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MQ12 Maths Methods / Final Pages / 18/11/05
1
Graphs and
polynomials
VCE coverage
Areas of study
Units 3 & 4 • Functions and graphs
• Algebra
In this chapter
1A
1B
1C
1D
1E
1F
1G
The binomial theorem
Polynomials
Division of polynomials
Linear graphs
Quadratic graphs
Cubic graphs
Quartic graphs
MQ12 Maths Methods / Final Pages / 18/11/05
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2
Maths Quest 12 Mathematical Methods
The binomial theorem
In Maths Quest 11 Mathematical Methods we learned 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
5
(x + a) = (x4 + 4x3a + 6x2a2 + 4xa3 + a3)(x + a)
= x5 + 5x4a + 10x3a2 + 10x2a3 + 5xa4 + a5
The coefficients associated with each term can be arranged in a triangular shape as
shown:
(x + a)0
1
(x + a)1
1
(x + a)2
1
(x + a)
3
1
(x + a)4
1
(x + a)5
1
1
2
3
4
5
1
3
6
10
1
4
10
1
5
1
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.
Row
 0
 0
0
 1
 0
1
 2
 0
2
 3
 0
3
4
 4
 0
 1
 1
 2
 1
 3
 1
 4
 1
 2
 2
 3
 2
 4
 2
 3
 3
 4
 3
 4
 4
MQ12 Maths Methods / Final Pages / 18/11/05
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Chapter 1 Graphs and polynomials
n
Note:   =
 r
nC
r
3
n!
= ----------------------( n – r )!r!
Remember that nCr is another way of writing  n .
 r
For example, the expansion of (x + a)6 can be written using combinations and then
evaluated:
(x + a)6 =  6 x6a0+  6 x5a1 +  6 x4a2 +  6 x3a3 +  6 x2a4 +  6 x1a5 +  6 x0a6
 0
 1
 2
 3
 4
 5
 6
= x6 + 6x5a + 15x4a2 + 20x3a3 + 15x2a4 + 6xa5 + a6
Now the binomial theorem can be formally stated.
(ax + b)n =  n (ax)n b0 +  n (ax)n − 1b1 + . . . +  n  (ax)1bn − 1 +  n (ax)0 bn
 0
 1
 n – 1
 n
Notes:
1. The indices always sum to n, that is, the powers of (ax) and b 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.
4. The (r + 1)th term is  n (ax)n−r br.
 r
The binomial theorem can also be stated using summation notation:
n
( ax +
b )n
=
∑  r ( ax ) n – r b r
n
r=0
WORKED Example 1
Use the binomial theorem to expand (2x − 3)4.
THINK
1
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.
WRITE
(2x − 3)4 =  4 (2x)4(−3)0 +  4 (2x)3(−3)1
 0
 1
(2x − 3)4 = +  4 (2x)2(−3)2 +  4 (2x)1(−3)3
 2
 3
+  4 (2x)0 (−3)4
 4
2
3
Evaluate the combinations and the
powers.
= 1(16x4) + 4(8x3)(−3) + 6(4x2)(9)
Simplify each term.
= 16x4 − 96x3 + 216x2 − 216x + 81
+ 4(2x)(−27) + 1(81)
Continued over page
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4
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
WORKED Example 2
5
2
Expand the binomial expression  ----2- + x .
x

THINK
1
2
Complete the binomial expansion
2
where ax = ----2- , b = x and n = 5, using
x
row 5 of Pascal’s triangle to assist.
WRITE
5
2
2 5
2 4
2 3
 ---- + x =  ----- + 5  ----- x + 10  ----- x 2
 x2

 x 2
 x 2
 x 2
2 2 3
2
= + 10  ----2- x + 5  ----2- x 4 + x 5
 
x 
x
32
16
8
- + 5  ------ x + 10  ----- x 2
= ----- x8
 x 6
x 10
Evaluate the powers.
4
2
= + 10  ----4- x 3 + 5  ----2- x 4 + x 5
x 
x 
3
32 80 80 40
- + ------ + ------ + ------ + 10x 2 + x 5
= -----x 10 x 7 x 4 x
Simplify each term.
WORKED Example 3
State the coefficient of i x2 and ii x4 in (3 − 2x)8.
THINK
WRITE
ii
ii x0, x1, x2
1
2
3
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.
Find the appropriate term by using the
binomial theorem.
Evaluate the term.
The third term gives a power of x2.
Third term =  8 36 (−2x)2
 2
= 28 × 729 × 4x2
= 81 648x2
ii
4
State the coefficient.
1
Find which term gives a power of x4.
The coefficient of x2 is 81 648.
ii x0, x1, x2, x3, x4
The fifth term gives a power of x4.
2
Evaluate the term.
Fifth term =  8 34 (−2x)4
 4
= 70 × 81 × 16x4
= 90 720x4
3
State the coefficient.
The coefficient of the fifth term is 90 720.
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
WORKED Example 4
Find the fourth term in the expansion of (x − 2y)5.
THINK
1
Find the fourth term by using the
binomial theorem.
2
Evaluate the term.
WRITE
Fourth term =  5 x2(2y)3
 3
= 10 × x2 × 8y3
= 80x2y3
WORKED Example 5
1 5
Find and evaluate the term that is independent of x in the expansion of  x 3 + ----2- .

x
THINK
1
Find how the powers of x are generated
in the expansion from left to right.
WRITE
1
Powers of x are (x3)5 = x15, ( x 3 ) 4  ----2- = x 10 ,
x 
1 2
1 3
( x 3 ) 3  ----2- = x 5 , ( x 3 ) 2  ----2- = x 0 . . .
x 
x 
that is, x15, x10, x5, x0.
2
Find the required term.
3
Evaluate.
The fourth term is independent of x.
5
1 3
Fourth term =   ( x 3 ) 2  ----2-
 3
x 
1
Fourth term = 10x6  ----6-
x 
4
State the solution.
Fourth term = 10
The term that is independent of x is the fourth
term, 10.
WORKED Example 6
Find the coefficient of y4 in the expansion of (y + 3)3 (2 − y)5.
THINK
1
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
Continued over page
5
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6
Maths Quest 12 Mathematical Methods
THINK
2
3
WRITE
y4 terms = y3[5(2)4(−y)] + 3y2(3)[10(2)3(−y)2]
Write down the sum of these 4 products,
using Pascal’s triangle to assist.
+ 3y(3)2[10(2)2(−y)3] + 33[5(2)(−y)4]
= −80y4 + 720y4 − 1080y4 + 270y4
Evaluate.
= −170y4
4
The coefficient of y4 is −170.
State the solution.
remember
remember
1. Pascal’s triangle:
1
1
1
1
1
1
2
3
4
5
1
1
3
6
10
1
4
1
10
5
1
2. Binomial theorem:
(ax + b)n =  n (ax)nb0 +  n (ax)n − 1b + . . . +  n  (ax)bn − 1 +  n (ax)0bn
 0
 1
 n – 1
 n
Notes:
1. The powers of (ax) and b sum to n.
2. There are n + 1 terms in the expansion.
3. The (r + 1)th term is  n (ax)n − r br.
 r
1A
1.1
SkillS
HEET
WORKED
Example
1
Binomial
expansions
d
WORKED
Mat
hca
Example
Expanding
2
The binomial theorem
1 Use the binomial theorem to expand each of the following.
a (x + 3)2
b (x + 2)3
c
8
d (x − 1)
e (x − 5)3
f
g (3x − 4)5
h (1 + x)7
i
5
3
j (2 − 3x)
k (x + 2y)
l
(x + 4)5
(2x + 3)4
(7 − x)4
(3y − 2x)6
2 Expand each of the following binomial expansions.
1 3
a  x + ---

x
1 5
b  x – ---

x
c
4
 2x + 1
---

x
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Chapter 1 Graphs and polynomials
Example
3
e
6
 x 2 + 3---


x
8
1
g  ----2- + x
x

5
3
h  ----- – 2x
 x2

f
 5--- – x 2
x

4
3 State the coefficient of i x2 ii x3 and iii x4 in each of the following.
a (x − 7)3
b (2x + 1)5
c (3x − 4)8
 2--- + 3x
x

