1
VCE
VCE
co
covverage
erage
Areas of study
Units 3 & 4 • Coordinate geometry
• Algebra
In this
In this
cha
chapter
pter
1A The binomial theorem
1B Polynomials
1C Division of polynomials
1D Linear graphs
1E Quadratic graphs
1F Cubic graphs
1G Quartic graphs
2
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4The binomial theorem
In Mathematical Methods Units 1 and 2 we learnt the following binomial expansions:
(x+a)2=x2+ 2xa +a2
(x+a)3=x3+ 3x2a+ 3xa2+a3
These are called binomial expansions because the expressions in the brackets contain 2 terms, ‘bi’ meaning 2.
By continuing to multiply successively by a further (x+ a), the following expansions would be obtained:
(x+a)4= (x3+ 3x2a+ 3xa2+a3)(x+a) =x4+ 4x3a+ 6x2a2+ 4xa3+a3
(x+a)5= (x4+ 4x3a+ 6x2a2+ 4xa3+ a3)(x + a) = x5+ 5x4a + 10x3a2+ 10x2a3+ 5xa4+ a5 The coefficients can be arranged in a ‘triangle’ as shown:
Notes
1. The first and last numbers of each row are 1.
2. Each other number is the sum of the two numbers immediately above it.
This triangle is known as Pascal’s triangle. Each number can also be obtained using combinations, as follows.
Where
(x + a)0 1
(x + a)1 1 1
(x + a)2 1 2 1
(x + a)3 1 3 3 1
(x + a)4 1 4 6 4 1
(x + a)5 1 5 10 10 5 1
Row
0
1
2
3
4
0 0
1 0
1
1
2 0
2
1
2
2
3 0
3
1
3
2
3
3
4 0
4
1
4
2
4
3
4
4
n
r nC
r
n! n–r ( )!r!
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
3
Remember that nCr is another way of writing .
For example, the expansion of (x + a)6 can be written using combinations and then evaluated:
(x + a)6= x6+ x5a + x4a2+ x3a3+ x2a4+ xa5+ a6 = x6+ 6x5a + 15x4a2+ 20x3a3 + 15x2a4+ 6xa5+ a6
Now the binomial theorem can be formally stated.
(ax+b)n= (ax)n+ (ax)n− 1b+ (ax)n− 2b2+ . . . + (ax)bn− 1+ bn
Notes
1. The indices always sum to n.
2. The power of ax decreases from left to right while the power of b increases. 3. The number of terms in the expansion is always n + 1.
The binomial theorem can also be stated using summation notation:
where the (r+ 1)th term is n
r
6 0
6
1
6
2
6
3
6
4
6
5
6
6
n
0
n
1
n
2
n
n–1
n
n
ax+b
( )n n
r
( )ax n–rbr r=0
n
∑
= nr ( )ax n–rbr
Use the binomial theorem to expand (2x− 3)4.
THINK WRITE
Complete the binomial theorem expansion where ax is the 1st term, b is the 2nd term and n is the index, using the appropriate row of Pascal’s triangle to assist.
(2x− 3)4 = (2x)4+ (2x)3(−3)1
(2x− 3)4 =+ (2x)2(−3)2+ (2x)1(−3)3
+ (−3)4
Evaluate the combinations and the powers.
= 1(16x4) + 4(8x3)(−3) + 6(4x2)(9)
+ 4(2x)(−27) + 1(81)
Simplify each term. = 16x4− 96x3+ 216x2− 216x+ 81
1 4
0
4
1
4 2
4
3
4 4
2
3
1
4
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Expand the binomial expression .
THINK WRITE
Complete the binomial expansion where , b = x and n = 5 and use row 5 of Pascal’s triangle to assist.
=
=
Evaluate the powers. =
=
Simplify each term. =
2
x2
---+x
5
1 ax 2 x2 ---= 2 x2
---+x
5 2
x2
---
5
5 2
x2
---
4
x 10 2
x2
---
3
x2
+ +
+ 10 2
x2
---
2
x3 5 2
x2
---
x4 x5
+ +
2 32
x10
--- 5 16
x8
---
x 10 8
x6
--- x2
+ +
+ 10 4
x4
---
x3 5 2
x2
---
x4 x5
+ + 3 32 x10 --- 80 x7 --- 80 x4 --- 40 x
--- 10x2 x5
+ + + + +
2
WORKED
E
xample
State the coefficient of i x2 and ii x4 in (3 − 2x)8.
THINK WRITE
ii The powers of the 1st term decrease and the powers of the 2nd term increase 0, 1, 2, . . . Use this to find which term gives a power of x2.
ii x0, x1, x2
The third term gives a power of x2.
Find the appropriate term by using the binomial theorem.
Third term = 36 (−2x)2
Evaluate the term. = 28 × 729 × 4x2
= 81 648x2
State the coefficient. The coefficient of x2 is 81 648.
ii Find which term gives a power of x4. ii x0, x1, x2, x3, x4
The fifth term gives a power of x4.
Evaluate the term. Fifth term = 34 (−2x)4
= 70 × 81 × 16x4
= 90 720x4
State the coefficient. The coefficient of the fifth term is 90 720.
1 2 8 2 3 4 1 2 8 4 3
3
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
5
Find and evaluate the term which is independent of x in the expansion of .
THINK WRITE
Find how the powers of x are generated in the expansion from left to right.
Powers of x are (x3)5=x15, ,
, . . .
that is, x15, x10, x5, x0
Find the required term. The fourth term is independent of x.
Evaluate. Fourth term=
Fourth term= 10x6 Fourth term= 10
State the solution. The term which is independent of x is the fourth term, 10.
x3 1
x2
---+
5
1 ( )x3 4 1
x2
---
= x10
x3
( )3 1
x2
---
2
x5
= ( )x3 2 1
x2
---
3
x0
=
2
3 5
3
( )x3 2 1
x2
---
3
1
x6
---
4
4
WORKED
E
xample
Find the coefficient of y4 in the expansion of (y+ 3)3 (2 −y)5.
THINK WRITE
y4 terms will result when multiplying from the first and second brackets respectively: terms 1 and 2, terms 2 and 3, terms 3 and 4 and terms 4 and 5. Write down the sum of these 4 products using Pascal’s triangle to assist.
y4 terms=y3[5(2)4(−y)] + 3y2(3)[10(2)3(−y)2]
+ 3y(3)2[10(2)2(−y)3] + 33[5(2)(−y)4]
Evaluate. = −80y4+ 720y4− 1080y4+ 270y4
= −170y4
State the solution. The coefficient of y4 is −170.
