2a------=

yt cb2

4a------–=

Y=2nd


Multiple choice1 Which of the following is the expansion of (3x − 7)(x + 5)?

2 Which of the following is a perfect square quadratic expression?

3 The expression 9x2 − 64 is an example of:

(One or more answers.)

4 Which of the following is equivalent to 36x2 − 49?

5 (x − 2)2 + 8 can be factorised to:

6 Which of the following gives all solutions of the equation (2x − 7)(x + 4) = 0?

7 Which of the following represent the full solution to x2 − 6x + 9 = 0?

8 A solution to (x − 5)2 − 7 = 0 is:A x = −5 − B x = 5 + C x = 7 + D x = −7 + E x = −5 +

9 If the equation 3x2 = 9 + x is to be solved using the quadratic formula, which set of values should be substituted into the formula?

10 The value of the discriminant in 2x2 + 3x + 4 = 0 is:

11 For a quadratic equation to have at least one real solution, the discriminant must be:

12 The turning point of the graph of y = 5(x + 2)2 − 1 is at:

A 3x2 + 8x − 35 B 3x2 − 7x − 35 C 3x2 − 2x − 35D 4x2 − 2x − 2 E 4x − 2

A x2 + 10x + 16 B x2 − 8x + 16 C x2 − 25D (x + 6)(x − 6) E (x2 + 7)2

A a perfect square B a difference of squares C a quadratic equationD a quadratic term E a factorised expression

A (6x + 7)2 B (6x − 7)2 C (6x + 7)(6x − 7)D (36x + 1)(x − 49) E (9x + 7)(4x − 7)

A (x + 6)2 B (x + 6)(x − 10) C (x − 2 + 4)(x − 2 − 4)D (x − 2 + )(x − 2 + ) E the expression cannot be factorised using real numbers.

A 7, −4 B −2, −1 C 2, 1 D − , 4 E , −4

A −3 B −3, +3 C 3 D −3, −6 E no real solution

A a = 3, b = −1 and c = 0 B a = 3, b = −1 and c = −9 C a = 3, b = 9 and c = 1D a = 1, b = 3 and c = E a = 9, b = 1 and c = −3

A −23 B −7 C 25 D 32 E 41

A negative B zero C positiveD either positive or negative E a perfect square

A (1, 2) B (1, −2) C (2, −1) D (−2, 1) E (−2, −1)

CHAPTERreview

2A

2B

2B

2B

2C8 8

2D 72--- 7

2---

2D

2E 7 7 5 5 7

2F19---

2G

2G

2H

C h a p t e r 2 Q u a d r a t i c f u n c t i o n s 10713 The maximum value of y, where y = −3(x − 2)2 + 8 is:

14 The graph shown could be for the equation:

A y = x2 + 5x − 30B y = x2 − 11x + 30C y = x2 + 11x + 30D y = −x2 + 11x + 30E y = −x2 + 11x − 30

15 The turning point of y = x2 + bx + 40 is at x = −7. The value of b is:

16 One approximate solution to the quadratic equation whose graph appears at right is x =

A −2 B −1 C 1.8D 3.2 E 3.9

17 The system of 2 simultaneous equations pictured below has:

A no solutions for xB 1 solution for xC 1 positive and one negative solution for xD two negative solutions for xE insufficient information for us to find a solution.

18 A solution to the system of equations y = 2x − 1 and y = x2 + 8x + 8 is:

Short answer1 Expand:

2 Factorise:

3 Factorise x2 − 6x − 14.

4 Solve the following.

5 Solve:

6 Solve x2 + 6x − 2 = 0.

7 Use the quadratic formula to solve 5x2 − 7x + 1 = 0.

A −3 B −4 C 2 D 8 E 20

A −14 B −7 C D 7 E 14

A (−3, −7) B (0, −1) C (0, 8) D (3, 5) E (2, 28)

a (7x + 8)(7x − 8) b (2x − 9)2 c (5x − )(5x + ) d (4x − 9)(2x − 13)

a 25x2 + 110x + 121 b 6x2 + 37x + 6 c 12x2 − 37x + 21 d 36x2 − 49

a (4x + 1)(3x − 9) = 0 b 3x2 − 40x − 75 = 0

a 4x2 − 5 = 0 b x2 − 14x + 42 = 0

2H

2I

x

y

30

–5–6

2I72---

2J

2K

x

y

2K

2A3 3

2B

2C2D

2D2E2F


8 Evaluate the discriminant for 4x2 − 3x + 9 = 0.

9 Find the value/s of k for which the equation 2x2 + 3kx + 6 = 0 has:

10 Sketch, showing the turning point and y-intercept in each case (x-intercepts not required).

11 Convert y = x2 + 8x + 3 to turning point form.

12 Sketch the following.

13 Use a graphics calculator to find both roots of the equation 6.29x2 + 0.73x − 1.661 = 0.

14 Find any points of intersection of the line y = −2x − 6 and the parabola y = x2 + 7x + 12.

Analysis1 MacBurger’s restaurants have employed a mathematician

to design a new logo based on an ‘M’ made up of two parabolas as shown in the sketch at right.The parabolas both have the form y = −2x2 + bx + c.a Give the coordinates of two points on each parabola.b Find the equation of each parabola.c Find the height, h, of the sign.d Which domain of x-values should each graph be restricted to so that the parabolas form

the logo shown above?e Determine the equations of two parabolas that may be

used to form the ‘W’ shape below, given they are both of the form y = x2 + bx + c.

2 Find the equation of a parabola that goes through the points listed in each case. (Hint: Assume equations of the form y = ax2 + bx + c, and form simultaneous equations by substituting coordinate values.)a (0, 1), (1, 0) and (2, 3)b (0, −1), (1, 4) and (2, 15)c (0, 5), (−1, 11) and (1, −3)d Find the equations of two parabolas which form a path similar to

the one below depicting a proposed water slide, given the x2 coefficient of each curve is 1 or −1.

3 a Using algebra, sketch the graphs of the quadratic equations y = x2 − 6x + 8 and y = 3x2 + 5x − 28 showing intercepts and turning points.

b Why was one equation easier to sketch than the other?c Determine another quadratic function whose graph has integer intercepts and turning point

coordinates. How can this be ensured?

a no solution b one solution c two solutions.

a y = −4(x + 1)2 + 2 b y = 6x2 − 1

a y = (x − 6)(x − 14) b y = x2 + 2x − 80 c y = 3x2 − 26x + 48 d y = −2x2 − 5x − 3

2G2G

2H

2H2I

2J2K

(2, 0) x

y

(–2, 0)

Parabola 1 Parabola 2

h

(3, 0) x

y

(–3, 0)

9

x

y

(4, 4)

(8, 8)

testtest

CHAPTERyyourselfourself

testyyourselfourself

2

Quadratic functionsmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 02... · 2015-08-17 · 2...

Documents

Transcript of Quadratic functionsmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 02... · 2015-08-17 · 2...