5
e
2 3
g  3x 2 + ---

x
9
5
h  --- – x 2
x

f
i
6
 x2 – 3
---


x
36
 7x + ---
x 2
B 2744
C −784
D −9604
E 11 760
D 45
E 135
5 multiple choice
2 5 3
The coefficient of x3 in  3x – --- is:

x
A −135
B −45
C −75
6 multiple choice
Which of the following does not have an x5 term when expanded?
A (x + 6)8
1 7
B  3x 2 – ---

x
D (8 − 3x)5
1 8
E  2x – -----

x 2
5 8
C  6x + ---

x
7 multiple choice
2 5
e
f
- , then a + b + c + d + e + f equals:
If  x 3 + ----2- = ax 15 + bx 10 + cx 5 + d + ----5- + -----

x
x
x 10
A 15
B 31
C 63
D 243
E 127
8 multiple choice
Which one of the following expressions is not equal to (2x − 3)4?
WORKED
Example
4
Binomial
theorem
GC
The coefficient of x4 in (2x − 7)6 is:
A 27 440
program
GC
A (3 − 2x)4
( 2x – 3 ) 6
C ---------------------2( 3 – 2x )
E 16x4 − 96x3 + 216x2 − 216x + 81
B (2x − 3)(2x − 3)3
D 16x4 − 24x3 + 36x2 − 54x + 81
9 Find the fourth term in the expansion (x + 3y)6.
am
progr –C
Binomial
theorem
asio
4 multiple choice
Binomial
theorem
–TI
d (5 − 4x)7
Math
cad
WORKED
2 7
d  3x – ---

x
7
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8
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
x 9
10 Find the third term in the expansion of  3 – --- , assuming ascending powers of x.

4
WORKED
Example
5
2 6
11 Find and evaluate the term that is independent of x in the expansion of  3x + ----2- .

x 
4 5
12 Find and evaluate the term independent of x in the expansion of  x 2 – ----- .

x 3
5 8
13 Find and evaluate the term that is independent of x in the expansion of  x 3 – --- .

x
3 4
14 Find and evaluate the term that is independent of x in the expansion of  x 2 + ----2- .