1
2
3
4
5
6
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4The binomial theorem
1 Use the binomial theorem to expand each of the following:
2 Expand each of the following binomial expansions:
3 State the coefficient of i x2 ii x3 and iii x4 in each of the following:
a (x + 3)2 b (x + 2)3 c (x + 4)5
d (x − 1)8 e (x − 5)3 f (2x + 3)4
g (3x − 4)5 h (1 + x)7 i (7 − x)4
j (2 − 3x)5 k (x + 2y)3 l (3y − 2x)6
a b c
d e f
g h
a (x − 7)3 b (2x + 1)5 c (3x − 4)8
d (5 − 4x)7 e f
g h i
j
remember
1. Pascal’s triangle
2. Binomial theorem
(ax+b)n= (ax)n+ (ax)n− 1b+ . . . + (ax)bn− 1+ bn
where the (r+ 1)th term is 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
n
0
n
1
n
n–1
n
n
ax+b
( )n n
r
( )ax n–rbr r=0
n
∑
= nr ( )ax n–rbr
remember
1A
Skill
SHEET
1.1
Mathc
ad
Expanding
Mathc
ad
Binomial theorem
G Cprogr
am Binomial theorem WORKED Example 1 WORKED Example
2 x 1
x ---+
3 x 1
x ---–
5
2x 1
x ---+
4
3x 2 x ---–
7
x2 3
x ---+
6 5
x ---–x2
4
1 x2
---+x
8 3
x2
---–2x
5
WORKED
Example 3
2 x ---+3x
5 x2 3
x ---–
6
3x2 2
x ---+
3 5
x ---–x2
9 7x 3
x2
---+
6
x3 4
x ---–
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
7
4
The coefficient of x4 in (2x − 7)6 is:
5
The coefficient of x3 in is:
6
Which of the following does not have an x5 term when expanded?
7
If , then a + b + c + d + e + f equals:
8
Which one of the following expressions is not equal to (2x − 3)4?
9 Find the third term in the expansion of , assuming ascending powers of x.
10 Find and evaluate the term that is independent of x in the expansion of .
11 Find and evaluate the term that is independent of x in the expansion of .
12 Find the coefficient of p4 in the expansion of (p + 3)5 (2p − 5).
13 Find the coefficient of m5 in the expansion of (1 − m)6 (2m + 3)4.
A 27 440 B 2744 C −784 D −9604 E 11 760
A −135 B −45 C −75 D 45 E 135
A (x + 6)8 B C
D (8 − 3x)5 E
A 15 B 31 C 63 D 243 E 127
A (3 − 2x)4 B (2x − 3)(2x − 3)3
C D 16x4− 24x3+ 36x2− 54x + 81
E 16x4− 96x3+ 216x2− 216x + 81 m
multiple choiceultiple choice
m
multiple choiceultiple choice
3x2 5 x ---–
3
m
multiple choiceultiple choice
3x2 1
x ---–
7
6x 5
x ---+
8
2x 1
x2
---–
8
m
multiple choiceultiple choice
x3 2
x2
---+
5
ax15 bx10 cx5 d e
x5
--- f x10
---+ + + + +
=
m
multiple choiceultiple choice
2x–3 ( )6
3–2x ( )2
---3 x
4 ---–
9
W WORKEDORKED
E Examplexample
4
3x 2
x2
---+
6
x3 5
x ---–
8
WORKED
8
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Polynomials
A polynomial in x is an expression which consists of terms which have non-negative integer powers of x only.
P(x) is a polynomial in x if:
P(x) = anxn+ an− 1xn− 1+ . . . + a2x2+ a1x + a0
where n is the degree (or highest power) of the polynomial and is a non-negative integer. The values of an, an− 1, . . ., a2, a1 and a0 are called the coefficients of their respective power of x terms.
Polynomials can be added and subtracted by collecting like terms.
Evaluating polynomials
A value for a polynomial, P(x), can be found for a particular value of x by simply sub-stituting the given value of x into the polynomial expression and evaluating. That is, polynomial functions are evaluated in the same way as any function.
Which of the following expressions are not polynomials? A x6 − 4x4+ 2x3+ 7x B
C 7 − 3xy+ 4x2−x3+ D 8 + 2x− 3x2+ 9x3−x4
THINK WRITE
A and D are polynomials because they are expressions with non-negative integer powers of x only.
B is not a polynomial as it has a power of which is not an integer.
C is not a polynomial as it has a power of which is not an integer and it also has one term, −3xy, which is not a power of x only.
B and C are not polynomials. x
9 2
---x3–3x2+6x–5
+
x
1
2
9 2
---3
1 2 --- ( )
6
WORKED
E
xample
Given that P(x) = 6 − 2x+ 3x2+x4, Q(x) =x5− 2x4+x2− 5x− 2 and R(x) = x2− 4, find:
a P(x) +Q(x) b P(x) −R(x)
.
THINK WRITE
a Add the polynomials. a P(x) +Q(x) = 6 − 2x+ 3x2+x4+x5− 2x4
P(x) +Q(x) =+x2− 5x− 2
Collect like terms. P(x) +Q(x)=x5−x4+ 4x2− 7x + 4
b Subtract the polynomials. b P(x) −R(x)= 6 − 2x+ 3x2+x4− (x2− 4)
Remove brackets. = 6 − 2x+ 3x2+x4−x2+ 4
Collect like terms. =x4+ 2x2− 2x+ 10
1
2
1 2 3
7
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
9
For the polynomial P(x) = 2x4−x3+ 5x2− 6x + 4, find:
a its degree b P(1) c P(−2).
THINK WRITE
a The degree of the polynomial is the highest power of x.
a Degree of P(x) is 4.
b Substitute the given value of x into the polynomial expression.
b P(1) = 2(1)4− (1)3+ 5(1)2− 6(1) + 4
Evaluate. = 2 − 1 + 5 − 6 + 4
= 4
c Substitute the given value of x into the polynomial expression.
c P(−2) = 2(−2)4− (−2)3+ 5(−2)2− 6(−2) + 4
Evaluate. = 32 + 8 + 20 + 12 + 4
= 76
1
2
1
2
8
WORKED
E
xample
If P(x) =ax5+x4− 3x3+bx− 5, P(−1) = −5 and P(2) = −65, find the values of a and b.
THINK WRITE
Substitute a given value of x into the polynomial and equate it to the given answer.
P(−1)=a(−1)5+ (−1)4− 3(−1)3+b(−1) − 5
= −5
Simplify the equation. −a+ 1 + 3 −b− 5 = −5
−a+ 4 −b= 0
Make a the subject of the equation and call this equation [1].
a= 4 −b [1]
Substitute a given value of x into the polynomial and equate it to the given answer.
P(2)=a(2)5+ (2)4− 3(2)3+b(2) − 5
= −65
Simplify the equation. 32a+ 16 − 24 + 2b− 5 = −65 32a+ 2b− 13 = −65
Substitute the value of a into this equation.