x 
WORKED
Example
15 Find the coefficient of p4 in the expansion of ( p + 3)5 (2p − 5).
6
16 Find the coefficient of m5 in the expansion of (1 − m)6 (2m + 3)4.
17 In the expansion of (2a − 1)n, the coefficient of the second term is −192. Find the
value of n.
Polynomials
A polynomial in x is an expression that consists of terms which have non-negative
integer powers of x only.
P(x) is a polynomial in x if:
P(x) = a n x n + a n − 1 x n − 1 + . . . + a 2 x2 + a 1 x + a 0
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.
WORKED Example 7
Which of the following expressions are not polynomials?
9
---
a x6 - 4x4 + 2x3 + 7x
c 7 − 3xy + 4x2 − x3 +
e
b x2 + x3 – 3 x2 + 6 x – 5
2
 3 x 2 – ---
x 2
THINK
1
d 8 + 2x − 3x2 + 9x3 − x4
x
a and d are polynomials because they
are expressions with non-negative
integer powers of x only.
WRITE
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Chapter 1 Graphs and polynomials
THINK
2
which is not an integer.
b, c and e are not polynomials.
c is not a polynomial as it has a power of
1
--2
4
WRITE
b is not a polynomial as it has a power of
9
--- ,
2
3
9
(
) , which is not an integer, and it
also has one term, −3xy, which is not a
power of x only.
2
e is not a polynomial because ----2- = 2x −2 and
x
so has a power that is not a positive integer.
Polynomials can be added and subtracted by collecting like terms.
WORKED Example 8
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
b
1
Add the polynomials.
2
Collect like terms.
a P(x) + Q(x) = 6 − 2x + 3x2 + x4 + x5 − 2x4
P(x) + Q(x) = + x2 − 5x − 2
P(x) + Q(x) = x5 − x4 + 4x2 − 7x + 4
1
Subtract the polynomials.
Remove brackets.
Collect like terms.
b P(x) − R(x) = 6 − 2x + 3x2 + x4 − (x2 − 4)
= 6 − 2x + 3x2 + x4 − x2 + 4
= x4 + 2x2 − 2x + 10
2
3
Evaluating polynomials
A value for a polynomial, P(x), can be found for a particular value of x by simply substituting the given value of x into the polynomial expression and evaluating. That is,
polynomial functions are evaluated in the same way as any function.
WORKED Example 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 The degree of P(x) is 4.
Continued over page
MQ12 Maths Methods / Final Pages / 18/11/05
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10
Maths Quest 12 Mathematical Methods
THINK
WRITE
b
Substitute the given value of x into
the polynomial expression.
Evaluate.
b P(1) = 2(1)4 − (1)3 + 5(1)2 − 6(1) + 4
Substitute the given value of x into
the polynomial expression.
Evaluate.
c P(−2) = 2(−2)4 − (−2)3 + 5(−2)2 − 6(−2) + 4
1
2
c
1
2
=2−1+5−6+4
=4
= 32 + 8 + 20 + 12 + 4
= 76
WORKED Example 10
If P(x) = ax5 + x4 − 3x3 + bx − 5, P(−1) = −5 and P(2) = −65, find the values of a and b.
THINK
1
2
3
4
5
Substitute a given value of x into the
polynomial and equate it to the given
answer.
Simplify the equation.
Make b the subject of the equation and
call this equation [1].
Substitute a given value of x into the
polynomial and equate it to the given
answer.
Simplify the equation.
6
Substitute [1] into [2].
7
Solve this equation for a.
8
Substitute the value of a into equation
[1].
Find the value of b.
State the solution.
9
10
WRITE
P(−1) = a(−1)5 + (−1)4 − 3(−1)3 + b(−1) − 5
= −5
−a + 1 + 3 − b − 5 = −5
−a + 4 − b = 0
b=4−a
[1]
P(2) = a(2)5 + (2)4 − 3(2)3 + b(2) − 5
= −65
32a + 16 − 24 + 2b − 5 = −65
32a + 2b − 13 = −65
32a + 2b = −52
Substituting b = 4 − a:
32a + 2(4 − a) = −52
32a + 8 − 2a = −52
30a = −60
a = −2
Substituting a = −2 into equation [1]:
b = 4 − −2
=6
Therefore, a = −2 and b = 6.
Note: The simultaneous equations b = 4 − a and 32a + 2b = −52 could be solved using a
graphics calculator. Rewrite the second equation as b = −16a −26.
[2]
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Chapter 1 Graphs and polynomials
11
remember
remember
1. If P(x) = a n x n + a n − 1 x n − 1 + . . . + a 2 x2 + a 1 x + a 0 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.
1B
WORKED
Example
7
WORKED
Example
8
WORKED
9
1 Which of the following are not polynomial expressions?
viii x3 − 2x
viii x4 + 3x2 − 2x + x
7
6
viii x + 3x − 2xy + 5x
iiiv 3x8 − 2x5 + x2 − 7
2
6
3
iiiv 4x − x + 2x − 3
iivi 2x57 + x4 − x3 + x2 + 3x − ---x
ivii 5x10 − x7 + x4
viii 3x 2 + 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:
a P(x) + Q(x)
b Q(x) − R(x)
c 3P(x) − 2R(x) d 2P(x) − Q(x) + 3R(x)
3 For each of the following polynomials, find: i its degree ii P(0) iii P(2) and iv P(−1).
a P(x) = x6 + 2x5 − x3 + x2
b P(x) = 3x7 − 2x6 + x5 − 8
5
3
2
Evaluating
c P(x) = x − 4x + 3x + 2x + 1
d P(x) = 4x4 − 2x3 + x2 − 7x − 10
polynomials
6
4
3
2
e P(x) = 5x + 3x − 2x − 6x + 3
f P(x) = −7 + 2x − 5x2 + 2x3 − 3x4
4 multiple choice
Example
E
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) = ax + bx − 3x − 4x + 7, f (1) = −2 and f (2) = −5.
4
3
4
1.2
2
7 For Q(x) = x + 2x + ax − 6x + b, Q(2) = 45 and Q(0) = −7. Find a and b.
5
Evaluating
polynomials
3
8 Find a and b if P(x) = ax6 + bx4 + x3 − 6, 3P(1) = −24 and 3P(−2) = 102.
9 multiple choice
a If P(x) = ax4 − x3 + 3x2 − 5 and P(1) = −1, then a is equal to:
A 1
B 0
C 2
D −3
E −2
b If f (x) = xn − 2x3 + x2 − 5x and f (2) = 10, then n is equal to:
A 4
B 6
C 7
D 5
E −1
SkillS
HEET
10
E 103
sheet
L Spre
XCE ad
If P(x) = x8 − 3x6 + 2x4 − x2 + 3, then P(−2) is equal to:
A 479
B 95
C 31
D 481
WORKED
Math
cad
Example
Polynomials
Simultaneous
equations
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Maths Quest 12 Mathematical Methods
O
several values
Graphics Calculator tip! Finding
of a function
CASI
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 ENTER .
Finding several
values of a function
Division of polynomials
When sketching cubic or higher order graphs, it is necessary to factorise the polynomials in order to find the x-intercepts. As will be shown later in this section, division of
polynomials can be used to factorise an expression.
When one polynomial, P(x), is divided by another, D(x), the result can be expressed
as:
R( x)
P(x) = Q(x) + -----------D( x)
where Q(x) is called the quotient,
R(x) is called the remainder, and
D(x) is called the divisor.
WORKED Example 11
Find the quotient, Q(x), and the remainder, R(x), when x4 − 3x3 + 2x2 − 8 is divided by the
linear expression x + 2.
THINK
1
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.
4
2
Divide x into x and write the result above.
3
Multiply the result x3 by x + 2 and write the
result underneath.
4
Subtract and bring down the remaining terms
to complete the expression.
x + 2 ) x4 − 3x3 + 2x2 + 0x − 8
x3
x + 2 ) x4 − 3x3 + 2x2 + 0x − 8
x3
x + 2 ) x4 − 3x3 + 2x2 + 0x − 8
x4 + 2x3
x3 − 5x2
x + 2 ) x4 − 3x3 + 2x2 + 0x − 8
−(x4 + 2x3)
− 5x3 + 2x2 + 0x − 8
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
THINK
5
6
7
8
13
WRITE
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 result of the final subtraction, 40, is the
remainder.
x3 − 5x2 + 12x − 24
x + 2 ) x − 3x3 + 2x2 + 0x − 8
−(x4 + 2x3)
− 5x3 + 2x2 + 0x − 8
−(− 5x3 − 10x2)
12x2 + 0x − 8
−(12x2 + 24x)
−24x − 8
−(−24x − 48)
40
4
The quotient is x3 − 5x2 + 12x − 24.
The remainder is 40.
Note: 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
b
when P(x) is divided by (ax + b), the remainder is P  – --- .
 a
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
b
if (ax + b) is a factor of P(x), then P  – --- = 0.
 a
Note: If (x − a) is a factor of P(x) and a is an integer, then a must be a factor of the
term independent of x. For example, if (x − 2) is a factor of P(x), then the term independent of x must be divisible by 2. Therefore, (x − 2) could be a factor of x3 − 2x2 −
x + 2, but (x + 3) could not be a factor.
WORKED Example 12
Determine whether or not D(x) = (x − 3) is a factor of P(x) = 2x3 − 4x2 − 3x − 8.
THINK
1
2
Evaluate P(3).
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).
WRITE
P(3) = 2(3)3 − 4(3)2 − 3(3) − 8
= 54 − 36 − 9 − 8
=1
P(3) ≠ 0 so (x − 3) is not a factor of P(x).
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Maths Quest 12 Mathematical Methods
WORKED Example 13
a Factorise P(x) = 2x3 − x2 − 13x − 6.
b Solve 2x3 − x2 − 13x − 6 = 0.
THINK
WRITE
a
a
1
2
b
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, −1, 2, −2, 3,
−3, 6, −6 until a factor is found.
Divide P(x) by the divisor (x + 2) using
long division.
3
Express P(x) as a product of linear and
quadratic factors.
4
Factorise the quadratic, if possible.
1
Rewrite the equation in factorised form,
using the answer to part a.
2
Use the Null Factor Law to state the
solutions.
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.
2x2 − 5x − 3
x + 2 ) 2x − x2 − 13x − 6
−(2x3 + 4x2)
−5x2 − 13x − 6
−(−5x2 − 10x)
−3x − 6
−(−3x − 6)
0
3
P(x) = (x + 2)(2x2 − 5x − 3)
= (x + 2)(2x + 1)(x − 3)
b 2x3 − x2 − 13x − 6 = 0
(x + 2)(2x + 1)(x − 3) = 0
x = −2, − 1--- or 3
2
Note: 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 found in
part b.
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
15
remember
remember
R( x)
1. P(x) = Q(x) + -----------D( 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 P  – b--- = 0.
 a
4. If (x − a) is a factor of P(x) then a must be a factor of the term independent
of x.
1C
Division of polynomials
–TI
program
GC
11
WORKED
Example
12
Math
Division of
polynomials
Math
Evaluating
polynomials
L Spre
XCE ad
Finding
factors of
polynomials
E x−4
sheet
4 multiple choice
Examine the equation f (x) = x4 − 4x3 − x2 + 16x − 12.
a Which one of the following is a factor of f (x)?
A x+1
B x
C x+2
D x+3
Division of
polynomials
cad
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) = 4 − 9x + 6x2 − 13x3 − 12x4 + 3x5, D(x) = 4 − x
g P(x) = 4x6 + 2x5 − 8x4 − 4x3 + 6x2 − 9x − 6, D(x) = 2x + 1
am
progr –C
cad
2 a For each corresponding polynomial in question 1, evaluate:
i P(4)
ii P(1)
iii P(−3)
iv P(−2)
v P(3)
vi P(2)
vii P(− 1--- )
viii P( 3--- )
3
2
b Compare these values to R(x) in question 1 and comment on the result.
Division of
polynomials
asio
1 Find the quotient, Q(x), and the remainder, R(x), when each of the following
polynomials are divided by the given linear expression.
a x3 − 2x2 + 5x − 2, x − 4
b x4 + x3 + 3x2 − 7x, x − 1
5
3
c x − 3x + 4x + 3, x + 3
d 2x6 − x4 + x3 + 6x2 − 5x, x + 2
4
3
2
e 6x − x + 2x − 4x, x − 3
f x4 − 13x2 + 36, x − 2
4
3
g 3x − 6x + 12x, 3x + 1
h x5 + 3x3 − 4x2 + 6x − 8, 3 − 2x
GC
Example
E
WORKED
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MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
b When factorised, f (x) is equal to:
A (x + 1)(x − 3)(x + 4)
C (x + 2)(x − 4)(x + 3)(x + 1)
E x(x − 1)(x + 2)(x + 3)
WORKED
Example
13a
WORKED
Example
13b
Work
T
SHEE
1.1
B (x + 2)(x − 2)(x − 3)(x − 1)
D (x − 1)(x + 1)(x − 3)(x − 4)
5 Factorise the following polynomials.
a P(x) = x3 + 4x2 − 3x − 18
c P(x) = −x3 + 12x − 16
e P(x) = x4 + 2x3 − 13x2 − 14x + 24
g P(x) = x4 + 2x3 − 7x2 − 8x + 12
b
d
f
h
6 Solve each of the following equations.
a x3 − 3x2 − 6x + 8 = 0
c 3x3 + 3x2 − 18x = 0
e 2x4 + x3 − 14x2 − 4x + 24 = 0
b x3 + x2 − 9x − 9 = 0
d 2x4 + 10x3 − 4x2 − 48x = 0
f x4 − 2x2 + 1 = 0
P(x) = 3x3 − 13x2 − 32x + 12
P(x) = 8x3 + 10x2 − 38x + 20
P(x) = −72 − 42x + 19x2 + 7x3 − 2x4
P(x) = 4x4 + 12x3 − 24x2 − 32x
7 If (x − 2) is a factor of x3 + ax2 − 6x − 4, then find a.
8 If (x − 1) is a factor of x3 + x2 − ax + 3, then find a.
9 Find the value of a if (x + 3) is a factor of 2x4 + ax3 − 3x + 18.
10 Find the value of a and b if (x + 1) and (x − 2) are factors of ax3 − 4x2 + bx − 12.
11 If (2x − 3) and (x + 2) are factors of 2x3 + ax2 + bx + 30, find the values of a and b.
Linear graphs
Linear graphs are polynomials of degree 1. Graphs of linear functions are straight lines
and may be sketched by finding the intercepts.
Revision of properties of straight line graphs
1. The gradient of a straight line joining two points is:
y2 – y1
m = --------------x2 – x1
y
B (x2, y2)
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.
A (x1, y1)
x
0
y
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)
Gradient = m
A (x1, y1)
x
0
4. The intercept form of the equation of a straight line is:
x y
--- + --- = 1 or bx + ay = ab
a b
y
(0, b)
5. Parallel lines have the same gradient.
(a, 0)
0
x
MQ12 Maths Methods / Final Pages / 18/11/05
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Chapter 1 Graphs and polynomials
17
6. The product of the gradients of two lines that are
perpendicular equals −1.
1
That is,
m1 × m2 = −1 or m1 = − -----m2
WORKED Example 14
Sketch the graph of the linear function 3x − 2y = 6 by indicating the intercepts.
THINK
1
2
3
4
5
6
WRITE/DRAW
Substitute y = 0 into the equation.
Solve the equation for x to find the
x-intercept.
Substitute x = 0 into the equation.
Solve the equation for y to find the
y-intercept.
Draw a set of axes.
Indicate the x-intercept and y-intercept
and rule a line through these points.
When y = 0, 3x − 2 × 0 = 6
x=2
Therefore, the x-intercept is 2.
When x = 0, 3 × 0 − 2y = 6
y = −3
Therefore, the y-intercept is −3.
y
0
2
3x – 2y = 6
x
–3
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
[a, b] = {x : a ≤ x ≤ b}
a
b
(a, b) = {x : a < x < b}
a
b
[a, b) = {x : a ≤ x < b}
If the domain or range is unrestricted, it can be denoted as R or (−∞, ∞).
R+ ∪ {0} = [0, ∞)
R+ = (0, ∞)
−
R = (−∞, 0)
R− ∪ {0} = (∞, 0]
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Maths Quest 12 Mathematical Methods
WORKED Example 15
Find 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
d The line which passes through (1, 3) and is perpendicular to the line with equation
y + 2x − 3 = 0
THINK
WRITE
a
Write the rule for the point–gradient
form of the equation of a straight
line, y − y1 = m(x − x1).
Substitute the value of the gradient,
m, and the coordinates of the point
(x1, y1), into the equation.
Expand the brackets.
Express the equation in the form
required.
a
1
Write the rule for the gradient, m, of
a straight line, given 2 points.
b
2
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.)
Expand the brackets.
Express the equation in the form
required.
1
2
3
4
b
4
5
c
1
2
3
4
State the gradient of the given line,
which is the same as the gradient of
the parallel line.
Write the rule for the point–gradient
form of the equation of a straight line.
Substitute the values of m and the
coordinates (x1, y1) = (3, 4).
Simplify and write in the required
form.
y − y1 = m(x − x1)
y − (−2) = 2(x − 3)
y + 2 = 2x − 6
y − 2x + 8 = 0
or 2x − y − 8 = 0
y2 – y1
m = --------------x2 – x1
⇒
2–8
= --------------−2 – 0
–6
= -----–2
=3
y − 8 = 3(x − 0)
y − 8 = 3x
3x − y + 8 = 0
c y − 2x − 5 = 0 becomes y = 2x + 5.
The gradient of the parallel lines is 2.
y − y1 = m(x − x1)
y − 4 = 2(x − 3)
y − 4 = 2x − 6
2x − y − 2 = 0
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
THINK
WRITE
d
d
1
Find the gradient of the given line.
2
Find the gradient of the perpendicular line.
3
Write the rule for the point–gradient
form of the equation of a straight line.
Substitute the values of m and the
coordinates (x1, y1) = (1, 3).
Simplify and write in the required form.
4
5
19
y = −2x + 3
The gradient of the line is −2
The gradient of the perpendicular line
is 1--- .
2
y − y1 = m(x − x1)