Substituting a= 4 −b:
32(4 −b) + 2b− 13 = −65
Solve this equation for b. 128 − 32b+ 2b− 13 = −65
−30b+ 115 = −65 30b= 180
b= 6
Substitute the value of b into equation [1].
Substituting b= 6 into equation [1]:
a= 4 − 6
Find the value of a. a= −2
State the solution. Therefore, a= −2 and b= 6.
1
2
3
4
5
6
7
8
9 10
9
10
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Polynomials
1 Which of the following are not polynomial expressions?
viii x3− 2x viii x4+ 3x2− 2x +
viii x7+ 3x6− 2xy + 5x iiiv 3x8− 2x5+ x2− 7
iiiv 4x6 − x3+ 2x − 3 iivi 2x5+ x4− x3+ x2+ 3x −
ivii 5x10− x7+ x4 viii + x2− 8x + 9 2 Given that P(x) = 8 − 3x + 2x2+ x4, Q(x) = x5− 3x4− 4x2− 1 and
R(x) = 8x3+ 7x2 − 4x then find:
3 For each of the following polynomials find: i its degree ii P(0) iii P(2) and iv P(−1).
4
If P(x) = x8− 3x6+ 2x4− x2+ 3, then P(−2) is equal to:
5 If P(x) = 2x7+ ax5+ 3x3+ bx − 5, P(1) = 4 and P(2) = 163, find a and b. 6 Find a and b, given that f(x) = ax4+ bx3− 3x2− 4x + 7, f(1) = −2 and f(2) = −5. 7 For Q(x) = x5+ 2x4+ ax3− 6x + b, Q(2) = 45 and Q(0) = −7. Find a and b. 8 Find a and b if P(x) = ax6+ bx4+ x3− 6, 3P(1) = −24 and 3P(−2) = 102. 9
a If P(x) = ax4− x3+ 3x2− 5 and P(1) = −1, then a is equal to:
b If f(x) = xn− 2x3+ x2− 5x and f(2) = 10, then n is equal to:
To evaluate several values of a function at once, type Y1({−3, −2, −1, 1, 2, 3}) (for
example) at the home screen, and press .
a P(x) + Q(x) b Q(x) − R(x) c 3P(x) − 2R(x) d 2P(x) − Q(x) + 3R(x)
a P(x) = x6+ 2x5− x3+ x2 b P(x) = 3x7− 2x6+ x5− 8 c P(x) = x5− 4x3+ 3x2+ 2x+ 1 d P(x) = 4x4− 2x3+ x2− 7x − 10 e P(x) = 5x6+ 3x4− 2x3− 6x2+ 3 f P(x) = −7 + 2x − 5x2+ 2x3 − 3x4
A 479 B 95 C 31 D 481 E 103
A 1 B 0 C 2 D −3 E −2
A 4 B 6 C 7 D 5 E −1
remember
1. If P(x) =anxn+an− 1xn− 1+ . . . +a2x2+a1x+a0 and n is a non-negative integer then P(x) is a polynomial of degree n and an, an− 1, . . . a2, a1 are called coefficients and ∈R.
2. A polynomial P(x) is evaluated in the same way as any function.
remember
1B
WORKED
Example
6 x
2 x ---3x
7 2
---WORKED
Example 7
Mathc
ad
Evaluating polynomials
E XCEL S
preadsheet
Evaluating polynomials
WORKED
Example 8
m
multiple choiceultiple choice
SkillS HEET
1.2
WORKED
Example 9
m
multiple choiceultiple choice
Graphics Calculator
Graphics Calculator
tip!
tip!
Finding several values
of a function
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
11
Division of polynomials
When one polynomial, P(x) is divided by another, D(x), then the result can be expressed as:
P(x) = Q(x) × D(x) + R(x) where Q(x) is called the quotient,
R(x) is called the remainder, D(x) is called the divisor.
Find the quotient, Q(x), and the remainder, R(x), when x4− 3x3+ 2x2− 8 is divided by the linear expression x+ 2.
THINK WRITE
Set out the long division with each
polynomial in descending powers of x. If one of the powers of x is missing, include it with 0 as the coefficient.
Divide x into x4 and write the result above. Multiply the result x3 by x + 2 and write the result underneath.
Subtract and bring down the remaining terms to complete the expression.
Divide x into −5x3 and write the result above. Continue this process to complete the long division.
The polynomial x3− 5x2+ 12x − 24, at the top, is the quotient.
The quotient is x3− 5x2+ 12x− 24.
The result of the final subtraction, 40, is the remainder.
The remainder is 40.
1 x+ 2
)
x4− 3x3+ 2x2+ 0x− 82
3
x + 2
)
x4− 3x3+ 2x2+ 0x − 8x3
x + 2
)
x4− 3x3+ 2x2+ 0x − 8 x3x4+ 2x3 4
x + 2
)
x4− 3x3+ 2x2+ 0x − 8 x3− 5x2−(x4+ 2x3)
− 5x3+ 2x2+ 0x− 8
5
6 x + 2
)
x4− 3x3+ 2x2+ 0x − 8x3− 5x2+ 12x − 24
−(x4+ 2x3)
− 5x3+ 2x2+ 0x − 8
−(− 5x3− 10x2) 12x2+ 0x − 8
−(12x2+ 24x)
−24x− 8
−(−24x− 48)
40
7
8
10
12
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Note: P(−2) = (−2)4− 3(−2)3+ 2(−2)2− 8 = 16 + 24 + 8 − 8
= 40
The remainder when P(x) is divided by (x + 2) is P(−2). This leads to the remainder theorem which states: When P(x) is divided by (x−a), the remainder is P(a) or
when P(x) is divided by (ax+b), the remainder is .
Furthermore, if the remainder is zero, then (x − a) is a factor of P(x). This leads to the factor theorem which states:
If P(a) = 0, then (x−a), is a factor of P(x) or
if (ax+b) is a factor of P(x), then = 0.
Note: If (x − a) is a factor of P(x) = anxn+ an− 1xn− 1+ . . . + a1x + a0, then a must be a factor of a0.
P b
a
---–
P b a
---–
Determine whether or not D(x) = (x− 3) is a factor of P(x) = 2x3− 4x2− 3x− 8.
THINK WRITE
Evaluate P(3). P(3)= 2(3)3− 4(3)2− 3(3) − 8
= 54 − 36 − 9 − 8
= 1
If P(3) = 0 then (x − 3) is a factor of P(x), but if P(x) ≠ 0, (x − 3) is not a factor of P(x).
P(3) ≠ 0 so (x− 3) is not a factor of P(x).
1
2
11
WORKED
E
xample
a Factorise P(x) = 2x3−x2− 13x− 6. b Solve2x3−x2− 13x− 6 = 0.