y − 3 = 1--- (x − 1)
2
2y − 6 = (x − 1)
x − 2y + 5 = 0
WORKED Example 16
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)
THINK
WRITE/DRAW
a
a When x = −3,
−12 − 2y = 8
−2y = 20
y = −10
(−3, −10) is a closed end of the line.
When x = 3,
12 − 2y = 8
−2y = −4
y=2
(3, 2) is the other closed end of the line.
1
2
3
4
5
6
7
8
Substitute the smallest value of x
into the equation.
Solve the equation for y, to find an
end point of the straight line.
State the coordinates of the end point.
Substitute the largest value of x into
the equation.
Solve the equation for y, to find the
other end point of the line.
State the coordinates of the 2nd end
point.
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.
y
2
–3
0
–4
(–3, –10)
(3, 2)
x
2 3
4x – 2y = 8,
x ∈ [–3, 3]
–10
9
Find the intercepts and mark them
on the graph.
When x = 0, y = −4
When y = 0, x = 2
The x-intercept is 2 and the y-intercept is −4.
10
State the domain, which is given
with the rule.
The domain is [−3, 3].
11
State the range from the graph.
The range is [−10, 2].
Continued over page
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MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
THINK
WRITE/DRAW
b
b When
1
There is no smallest value of x, so
substitute the largest value of x into the
equation and find y.
x = −1,
y = f (−1)
=3
2
State the coordinates of the upper end
point.
(−1, 3) is an open end of the line.
3
Substitute another value of x within the
domain into the equation (that is, a
value of x < −1, since x ∈ (−∞, −1)) and
find y.
When
4
State the coordinates of the point.
(−2, 5) is another point on the line.
5
Plot the 2 points on a set of axes and
mark the point (−1, 3) with an open
circle.
6
x = −2,
y = f (−2)
=5
f(x) = 1 – 2x,
x ∈ (–∞, –1)
(–2, 5)
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.
(–1, 3)
y
5
3
–2 –1 0
7
Note that there are no intercepts.
8
State the domain, which is given with
the rule.
The domain is (−∞, −1).
9
State the range by examining the graph.
The range is (3, ∞).
remember
remember
Linear graphs
1. Linear equations are polynomials of degree 1.
y2 – y1
2. Gradient, m = --------------x2 – x1
3. General equation is
ax + by + c = 0 or y = mx + c
where m = gradient and c = y-intercept.
4. Equation if a point and gradient is known:
y − y1 = m(x − x1)
5. Equation if the intercepts are known:
x y
--- + --- = 1
a b
6. Parallel lines have the same gradient.
7. If m1 and m2 are the gradients of perpendicular lines, then:
m1 × m2 = −1
or
1
m1 = − -----m2
x
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
1D
14
WORKED
Example
WORKED
Example
Linear
graphs
2 Find the equation, in the form ax + by + c = 0, of each straight line described below.
program
GC
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).
Finding a
c The line with a gradient of 4 and passing through (0, 3).
linear
equation
3 Find the equation, in the form ax + by + c = 0, of each straight line described below.
am
a The line passing through (0, 2) and (3, 6).
progr –C
b The line passing through (−3, −4) and (−1, −10).
Finding a
c The line passing through (7, 5) and (2, 0).
linear
asio
15b
Math
–TI
15a
1 Sketch the graph of each of the following linear functions by indicating the intercepts.
a 2x + 3y = 12
b x − 4y = 8
c 2y − 5x − 10 = 0
d 2x − y = 1
e 3y + x − 2 = 0
f y=5
GC
Example
Linear graphs
cad
WORKED
21
equation
WORKED
Example
15c
Example
c
15d
1.4
Using
Find the equation in the form ax + by + c = 0 that passes through (−2, 4) and is gradient to
find the
perpendicular to the line with equation 2y − x + 1 = 0.
value of a
parameter
6 Match each of the following graphs with the appropriate rule below.
a
b
c
y
y
y
(2, 4)
2
2
x
0
–1
d
0
e
iii x + 2y + 4 = 0
iv 3y + 2x = 6
f
y
–4
0
3
x
x
y
–2
0
x
0
–2
iii y = 3
iv y − 2x = 0
x
y
3
0
iii y − 2x − 2 = 0
vi x = −2
x
SkillS
HEET
WORKED
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.
SkillS
HEET
4 multiple choice
Which one of the following points does not lie on the straight line with equation 1.3
2y − 3x − 6 = 0?
Gradient
A (2, 6)
B (−2, 0)
C (0, 3)
D (1, 2)
E (4, 9)
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22
Maths Quest 12 Mathematical Methods
7 State the range for each function graphed below.
1.5
SkillS
HEET
MQ12 Maths Methods / Final Pages / 18/11/05
a
Interval
notation
b
y
y
x
0
x
0
1.6
(–5, –2)
SkillS
HEET
Domain and
range for
linear graphs
c
(6, –5)
y
(4, 3)
d
(–3, 3)
x
0
WORKED
Example
16
(5, –2)
f
y
0
x
0
(–4, –2)
e
y
x
4
y
(5, 6)
x
0
8 Sketch the graph of each of the following functions, stating i the domain and ii the
range of each.
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)
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 multiple choice
If the straight lines 3x − y = −2 and ax + 2y = 3 are parallel then a = :
A 6
B 2
C −2
D −3
E −6
12 multiple choice
If the straight lines 5x + y −3 = 0 and bx −y −2 = 0 are perpendicular, then b is equal
to:
A 5
B 1--C −5
D − 1--E 3
5
5
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Chapter 1 Graphs and polynomials
23
Quadratic graphs
Quadriatic functions are polynomials of degree 2. Graphs of quadratic functions are
parabolas and may be sketched by finding the turning point and intercepts.
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
b
(d) the equation of the axis of symmetry is x = – -----2a
(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 ± b 2 – 4ac
(b) using the quadratic formula, x = -------------------------------------- .
2a
4. The turning point can be found by ‘completing the square’ (see page 24). The turning
point 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 and is a perfect square, the solutions are rational; otherwise they are
irrational.
3. 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.
4. If b2 − 4ac < 0, there are no real solutions and there are no x-intercepts on the graph.
WORKED Example 17
Use the discriminant to determine the number of x-intercepts for the quadratic function
f (x) = 2x2 + 3x − 10.
THINK
1
2
3
Find the values of the quadratic
coefficients a, b and c using the general
quadratic function, y = ax2 + bx + c.
Evaluate the discriminant.
If the discriminant is greater than 0,
there are two x-intercepts.
WRITE
a = 2, b = 3, c = −10
b2 − 4ac = 32 − 4(2)(−10)
= 9 + 80
= 89
b2 − 4ac > 0
So there are two x-intercepts, which are both
irrational.
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24
Maths Quest 12 Mathematical Methods
WORKED Example 18
Sketch the graph of the function f (x) = 12 − 5x − 2x2, showing all intercepts. Give exact
answers.
THINK
WRITE/DRAW
1
Evaluate f (0) to find the y-intercept (or
state the value of c).
f (0) = 12 − 5(0) − 2(0)2
= 12
2
State the y-intercept.
The y-intercept is 12.
3
Set f (x) = 0 to find the x-intercepts.
f (x) = 12 − 5x − 2x2 = 0
4
Factorise the quadratic (or use the
quadratic formula).
(4 + x)(3 − 2x) = 0
5
Solve the equation using the Null
Factor Law.
⇒ 4 + x = −4 or 3 − 2x =
6
State the x-intercepts.
The x-intercepts are −4 and 3--- .
7
Draw a set of axes and mark the
intercepts or the coordinates of the
points where the graph crosses the axes.
8
4 + x = 0− or 3 − 2x = 0
3
--2
2
y
12 (0, 12)
f(x) = 12 – 5x – x2
Sketch a parabola through the
intercepts.
( 3–2 , 0)
(–4, 0)
–4
0 1 2
x
The x-coordinate of the turning point of a quadratic function is exactly halfway
– 4 + 3--between the two x-intercepts, so for worked example 18, x = ---------------2- = − 5--- (or −1 1--- ).
4
4
2
Substitute x = − 5--- into the original equation to find the y-coordinate of the turning point.
4
The x-coordinate of the turning point can also be found by using the formula
–b
x = ------ , where ax2 + bx + c = 0.
2a
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 − h)2 + k
where the turning point is (h, k).
This way of writing the function is known as turning point form.
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Chapter 1 Graphs and polynomials
25
WORKED Example 19
For the function y = −2 (y + 3) −4, find:
ii the coordinates of the turning point
THINK
Write the general formula.
Write the function.
ii 1 Identify the values of a, h and k.
2 State the coordinates of the turning
point (h, k).
ii 1 Write the domain of the parabola.
2 Write the range y ≤ k (as a < 0).
ii the domain and range.
WRITE
y = a(x − h)2 + k
y = −2(x + 3)2 − 4
ii a = −2, h = −3, k = −4
The turning point is (−3, −4).
ii The domain is R.
The range is y ≤ −4.
WORKED Example 20
The function graphed at right 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.
y
(–5, 5)
0
(–1, –6)
THINK
WRITE
a
a y = a (x − h)2 + k
b
c
1
Write the general rule for a quadratic
in turning point form.
2
Find the values of h and k using the
given turning point.
Since the turning point is (−1, −6):
3
State the value of a (given).
a=1
4
Substitute these values in the rule.
So y = 1(x + 1)2 − 6
5
Expand the brackets.
6
Simplify.
1
Use the graph to find the domain.
Look at all the values that x can take.
2
State the domain in interval notation.
1
Use the graph to find the range.
Look at all the values that y can take.
2
State the range in interval notation.
h = −1, k = −6
= x2 + 2x + 1 − 6
⇒ y = x2 + 2x − 5
The rule is y = x2 + 2x − 5.
b x ≥ −5
Domain = [−5, ∞)
c y ≥ −6
Range = [−6, ∞)
x
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Maths Quest 12 Mathematical Methods
WORKED Example 21
Sketch the graph of y = 1--2- (x − 1)2 + 2, clearly showing the coordinates of the turning point
and the intercepts with the axes.
THINK
1
2
3
4
5
6
Write the general equation of the
parabola.
Identify the values of the variables.
Write a brief statement on the transformation of the basic parabola.
State the shape of the parabola (that is,
positive or negative).
State the coordinates of the turning
point (h, k).
As both a and k are positive, only the yintercept needs to be determined. Find
the y-intercept by making x = 0.
WRITE/DRAW
y = a(x − h)2 + k
a = 1--- , h = 1, k = 2
2
The graph of y = x2 is dilated in the y direction by the factor of 1--- (that is, it is wider
2
than the basic curve); it is translated 1 unit
to the right and 2 units up.
a > 0; the parabola is positive.
The turning point is (1, 2).
y-intercept: x = 0
y = 1--- (0 − 1)2 + 2
2
y = 1--- (−1)2 + 2
2
y=
1
--2
+2
y = 2 1--2
7
Sketch the graph:
Draw a set of axes and label them. Plot
the turning point and the y-intercept.
Sketch the graph of the positive
parabola, so that it passes through the
points previously marked.
y
1
2 –2
2
0
x
1
y = 1–2 (x – 1)2 + 2
WORKED Example 22
Sketch the graph of y = 3 + 8x − 2x2, showing the turning point and all intercepts,
rounding answers to 2 decimal places where appropriate.
THINK
1
2
3
Find y when x = 0.
State the y-intercept.
Let the quadratic equal zero.
WRITE/DRAW
When x = 0, y = 3
The y-intercept is 3.
When y = 0,
3 + 8x − 2x2 = 0
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Chapter 1 Graphs and polynomials
THINK
4
27
WRITE/DRAW
Solve for x using the quadratic formula.
⇒
−8 ± 8 2 – 4 ( – 2 ) ( 3 )
x = --------------------------------------------------2 ( –2 )
−8 ± 88
= ----------------------–4
−8 ± 2 22
= -------------------------–4
−4 ± 22
= ----------------------–2
22
22
= 2 – ---------- or 2 + −−−−−
2
2
5
State the x-intercepts, rounding to 2
decimal places.
The x-intercepts are −0.35 and 4.35.
6
Write the original rule using decreasing
powers of x.
y = −2x2 + 8x + 3
7
Complete the square.
= −2(x2 − 4x − 3--- )
2
= −2[(x − 4x + 4) −
2
= −2[(x − 2)2 −
3
--2
− 4]
11
------ ]
2
y = −2(x − 2)2 + 11
8
State the turning point.
9
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.
10
The turning point is (2, 11).
y
12
f(x) = 3 + 8x – 2x2
(2, 11)
9
6
3 (0, 3)
(–0.35, 0)
(4.35, 0)
x
–1 0
4 5
In 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 determining the range of quadratic functions. Sketch graphs are also useful.
Intercepts and turning points can be found using a graphics calculator. This is useful
for multiple-choice questions, questions that are allocated only one mark, and questions
that do not require algebraic methods.
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Maths Quest 12 Mathematical Methods
WORKED Example 23
The weight of a person t months after a
gymnasium program is started is given by
t2
the function: W(t) = ---- − 3t + 80, where
2
t ∈ [0, 8] and W is in kilograms. Find:
a the minimum weight of the person
b the maximum weight of the person.
THINK
1
Complete the square to find the turning
point.
WRITE
t2
W = ---- – 3t + 80
2
1
= --- [t2 − 6t + 160]
2
=
1
--2
[t2 − 6t + 9 + 160 − 9]
=
1
--2
[(t − 3)2 + 151]
=
1
--2
(t − 3)2 + 75.5
2
State the minimum turning point.
The turning point is (3, 75.5).
3
Find the end point value for W when
t = 0.
When t = 0,
W = 80
4
State its coordinates.
One end point is (0, 80).
5
Find the end point value of W when
t = 8.
When t = 8,
W = 88
6
State its coordinates.
The other end point is (8, 88).
7
On a set of axes, mark the end points
and turning point.
W (kg)
8
Sketch a parabola between the end
points.
9
Locate the maximum and minimum
values of W on the graph.
Maximum (8, 88)
90
80 (0, 80)
70 Minimum (3, 75.5)
0
3
8 t (months)
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.
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Chapter 1 Graphs and polynomials
29
remember
remember
Quadratic graphs
1. Quadratic equations are polynomials of degree 2.
2. General equation is y = ax2 + bx + c.
3. The quadratic formula is given by the equation
4.
5.
6.
7.
−b ± b 2 – 4ac
x = -------------------------------------- .
2a
The discriminant is b2 − 4ac and if:
(a) b2 − 4ac > 0, there are two x-intercepts. If b2 − 4ac is a perfect square, the
intercepts are rational.
(b) b2 − 4ac = 0, there is one x-intercept, which is a turning point
(c) b2 − 4ac < 0, there are no x-intercepts.
The turning point form of the quadratic graph or parabola is:
y = a(x − h)2 + k
and the turning point is (h, k).
b
The axis of symmetry of a parabola is given by the expression – ------ .
2a
The axis of symmetry is halfway between the x-intercepts.
Quadratic graphs
Math
cad
1E
Discriminant
WORKED
Example
sheet
1 Use the discriminant to determine the number of x-intercepts for each of the following
quadratic functions.
L Spre
XCE ad
a f (x) = x2 − 3x + 4
b f (x) = x2 + 5x − 8
Discriminant
d f (x) = 2x2 + 7x − 11
c f (x) = 3x2 − 5x + 9
2
2
f f (x) = 3 + 6x + 3x
e f (x) = 1 − 6x − x
E
17
WORKED
Example
of the following functions, showing all intercepts. Give
b
d
f
h
Math
cad
2 Sketch the graphs of each
exact answers.
a f (x) = x2 − 6x + 8
c f (x) = x2 − 5x + 4
e f (x) = 10 + 3x − x2
g f (x) = 6x2 − x − 12
f (x) = x + 6x + 8
f (x) = 6 − x − x2
f (x) = 2x2 + 5x − 3
f (x) = 15 + x − 6x2
2
Quadratic
graphs
3 Find the turning point for each of the functions in question 2. Give exact answers.
WORKED
Example
19
2
d y = −(x + 2)2
g y = 2(2x + 1)2 − 5
e y = 2(x + 3)2 − 6
Quadratic
graphs
1.7
SkillS
HEET
4 For each of the following functions find:
i the coordinates of the turning point
ii the domain
iii the range.
2
1
-a y=x −
b y = 2 − x2
sheet