THINK WRITE
a Use the factor theorem to find a value for a where P(a) = 0 and a is a factor of the numerical term. Try a = ±1, ±2, ±3, ±6 until a factor is found.
a P(1) = 2(1)3− (1)2− 13(1) − 6
= −18
≠ 0
P(−1) = 2(−1)3− (−1)2− 13(−1) − 6
= 4
≠ 0
P(2) = 2(2)3− (2)2− 13(2) − 6
= −20
≠ 0
P(−2) = 2(−2)3− (−2)2− 13(−2) − 6
= 0
So(x+ 2) is a factor.
1
12
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
13
These solutions can be checked by drawing the graph of y = 2x3−x2− 13x − 6 on a graphics calculator. The x-intercepts should be the same as the solutions of worked example 12 b.
THINK WRITE
Divide P(x) by the divisor (x + 2) using long division.
Express P(x) as a product of linear and quadratic factors.
P(x)= (x+ 2)(2x2− 5x− 3)
Factorise the quadratic, if possible. = (x+ 2)(2x+ 1)(x− 3)
b Rewrite the equation in factorised form, using the answer to part a.
b 2x3−x2− 13x− 6 = 0 (x+ 2)(2x+ 1)(x− 3) = 0
Use the Null Factor Law to state the solutions.
x=−2, − or 3
2 x+ 2
)
2x3−x2− 13x− 62x2− 5x− 3
−(2x3+ 4x2)
−5x2− 13x− 6
−(−5x2− 10x)
−3x− 6
−(−3x− 6)
0
3
4
1
2 12
---remember
1. P(x) =Q(x) ×D(x) +R(x)
where Q(x) is called the quotient,
R(x) is called the remainder,
D(x) is called the divisor.
2. Remainder theorem:
If P(x) is divided by (x−a), then the remainder is P(a). 3. Factor theorem:
If P(a) = 0, then (x−a) is a factor of P(x). If (ax+b) is a factor of P(x) then = 0.
4. If (x−a) is a factor of P(x) =anxn+an− 1xn− 1+ . . . +a1x+ a0, then a must be a factor of a0.
P b a
---–
14
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Division of polynomials
1 Find the quotient, Q(x), and the remainder, R(x), when each of the following poly-nomials are divided by the given linear expression.
2 a For each corresponding polynomial in question 1 evaluate:
b Compare these values to R(x) in question 1 and comment on the result. 3 In each of the following determine whether or not D(x) is a factor of P(x).
a P(x) = x3+ 9x2+ 26x − 30, D(x) = x − 3 b P(x) = x4− x3− 5x2− 2x − 8, D(x) = x + 2 c P(x) = x5+ 2x4− 3x3+ 7x + 11, D(x) = x + 1 d P(x) = x6− 2x5− 8x4− x3+ 5x2− 4x, D(x) = x − 4 e P(x) = 2x4+ 3x3− 32x2+ 14x − 5, D(x) = x + 5 f P(x) = x7− 2x5+ x4− 3x2+ 7x − 9, D(x) = x − 2 g P(x) = 4 − 9x + 6x2− 13x3− 12x4+ 3x5, D(x) = 4 − x h P(x) = x4− 4x3− 19x2 + x − 15, D(x) = x + 3
i P(x) = 4x6+ 2x5− 8x4− 4x3 + 6x2− 9x − 6, D(x) = 2x + 1 j P(x) = 40 − 31x + 11x2− 22x3+ 8x4, D(x) = 5 − 2x 4
Examine the equation f(x) = x4− 4x3− x2+ 16x − 12. a Which one of the following is a factor of f(x)?
b When factorised, f(x) is equal to:
5 Factorise the following polynomials.
6 Solve each of the following equations.
7 If (x − 2) is a factor of x3+ ax2− 6x − 4, then find a.
8 Find the value of a if (x + 3) is a factor of 2x4+ ax3− 3x + 18.
9 Find the value of a and b if (x + 1) and (x − 2) are factors of ax3− 4x2+ bx − 12. a x3− 2x2+ 5x − 2, x − 4 b x4+ x3+ 3x2− 7x, x − 1
c x5− 3x3+ 4x + 3, x + 3 d 2x6− x4+ x3+ 6x2− 5x, x + 2 e 6x4− x3+ 2x2− 4x, x − 3 f x4− 13x2+ 36, x − 2
g 3x4− 6x3+ 12x, 3x + 1 h x5+ 3x3− 4x2+ 6x − 8, 3 − 2x
i P(4) ii P(1) iii P(−3) iv P(−2)
v P(3) vi P(2) vii P(− ) viiiP( )
A x + 1 B x C x + 2 D x + 3 E x − 4
A (x + 1)(x − 3)(x + 4) B (x + 2)(x − 2)(x − 3)(x − 1) C (x + 2)(x − 4)(x + 3)(x + 1) D (x − 1)(x + 1)(x − 3)(x − 4) E x(x − 1)(x + 2)(x + 3)
a P(x) = x3+ 4x2− 3x − 18 b P(x) = 3x3− 13x2− 32x + 12 c P(x) =−x3+ 12x − 16 d P(x) = 8x3+ 10x2− 38x + 20 e P(x) = x4+ 2x3− 13x2− 14x + 24 f P(x) =−72 − 42x + 19x2+ 7x3− 2x4 g P(x) = x4+ 2x3− 7x2− 8x + 12 h P(x) = 4x4+ 12x3− 24x2 − 32x
a x3− 3x2− 6x + 8 = 0 b x3+ x2− 9x − 9 = 0 c 3x3+ 3x2− 18x = 0 d 2x4+ 10x3− 4x2− 48x = 0 e 2x4+ x3− 14x2− 4x + 24 = 0 f x4− 2x2+ 1 = 0
1C
GCpr ogram
Division of polynomials
Mathc
ad
Division of polynomials
WORKED
Example 10
E XCEL S
preadshee t
Finding factors of polynomials
1 3
--- 3
2 ---Mathc
ad
Evaluating polynomials
WORKED
Example 11
m
multiple choiceultiple choice
WORKED
Example 12a
WORKED
Example 12b
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
15
Linear graphs
Revision of properties of straight line graphs
1. The gradient of a straight line joining two points is:2. The general equation of a straight line is: y = mx + c where m is the gradient
and c is the value of the y-intercept.
3. The equation of a straight line passing through the point (x1, y1) and having a gradient of m is:
y − y1= m(x − x1)
4. The intercept form of the equation of a straight line is: =1 or bx + ay = ab
5. The product of the gradients of two lines which are perpendicular equals −1.
That is, m1× m2= −1 or m1=−
y
x 0
B (x2, y2)
A (x1, y1)
m y2–y1
x2–x1 ---=
y
x 0
A (x1, y1)
Gradient = m
y
x 0
(0, b)
(a, 0)
x a --- y
b ---+
1 m2
---Sketch the graph of the linear function 3x− 2y= 6 by indicating the intercepts.
THINK WRITE
Substitute y = 0 into the equation. When y= 0, 3x− 2 × 0 = 6
Solve the equation for x to find the x-intercept.
3x= 6
x= 2 Therefore, the x-intercept is 2.
Substitute x = 0 into the equation. When x= 0, 3 × 0 − 2y= 6
Solve the equation for y to find the y-intercept.
−2y= 6
y= −3 Therefore, the y-intercept is −3.
Draw a set of axes.
Indicate the x-intercept and y-intercept and rule a line through these points. 1
2
3 4
5 6
y
x 0
3x – 2y = 6
–3
2
13
16
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Find the equation, in the form ax+by+c= 0, of each straight line described below.
a The line with a gradient of 2 and passing through (3, −2).
b The line passing through (0, 8) and (−2, 2).
c The line which passes through (3, 4) and is parallel to the line with equation y− 2x− 5 = 0.
THINK WRITE
a Write the rule for the point–gradient form of the equation of a straight line, y − y1= m(x − x1).
a y−y1=m(x−x1)
Substitute the value of the gradient, m, and the coordinates of the point (x1, y1), into the equation.
y− (−2) = 2(x− 3)
Expand the brackets. y+ 2 = 2x− 6
Express the equation in the form required.
y− 2x+ 8 = 0 or 2x−y − 8 = 0
b Write the rule for the gradient, m, of a straight line, given 2 points.
b m=
Substitute the values of m and (x1, y1), into the rule and evaluate the gradient.
=
= = 3
Substitute the value of m, and (x1, y1), into the rule for the point gradient form of the equation of a straight line. (Coordinates of either point given may be used.)
⇒ y− 8 = 3(x− 0)
Expand the brackets. y− 8 = 3x
Express the equation in the form required.
y− 3x− 8 = 0 or 3x−y+ 8 = 0
c State the gradient of the given line which is the same as the gradient of the parallel line.
c y− 2x− 5 = 0 becomes y= 2x+ 5. The gradient of the parallel lines is 2.
Write the rule for the point/gradient form of the equation of a straight line.
y−y1=m(x−x1)
Substitute the values of m and the coordinates (x1, y1) = (3, 4).
y− 4 = 2(x− 3)
Simplify and write in the required form.
y− 4 = 2x− 6
y= 2x− 2 2x−y− 2 = 0
1
2
3
4
1
y2–y1 x2–x1
---2 2–8
−2–0
---6 –
2 –
---3
4
5
1
2
3
4
14
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
17
The domain and range of functions
The domain of a function, y = f(x), is the set of values of x for which the function is defined (that is, all x-values that can be substituted into f(x) and an answer found).
The range of f(x) is the set of values of y for which the function is defined.
If the rule and the domain of a function are given, then the function is completely defined.
For example, y = −4x, x ≤ 0 f(x) = −4x, x ≤ 0
or f: (−∞, 0] → R, f(x) = −4x
Interval notation
Restricted domains or ranges can be represented by interval notation in three forms. 1. The closed interval. 2. The open interval. 3. The half-open interval.
[a, b] = {x: a ≤ x ≤ b} (a, b) = {x: a < x < b} [a, b) = {x: a ≤ x < b} If the domain or range is unrestricted, it can be denoted as R or (−∞, ∞).
R+ = (0, ∞) R− = (−∞, 0)
a b a b a b
Sketch the graph of each of the following functions, stating the domain and range of each.
a 4x− 2y= 8, x∈ [−3, 3] b f(x) = 1 − 2x, x∈ (−∞, −1)
Continued over page
THINK WRITE
a Substitute the smallest value of x into the equation.
a When x= −3, 4(−3) − 2y= 8
Solve the equation for y, to find an end point of the straight line.
−12 − 2y= 8
−2y= 20
y= −10
State the coordinates of the end point.
(−3, −10) is a closed end of the line.
Substitute the largest value of x into the equation.
When x= 3, 4(3) − 2y= 8
Solve the equation for y, to find the other end point of the line.
12 − 2y= 8
−2y= −4
y= 2
State the coordinates of the 2nd end point.
(3, 2) is the other closed end of the line.
Plot the two points on a set of axes with closed circles (since both points are included).
Draw a straight line between the two points.
1
2
3
4
5
6
7
8
y
x 0
–3
(–3, –10)
(3, 2)
–10 –4
3 2 2
4x – 2y = 8
x∈[–3, 3]
15
18
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4THINK WRITE
Find the intercepts and mark them on the graph.
x= 0, 4 × 0 − 2y= 8
−2y= 8
y= −4
y= 0, 4x− 2 × 0 = 8 4x= 8
x= 2
The x-intercept is 2 and the y-intercept is −4.
State the domain which is given with the rule.
Domain is [−3, 3].
State the range from the graph. Range is [−10, 2].
b There is no smallest value of x so substitute the largest value of x into the equation.
b When x= −1,
y=f(−1)
= 1 − 2(−1)
Evaluate f(−1). y= 1 + 2
= 3
State the coordinates of the upper end point.
(−1, 3) is an open end of the line.
Substitute another value of x within the domain into the equation (that is, a value of x < −1, since x ∈ (−∞, −1)).
When x= −2,
y=f(−2)
= 1 − 2(−2)
Evaluate f(−2). y= 1 + 4
= 5
State the coordinates of the point. (−2, 5) is another point on the line.
Plot the 2 points on a set of axes and mark the point (−1, 3) with an open circle.
Rule a straight line from (−1, 3) to (−2, 5) and beyond. An arrow may be placed on the other end to indicate that the line continues.
Note that there are no intercepts. State the domain that is given with the rule.
Domain is (−∞, −1).
State the range by examining the graph.
Range is (3, ∞).
9
10
11
1
2
3
4
5
6
7 y
x 0 –2 –1 (–1, 3) (–2, 5)
3 5 f(x) = 1 – 2x,
x∈(–∞, –1)
8
9
10
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
19
Linear graphs
1 Sketch the graph of each of the following linear functions by indicating the intercepts.
2 Find the equation, in the form ax + by + c = 0, of each straight line described below. a The line with a gradient of 3 and passing through (2, 1).
b The line with a gradient of −2 and passing through (−4, 3). c The line with a gradient of 4 and passing through (0, 3).
3 Find the equation, in the form ax + by + c = 0, of each straight line described below. a The line passing through (0, 2) and (3, 6).
b The line passing through (−3, −4) and (−1, −10). c The line passing through (7, 5) and (2, 0). 4
Which one of the following points does not lie on the straight line with equation 2y − 3x − 6 = 0?