L Spre
XCE ad
E
18
c
y = (x − 6)2
f
y = 1 − (x − 1)2
Domain
and range
for quadratic
graphs
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30
Maths Quest 12 Mathematical Methods
WORKED
Example
20
5 Each of the functions graphed below is of the form y = x2 + bx + c. For each function,
give: i the rule ii the domain iii the range.
Write the answers to b and c in interval notation.
a
b
c
y
y
(–3, 2)
x
0
0
x
0
(1, –2)
d
e
y
(0, 6)
f
y
(5, 0)
y
Example
Mat
21
b y=
d y = 2(x + 3) + 2
2
reads
L Sp he
− 1--- x2
2
e y = 1 − 4(2 − x)2
1
--4
c
f
y = −(x − 1)2
y = (2x − 3)2 − 8
7 multiple choice
et
EXCE
6 Sketch the graphs of the following, clearly showing the coordinates of the turning
point and the intercepts with the axes.
a y = 2x2 + 3
Function
grapher
x
(–4, –16)
(2, –3)
WORKED
4
x
0
x
0
d
(1, 9)
(–1, 6)
0
hca
x
Consider the function with the rule y = x2 − 2x − 3.
a It has x-intercepts:
A (1, 0) and (3, 0)
B (−1, 0) and (3, 0)
D (2, 0) and (−1, 0)
E (0, −1) and (0, 3)
Function
grapher
b It has a turning point with coordinates:
A (−1, 0)
B (2, −3)
C (1, −4)
C (1, 0) and (−3, 0)
D (−1, −4)
E (1, 0)
8 multiple choice
The function f (x) = −(x + 3)2 + 4 has a range given by:
A (3, ∞)
B (−∞, −3]
C [4, ∞)
D (−∞, 4]
E R−
9 multiple choice
The range of the function y = (x − 4)2, x ∈ [0, 6] is:
A [0, 16]
B [4, 16]
C [0, 4]
D (4, 12]
WORKED
EXCE
Example
et
reads
L Sp he
Quadratic graphs
— turning point
form
Mat
d
hca
Quadratic graphs
— turning point
form
22
E [0, 16)
10 Sketch the graph of each of the following functions, showing the turning point and all
intercepts. Round answers to 2 decimal places where appropriate.
b f (x) = 2(x − 3)2 − 2 c f (x) = (x + 1)2 + 3
a f (x) = (x − 2)2 − 4
2
d f (x) = −(x + 4) + 9 e y = x2 + 4x + 3
f y = 2x2 − 4x − 6
11 Sketch the graph of each of the functions below and state i the domain and ii the
range of each function.
b y = 7 + 8x − x2
a y = x2 + 6x − 5
2
c y = x − 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
2
g f (x) = 5 + 6x − 3x , x ∈ [−5, 3)
h f (x) = 5x2 − 5x + 3, x ∈ (−∞, 0]
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31
Chapter 1 Graphs and polynomials
WORKED
Example
23
12 The volume of water in a tank, V m3, over a 10 month period is given by the function
V(t) = 2t 2 − 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.
13 A ball thrown upwards from a tower attains
a height above the ground given by the
function h(t) = 12t − 3t 2 + 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.
Maximum height
Tower
Ball
h(t) = 12t – 3t2 + 36
Ground
14 A section of a roller-coaster at an amusement park follows the path of a parabola.
The function h(t) = t2 − 12t + 48, t ∈ [0, 11], models the height above the ground of
the front of one of the carriages, where t is the time in seconds and h is the height in
metres.
Find the lowest point of this section of the ride.
Find the time taken for the carriage to reach the lowest point.
Find the highest point above the ground.
Find the domain and the range of the function.
Sketch the function.
Cubic 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
T
SHEE
Work
a
b
c
d
e
1.2
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32
Maths Quest 12 Mathematical Methods
If a is positive (that is, a > 0), the function is called a ‘positive cubic’. Several positive
cubics appear below.
y
y
y
x
x
x
If a is negative (that is, a < 0), the function is called a ‘negative cubic’. Several negative
cubics appear below.
y
y
y
x
x
x
MM3&4 fig 1.204
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 form y = x3,
which has a point of inflection at the origin. These may be expressed in the form
y = a(x − h)3 + k
where (h, k) is the point of inflection.
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.
This form, called basic form, works in the same way as a quadratic equation expressed
in turning point form:
y = a(x − h)2 + k
where (h, k) is the turning point and a is the dilation factor.
Basic form and its transformations will be discussed in more detail in chapter 2.
y
y = x3
x
y
y = a(x – h)3 + k
(h, k)
x
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33
Chapter 1 Graphs and polynomials
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 function in general form may be factorised to express it in factor form.
y
y = a(x – b)(x – c)(x – d)
where a > 0
y = –(x + 2)(x – 1)(x – 3)
y
–2
b
c d
1
3 x
x
Repeated factors
A twice only repeated factor in a factorised cubic
function indicates a turning point that just touches
the x-axis.
y
a
b
x
Verify this for several cases using a graphics
calculator.
y = (x – a)2 (x – b)
WORKED Example 24
For each of the following graphs, find the rule and express it in factorised form.
Assume that a = 1 or a = −1.
a
b f(x)
y
y
f(x)
–4
0
3
x
–2 0
3
x
THINK
WRITE
a
a The graph is a positive cubic, so a = 1.
1
2
3
4
Find a by deciding whether the
graph is a positive or negative cubic.
Use the x-intercepts −4, 0 and 3 to
find the factors.
Express f (x) as a product of a and its
factors.
Simplify.
The factors are (x + 4), x and (x − 3).
f (x) = 1(x + 4)x(x − 3)
f (x) = x(x + 4)(x − 3)
Continued over page
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Maths Quest 12 Mathematical Methods
THINK
WRITE
b
b The graph is a negative cubic, so a = −1.
1
2
3
4
5
Find a by deciding whether the graph is
a positive or negative cubic.
Use the x-intercept −2, which is also a
turning point, to find the repeated factor.
Use the other x-intercept, 3, to find the
other factor.
Express f (x) as a product of a and its
factors.
Simplify.
(x + 2)2 is a factor.
(x − 3) is also a factor.
f (x) = −1(x + 2)2(x − 3)
f (x) = (3 − x)(x + 2)2
WORKED Example 25
Sketch the graph of y = x3 − x2 − 10x − 8, showing all intercepts.
THINK
1
2
3
4
5
Find y when x = 0.
State the y-intercept.
Let P(x) = y.
Use the factor theorem to find a factor
of the cubic
P(x) = x3 − x2 − 10x − 8.
Use long division, or otherwise, to find
the quadratic factor.
WRITE/DRAW
When
x = 0, y = −8
The y-intercept is −8.
Let
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.
By long division:
x2 − 2x − 8
x + 1 ) x − x2 − 10x − 8
x3 + x2
−2x2 − 10x − 8
−2x2 − 2x
−8x − 8
−8x − 8
3
6
7
8
9
Factorise the quadratic, if possible.
Express the cubic in factorised form
and let it equal 0 to find the
x-intercepts.
Solve for x using the Null Factor Law.
State the x-intercepts.
0
y = (x + 1)(x2 − 2x − 8)
= (x + 1)(x − 4)(x + 2)
If (x + 1)(x − 4)(x + 2) = 0
x = −1, 4 or −2
The x-intercepts are −2, −1, and 4.
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
THINK
10
35
WRITE/DRAW
Sketch the graph of the cubic.
y
y = x3 – x2 – 10x – 8
–2 –1 0
x
4
–8
11
Verify the graph and intercepts using a graphics calculator.
Restricting the domain of cubic functions
1. If the domain is R then the range is also R.
2. To find the range if the domain is restricted, it is necessary to look at the end points
and turning points, then find the highest and lowest y-values.
y
For example:
The range could not be stated here unless the
y-coordinate of the local minimum is known.
Recall that cubic functions that do not have any
turning points can have only one x-intercept.
(6, 8)
x
0
(–4, –3)
Coordinate of local
minimum required
WORKED Example 26
Sketch the graph of y = −x3 − 5x, where x ∈ (−2, −1], using the unrestricted function as a
guide. State the domain and range.
THINK
2
Decide whether it is a positive or negative
cubic by looking at the coefficient of x3.
Find the x-intercept/s.
3
Find the y-intercept.
4
Find y when x has the value of the lower end