5 Consider the points A(−2, 5) and B(1, b). a Find b if:
i the gradient of the straight line AB is −2 ii the equation of the straight line AB is y − x = 7.
b Find the general equation of the straight line which passes through (4, 5) and is parallel to the line with equation y − 3x + 4 = 0.
a 2x + 3y = 12 b x − 4y = 8 c 2y − 5x = 10 d 4x + 2y + 4 = 0 e 5y − 3x − 15 = 0 f 2x − y = 1 g 3y + x − 2 = 0 h y = 5
A (2, 6) B (−2, 0) C (0, 3) D (1, 2) E (4, 9)
remember
Linear graphs
1. Gradient, m =
2. General equation is
ax+by+c=0 or y=mx+c
where m= gradient and c=y-intercept. 3. Equation if a point and gradient is known:
y−y1=m(x−x1) 4. Equation if the intercepts are known:
= 1
5. If m1 and m2 are the gradients of perpendicular lines, then:
m1×m2= −1
or m1=−
y2–y1 x2–x1
---x a
--- y
b
---+
1
m2
---remember
1D
Math
cad Linear graphs
WORKED
Example 13
GC progr am Finding a linear equation
WORKED
Example 14a
SkillS HEET
1.3
WORKED
Example 14b
m
multiple choiceultiple choice
Skill
SHEET
1.4
WORKED
20
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 46 Match each of the following graphs with the appropriate rule below.
a b c
d e f
iii x+ 2y+ 4 = 0 iii y= 3 iii y− 2x− 2 = 0
iv 3y+ 2x= 6 iv y− 2x= 0 vi x= −2
7 State the range for each function graphed below.
a b c
d e f
8 Sketch the graph of each of the following functions, stating i the domain and ii the range of each.
9 Find the equation of the straight line which passes through the point (2, 5) and is:
a parallel to the line with equation y= 3 − 2x
b perpendicular to the line with equation y= 3x− 7. Write equations in the form ax+by+c= 0.
10 Find the equation of the straight line which passes through the point (−3, 1) and is:
a parallel to the line with equation 4x− 2y= 13
b perpendicular to the line with equation 4x− 2y= 13.
11
If the straight lines 3x−y= −2 and ax+ 2y= 3 are parallel then a= :
a x− 4y= 8 b y= 3x− 1, x≥ 0
c 4y+ 3x= 24, x∈ [−12, 12] d 2x− 5y= 10, x < 5
e 2y + 4x − 4 = 0, x ∈ [−8, ∞) f f(x) = 3x − 6, x ∈ (−6, 4)
g f(x) = 5x + 4, x ∈ (−∞, 3] h 4x − 3y − 6 = 0, x ∈ [2, 5)
A 6 B 2 C −2 D −3 E −6
y
x 0 2
–1
y
x 0
(2, 4)
y
x 0
2
3
y
x 0 –2
y
x 0 –4
–2
y
x 0
3
SkillS
HEET
1.5
SkillS
HEET
1.6
y
x 0
(–5, –2)
y
x 0
(6, –5)
y
x 0
(–4, –2)
(4, 3)
y
x 0
(5, –2)
(–3, 3) y
x
0 4
y
x 0
(5, 6)
W WORKEDORKED
E Examplexample
15
m
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
21
Quadratic graphs
Revision of quadratic functions
1. The general form of the quadratic function is y = ax2+ bx + c, x ∈ R. 2. The graph of a quadratic function is called a parabola and:
(a) for a > 0, the graph has a minimum value (b) for a < 0, the graph has a maximum value (c) the y-intercept is c
(d) the equation of the axis of symmetry is x =
(e) the x-intercepts are found by solving the equation ax2+ bx + c = 0. 3. The equation ax2+ bx + c = 0 can be solved by either:
(a) factorising or
(b) using the quadratic formula, .
4. The turning point can be found by ‘completing the square’ (see later), and it is located on the axis of symmetry which is halfway between the x-intercepts.
The discriminant
The value of (b2− 4ac), which is the value inside the square root sign in the quadratic formula, determines the number of solutions to a quadratic equation or the number of x-intercepts on a quadratic graph.
This value is called the discriminant.
1. If b2− 4ac > 0, there are two solutions to the equation and there are two x-intercepts on the graph.
2. If b2− 4ac = 0, the two solutions are equal and there is one x-intercept on the graph; that is, the graph has a turning point on the x-axis.
3. If b2− 4ac < 0, there are no real solutions and there are no x-intercepts on the graph. b
2a ---–
x −b± b2–4ac
2a ---=
Use the discriminant to determine the number of x-intercepts for the quadratic function:
f(x) = 2x2+ 3x− 10
THINK WRITE
Find the values of the quadratic coefficients a, b and c using the general quadratic function:
y = ax2+ bx + c
a= 2, b= 3, c= −10
Evaluate the discriminant. b2− 4ac= 32− 4(2)(−10)
= 9 + 80
= 89
If the discriminant is:
(a) greater than 0, there are two x-intercepts
(b) equal to 0, there is one x-intercept (c) less than 0, there are no x-intercepts.
b2− 4ac> 0
sothere are 2 x-intercepts.
1
2
3
16
22
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4The turning point of a quadratic function is exactly halfway between the two x-intercepts so it is possible to find the turning point for worked example 17.
Sketch the graph of the function f(x) = 12 − 5x− 2x2, showing all intercepts.
THINK WRITE
Evaluate f(0) to find the y-intercept (or state the value of c).
f(0)= 12 − 5(0) − 2(0)2
= 12
State the y-intercept. The y-intercept is 12.
Set f(x) = 0 to find the x-intercepts. f(x) = 12 − 5x− 2x2= 0
Factorise the quadratic (or use the quadratic formula).
(4 +x)(3 − 2x) = 0
Solve the equation using the Null Factor Law.
4 +x= 0− or 3 − 2x= 0
⇒ 4 +x= −4 or 3 − 2x=
State the x-intercepts. The x-intercepts are −4 and .
Draw a set of axes and mark the intercepts or the coordinates of the points where the graph crosses the axes. Sketch a parabola through the
intercepts. 1
2
3
4
5
3 2
---6 32
---7 y
x 0
–4 1 2
12 (0, 12)
(–4, 0) (3– , 0)
2
f(x) = 12 – 5x – x2
8
17
WORKED
E
xample
Find the turning point of the function drawn in worked example 17.
THINK WRITE
Find an x-value which is halfway between x = −4 and x = .
For the turning point:
Halfway is x=
Simplify. Halfway is x=− (or −1 )
Substitute the x-value into the original equation to find the y-coordinate of the turning point.
y= 12 − 5(− ) − 2(− )2
Simplify. y= 15
State the coordinates of the turning point and mark the point on the graph.
Turning point is (−1 , 15 ).
1
3 2
---4 – +32
---2
---2 54--- 1
4
---3 54--- 5
4
---4 18
---5 14--- 1
8
---18
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
23
Finding turning points by completing the square
Consider the general quadratic equation:y = ax2+ bx + c
By completing the square, this equation may be manipulated into the form y = A(x + B)2+ C
where the turning point is (−B, C) and A = a.
The function graphed is of the form y=x2+bx+c. Find:
a the rule
b the domain
c the range.
Write the answers to b and c in interval notation.
THINK WRITE
a Write the general rule for a quadratic in turning point form.
a y=A(x+B)2+C
Find the values of B and C using the given turning point.
Since the turning point is (−1, −6).
B= 1
C= −6
State the value of A (given). A= 1
Substitute these values in the rule. So y= 1(x+ 1)2− 6
Expand the brackets. =x2+ 2x+ 1 − 6
Simplify. ⇒ y=x2+ 2x− 5
The rule is y=x2+ 2x− 5
b Use the graph to find the domain. Look at all the values that x can take.
b x≥ −5
State the domain in interval notation. Domain = [−5, ∞)
c Use the graph to find the range. Look at all the values that y can take.
c y≥ −6
State the range in interval notation. Range = [−6, ∞)
y
x 0
(–1, –6) (–5, 5)
1
2
3
4
5
6
1
2
1
2
19
24
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4In general the turning point of a quadratic function is required if the range needs to be determined. However, the x-intercepts and y-intercept are not required in deter-mining the range of quadratic functions. Sketch graphs are also useful.
Sketch the graph of y= 3 + 8x− 2x2, showing the turning point and all intercepts.
THINK WRITE
Find y when x = 0. Whenx= 0,
y= 3
State the y-intercept. The y-intercept is 3.
Let the quadratic equal zero. Wheny= 0, 3 + 8x− 2x2= 0
Solve for x using the quadratic formula. ⇒ x=
=
=
=
= 2 − or 2 +
Give approximate answers for the surds.
x≈ −0.345 or 4.345
State the x-intercepts. The x-intercepts are −0.345 and 4.345.
Write the original rule using decreasing powers of x.
y= −2x2+ 8x+ 3
Complete the square. = −2(x2− 4x− )
=−2[(x2− 4x+ 4) − − 4]
= −2[(x− 2)2− ]
y= −2(x− 2)2+ 11
State the turning point. The turning point is (2, 11).
Draw a set of axes and mark the coordinates of the turning point and the points where the graph crosses the axes. Sketch a parabola through these points. 1
2 3
4 −8 8
2–4( )–2 ( )3
±
2( )–2
---−8± 88
4 –
---−8±2 22
4 –
---−4± 22
2 –
---22 2
--- 22 2
---5
6 7
8 32
---3 2 ---11
2
---9 10
11
y
x
0 4
–1 5
(2, 11)
(–0.345, 0) f(x) = 3 + 8x – 2x2
(4.345, 0) 3
6 9 12
(0, 3)
20
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
25
The weight of a person t months after a gymnasium program is started is given by
the function: W(t) = − 3t+ 80, where
t∈ [0, 8] and W is in kilograms. Find:
a the minimum weight of the person
b the maximum weight of the person.
THINK WRITE
Complete the square to find the turning point.
W=
= [t2 − 6t+ 160]
= [t2− 6t+ 9 + 160 − 9]
= [(t− 3)2+ 151]
= (t− 3)2+ 75.5
State the minimum turning point. The turning point is (3, 75.5).
Find the end point value for W when t = 0.
When t= 0
W=
= 80
State its coordinates. One end point is (0, 80).
Find the end point value of W when t = 8.
When t= 8
W= − 3(8) + 80
= 88
State its coordinates. The other end point is (8, 88).
Draw a set of axes and mark the points on it.
Sketch a parabola between the end points.
Locate the maximum and minimum values of W on the graph.
a State the minimum weight from the graph.
a The minimum weight is 75.5 kg.
b State the maximum weight from the graph.
b The maximum weight is 88 kg. t2
2
----1 t
2
2
----–3t+80
1 2 ---1 2 ---1 2 ---1 2
---2 3
02
2
---–3 0( )+80
4 5
82
2
---6
7 W (kg)
t (months)
0
3 8
90 80 70
Maximum (8, 88)
Minimum (3, 75.5) (0, 80)
8
9
21
26
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Quadratic graphs
1 Use the discriminant to determine the number of x-intercepts for each of the following quadratic functions.
2 Sketch the graphs of each of the following functions, showing all intercepts.
3 Find the turning point for each of the functions in question 2.
4 Each of the functions graphed below are of the form y = x2+ bx + c. For each func-tion, give: i the rule ii the domain and iii the range.
Write the answers to b and c in interval notation.
a b c
d e f
a f(x) = x2− 3x + 4 b f(x) = x2+ 5x − 8 c f(x) = 3x2− 5x + 9 d f(x) = 2x2+ 7x − 11 e f(x) = 1 − 6x − x2 f f(x) = 3 + 6x + 3x2
a f(x) = x2− 6x + 8 b f(x) = x2+ 6x + 8 c f(x) = x2− 5x + 4 d f(x) = 6 − x − x2 e f(x) = 10 + 3x − x2 f f(x) = 2x2+ 5x − 3 g f(x) = 6x2− x − 12 h f(x) = 15 + x − 6x2
remember
Quadratic graphs
1. General equation is y=ax2+bx+c.
2. The quadratic formula is given by the equation . 3. The discriminant is b2− 4ac and if:
(a) b2− 4ac> 0, there are two x-intercepts
(b) b2− 4ac= 0, there is one x-intercept which is a turning point (c) b2− 4ac< 0, there are no x-intercepts.
4. The turning point form of the quadratic graph or parabola is:
y=A(x+B)2+C and the turning point is (−B, C).
5. The axis of symmetry of a parabola is given by the expression . 6. The axis of symmetry is halfway between the x-intercepts.
x −b b
2–4ac
±
2a
---=
b
2a
---–
remember
1E
E XCEL S
preadshee t
Discriminant
Mathc
ad
Discriminant
Mathc
ad
Quadratic graphs
WORKED
Example 16
WORKED
Example 17
E XCEL S
preadshee t
Quadratic graphs
WORKED
Example 18
Skill
SHEET
1.7
WORKED
Example 19
y
x 0
(1, –2)
y
x 0 (–3, 2)
x 0
5
y
x 0
6
y
x 0
(–1, 6)
(2, –3)
y
x 0
(1, 9)
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
27
5
Consider the function with the rule y = x2− 2x − 3. a It has x-intercepts:
b It has a turning point with coordinates:
6
The function f(x) = −(x + 3)2+ 4 has a range given by:
7
The range of the function y = (x − 4)2, x ∈ [0, 6] is:
8 Sketch the graph of each of the following functions, showing the turning point and all intercepts.
9 Sketch the graph of each of the functions below and state i the domain and ii the range of each function.
10 The volume of water in a tank, V m3, over a 10 month period is given by the function V(t) = 2t2− 16t + 40, where t is in months and t ∈ [0, 10]. Find:
a the minimum volume of water in the tank b the maximum volume of water in the tank. 11 A ball thrown upwards from a tower attains
a height above the ground given by the function h(t) = 12t − 3t2+ 36, where t is the time in seconds and h is in metres. Find:
a the maximum height above the ground that the ball reaches
b the time taken for the ball to reach the ground
c the domain and range of the function.
A (1, 0) and (3, 0) B (−1, 0) and (3, 0) C (1, 0) and (−3, 0) D (2, 0) and (−1, 0) E (0, −1) and (0, 3)
A (−1, 0) B (2, −3) C (1, −4) D (−1, −4) E (1, 0)
A (3, ∞) B (−∞, −3] C [4, ∞) D (−∞, 4] E R−
A [0, 16] B [4, 16] C [0, 4] D (4, 12] E [0, 16)
a f(x) = (x − 2)2− 4 b f(x) = (x + 3)2− 16 c f(x) = 2(x − 3)2− 2 d f(x) = (x + 1)2+ 3 e f(x) = −(x + 4)2+ 9 f y = x2+ 4x + 3 g y = x2− 10x + 16 h y = 2x2− 4x − 6 i y = −3x2− 12x − 9 j y = 6 + x − x2
a y = x2+ 6x − 5 b y = 7 + 8x − x2
c y = x2− 2x + 2, x ∈ [−2, 2] d y = −x2+ x − 1, x ∈ R+ e f(x) = x2 − 3x − 2, x ∈ [−10, 6] f f(x) = 2x2+ 8x + 7
g f(x) = 5 + 6x − 3x2, x ∈ [−5, 3) h f(x) = 5x2− 5x + 3, x ∈ (−∞, 0] m
multiple choiceultiple choice
m
multiple choiceultiple choice
m
multiple choiceultiple choice
Math
cad Quadratic graphs
– turning point form
WORKED
Example 20
EXCEL Spreadsheet
Quadratic graphs – turning point form
WORKED
Example 21
Work SHEET
1.2
Maximum heightGround Ball
Tower
28
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4Cubic graphs
Cubic functions are polynomials of degree 3. In this section, we will look at how graphs of cubic functions may be sketched by finding intercepts and recognising basic shapes.
Forms of cubic functions
Cubic functions may take several forms. The three main forms are described below.
General form
The general form of a cubic function is
y = ax3+ bx2+ cx + d
If a is positive (i.e. a > 0), the function is called a ‘positive cubic’. Several positive cubics appear below.
If a is negative (i.e. a < 0), the function is called a ‘negative cubic’. Several negative cubics appear below.
You may wish to investigate in more detail the type of equation required to produce each of the above graphs.
Basic form
Some (but certainly not all) cubic functions are transformations of the basic function y = x3, and may be expressed in the form
y = a(x − b)3+ c
For example, y = 2(x − 3)3+ 5 is the graph of y = x3 translated +3 in the x direction, +5 in the y direction, and dilated by a factor of 2 in the y direction.
y
x y
x
y
x
y
x y
x
MM3&4 fig 1.204 y
x
y y = x3
x
y
y = a(x – b)3 + c
(b, c)
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
29
Factor form
Cubic functions of the type
y = a(x − b)(x − c)(x − d)
are said to be in factor form, where b, c and d are the x-intercepts. Often a cubic func-tion in general form may be factorised to express it in factor form.
Repeated factors
A twice only repeated factor in a factorised cubic function indicates a turning point that just touches the x-axis.
Verify this for several cases using a graphics calculator.
y = a(x – b)(x – c)(x – d) where a > 0
y
x c d b
y
1 3
–2 x
y = –(x + 2)(x – 1)(x – 3)
y
x
a b
y = (x – a)2 (x – b)
For each of the following graphs, find the rule and express it in factorised form. Assume that a= ±1.
a b
Continued over page
THINK WRITE
a Find a by deciding whether the graph is a positive or negative cubic.
a Positive cubic, so a= 1.
Use the x-intercepts −4, 0 and 3 to find the factors.
The factors are (x+ 4), x and (x− 3).
Express f(x) as a product of a and its factors.
f(x) = 1(x+ 4)x(x− 3)
Simplify. f(x) =x(x+ 4)(x− 3)
y
x 0
f(x)
–4 3
y
x 0
f(x)
–2 3
1
2
3
4
22
30
M a t h e m a t i c a l M e t h o d s U n i t s 3 a n d 4THINK WRITE
b Find a by deciding whether the graph is a positive or negative cubic.
b Negative cubic, so a= −1.
Use the x-intercept −2, which is also a turning point, to find the repeated factor.
(x+ 2)2 is a factor.
Use the other x-intercept, 3, to find the other factor.
(x− 3) is also a factor.
Express f(x) as a product of a and its factors.
f(x) = −1(x+ 2)2(x− 3)
Simplify. f(x) = (3 −x)(x+ 2)2
1
2
3
4
5
Sketch the graph of y=x3−x2− 10x− 8, showing all intercepts.
THINK WRITE
Find y when x = 0. When x= 0,
y= −8
State the y-intercept. The y-intercept is −8.
Let P(x) = y. Let P(x) =x3−x2− 10x− 8
Use the factor theorem to find a factor of the cubic
P(x) = x3− x2− 10x − 8.
P(1) = 13− 12− 10(1) − 8
= −18
≠ 0
P(−1) = (−1)3− (−1)2 − 10(−1) − 8
= 0 so (x+ 1) is a factor.
Use long division, or otherwise, to find the quadratic factor.
By long division:
y= (x+ 1)(x2− 2x− 8)
Factorise the quadratic, if possible. = (x+ 1)(x− 4)(x+ 2)
Express the cubic in factorised form and let it equal 0 to find the
x-intercepts.
If (x+ 1)(x− 4)(x+ 2) = 0
Solve for x using the Null Factor Law. x= −1, 4 or −2
State the x-intercepts. The x-intercepts are −2, −1, and 4.
1
2 3 4
5
x+ 1
)
x3−x2− 10x− 8x2− 2x− 8
x3+x2
−2x2− 10x− 8
−2x2− 2x
−8x− 8
−8x− 8 0
6 7
8 9
23
C h a p t e r 1 G r a p h s a n d p o l y n o m i a l s
31
Restricting the domain of cubic functions
1. If the domain of a cubic function is unrestricted the range is always R.
2. If the domain of a cubic function is restricted, the range can be found by looking at the end points of the domain if the turning points are not included in the domain. 3. If the domain of a restricted cubic function with two turning points includes at least
one of the turning points, the