point of the domain.
1
WRITE/DRAW
Negative cubic
When
y = 0,
−x3 − 5x = 0
−x(x2 + 5) = 0
x = 0 (x2 + 5 ≠ 0)
The x-intercept is 0.
When
x = 0,
y = −(0)3 − 5(0)
=0
The y-intercept is 0.
When x = −2, y = −(−2)3 − 5(−2)
= 18
Continued over page
MQ12 Maths Methods / Final Pages / 18/11/05
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36
Maths Quest 12 Mathematical Methods
THINK
5
6
7
8
9
10
WRITE/DRAW
State the coordinates of this end point and
decide whether it is open or closed.
Find y when x has the value of the upper end
point.
State the coordinates of this end point and
decide whether it is open or closed.
Mark these points on a set of axes.
Sketch the part of the cubic between the end
points.
Verify this graph using a graphics calculator.
The open end point is (−2, 18).
When x = −1, y = −(−1)3 − 5(−1)
=6
The closed end point is (−1, 6).
(–2, 18)
y
(–1, 6)
x
0
11
12
The domain is (−2, −1].
The range is [6, 18).
State the domain, which is given with the rule.
From the graph, state the range. Note that the
intercept is not included in the domain.
remember
remember
Cubic graphs
1. The general equation is y = ax3 + bx2 + cx + d.
2. Basic shapes of cubic graphs:
Positive cubic
Negative cubic
Basic form
y
y
y
y = a(x – h)3 + k
x
x
(h, k)
x
Factor form
Repeated factor
y
y
y = a(x – b)(x – c)(x – d)
where a > 0
b
c d
x
a
b
x
y = (x – a)2 (x – b)
If a < 0, the reflections through the x-axis of the types of graph in the above
figures are obtained.
MQ12 Maths Methods / Final Pages / 18/11/05
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37
Chapter 1 Graphs and polynomials
1F
Example
24
1 For each of the following graphs, find the rule and express it in factorised form.
Assume that a = 1 or a = −1.
Cubic graphs
a
b
y
Math
cad
WORKED
Cubic graphs
— factor
form
y
E
0
–6
0
–2
x
5
1
x
sheet
L Spre
XCE ad
Cubic graphs
— factor
form
–4
2 Match each of the following graphs to the most appropriate rule below.
b
y
0 1
–3
d
4
c
y
–2 0
x
e
y
0
–3
x
5
y
f
y
–4
g
3
h
WORKED
Example
25
x
1
x
5
x
–2
0
x
y = (x − 3)3
y = (4 − x)(x + 2)(x − 1)
y = (x + 3)(x − 1)(x − 4)
y = (3 − x)3
viii
iiiv
iivi
viii
y = (x + 3)(1 − x)(x − 4)
y = (x + 2)2(5 − x)
y = (x + 4)(x + 2)(x − 1)
y = (x + 2)2(x − 5)
Cubic graphs
L Spre
XCE ad
Cubic graphs
sheet
3 Sketch the graph of each of the following, showing all intercepts.
a y = x3 + x2 − 4x − 4
b y = 2x3 − 8x2 + 2x + 12
3
c y = 24 + 26x − 2x
d y = 18 − 21x + 8x2 − x3
2
3
e y = 12 + 8x − x − x
f y = 3x3 − 15x2 + 9x + 27
3
g y=8−x
Verify your answers by using a graphics calculator.
Math
cad
vii
iiii
iiv
vii
1
0
y
–4
3
0
x
x
y
0
–2
4
y
–2
0
1
E
a
5_61_02561_MQV12MM - 01_tb Page 38 Friday, November 18, 2005 2:15 AM
38
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
4 multiple choice
a Fully factorised, x3 + 6x2 + 12x + 8 is equal to:
B (x + 2)3
A (x + 3)3
D (x − 3)3
E (x + 2)(x − 2)2
C (x − 2)3
b The graph of y = x3 + 6x2 + 12x + 8 is:
A
B
y
0
D
d
Cubic graphs —
y = a(x − b)3 + c
form
et
reads
L Sp he
EXCE
2
x
2
x
y
–2
x
0
5 multiple choice
Mat
hca
E
0
0
x
2
y
–2
y
–2
x
0
–2
–2
C
y
Cubic graphs —
y = a(x − b)3 + c
form
y
The function graphed in the figure could have the
following rule:
A y = (x − 2)3 + 2
B y = (x + 2)3 + 2
C y = (2 − x)3 + 2
D y = (x + 2)3 − 2
E y = (x − 2)3
10
(2, 2)
x
0
6 multiple choice
The graph of f (x) = 5(x + 1)3 − 3 is best represented by:
A
B
y
0
x
0
(–1, –3)
D
C
y
x
E
y
(1, 3)
(–1, 3)
0
x
0
x
0
(1, –3)
(–1, –3)
y
y
x
5_61_02561_MQV12MM - 01_tb Page 39 Friday, November 18, 2005 2:15 AM
MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
39
7 multiple choice
The graph of f (x) = 2 (x − 1) (x + 3) is best represented by:
y
A
B
y
(0, 6)
(0, 6)
(–1, 0) 0
C
x
(3, 0)
(–3, 0)
0
D
y
x
(1, 0)
y
(1, 0)
x
(–3, 0) 0
(–3, 0) 0
x
(0, –6)
(0, –6)
E
(1, 0)
y
(3, 0)
(–1, 0)
0
x
(0, –6)
y
8 multiple choice
The graph shown is best represented by the equation:
A y = (x − a)3 + b
B y = −(x − a)3 + b
C y = (a − x)3 + b
D y = −(x + a)3 + b
E y = (x + a)3 + b
(a, b)
(0, c)
0
x
MQ12 Maths Methods / Final Pages / 18/11/05
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40
Maths Quest 12 Mathematical Methods
9 multiple choice
y
If a < 0 and b, c > 0 then the graph shown is best
represented by the equation:
b
b
- (x + a)2 (x − c)
A y = ------a2c
b
- (x + a)2 (c − x)
B y = – ------a2c
a
0
c
x
b
- (x − a)2 (x + c)
C y = ------a2c
b
- (x + a)2 (x − b) (c − x)
D y = – ------a2c
b
E y = – ------- (x + a)2 (x − c)
a2c
WORKED
Example
26
10 Sketch the graph of each of the following restricted functions, using the unrestricted
function as a guide. State i the domain and ii the range in each case.
a f (x) = x3 + x2 − 10x + 8, x ∈ [2, ∞)
b f (x) = 3x3 − 5x2 − 4x + 4, x ∈ [−2, −1]
c
f (x) = −3x3 + 4x2 + 27x − 36, x ∈ (0, 1]
d f (x) = x3 + 6x, x ∈ [−2, 2]
e f (x) = −4x − 5x3, x ∈ (0, 1)
f
f (x) = −3x − x3, x ∈ [−1, 2)
g f (x) = x3 + 2x, x ∈ [−2, −1) ∪ (0, 3]
h f (x) = −2x3 − x, x ∈ (−1, 1) ∪ [2, 3)
Verify your answers by using a graphics calculator.
11 The function f (x) = x3 + ax2 + bx − 64 has x-intercepts (−2, 0) and (4, 0). Find the
values of a and b.
12 The functions y = x3 − 2x2 + ax + 10 and y = 6 + (a + b)x − 4x2 − x3 both have (−1, 0)
as an x-intercept. Find the values of a and b.
13 The cross-section of a glass vessel that is 6 cm high can be modelled by the cubic
function f (x) and its reflection through the y-axis, g(x), as shown below.
y
f(x) = a(x + b)3 + c
(4, 6)
g(x)
(3, 3)
0
(2, 0)
x
a Find the values of a, b and c, and hence state the rule of f (x).
b Find the rule for g(x) and state its domain and range.
c
What is the width of the vessel when the height is 3.375 cm?
5_61_02561_MQV12MM - 01_tb Page 41 Friday, November 18, 2005 2:15 AM
MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
41
14 The distance of a group of hikers, d km, from their starting point t hours after setting
off on a hike can be modelled by the function with the rule:
d(t) = at 2 (b − t)
The hikers are 3 km from the start after 2 hours and return to the starting point after
5 hours.
a Find the values of a and b.
b Hence, give the rule for d(t) stating its domain and range.
c Sketch the graph of d(t).
d Find to the nearest 100 metres the maximum distance of the hikers from their
starting point and the time, to the nearest minute, that it occurs.
Quartic graphs
Quartic functions are polynomials of degree 4. The general form of a quartic is:
y = ax4 + bx3 + cx2 + dx + e
When sketching the graphs of quartic functions, all axes’ intercepts can be found by
factorisation and a sign diagram used to check the shape. If a sign diagram is not sufficient and the basic shape is not recognised, then a graphics calculator could be used to
establish the shape of the graph.
Basic shapes of quartic graphs
Positive quartics (a > 0)
1. y = ax4
2. y = ax4 + cx2, c ≥ 0
y
0
y
x
3. y = ax2(x − b)(x − c)
4. y = a(x − b)2(x − c)2
y
b
0
x
0
y
c x
The repeated factor x2 shows
there is a turning point at the
origin. The factors (x − b) (x − c)
show x-intercepts at x = b and
x = c.
b
0
c
x
The repeated factors (x − b)2 and
(x − c)2 show the graph touches
the x-axis at x = b and x = c.
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42
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
6. y = a(x − b)(x − c)(x − d)(x − e)
5. y = a(x − b)(x − c)3
y
b
0
y
c
x
b
The cubed factor (x − c)3 show the
graph as a point of inflection at
x = c.
c
0
d
e x
The factors show intercepts
at x = b, c, d and e.
y
Negative quartics
x
0
If a < 0, that is, each of the above rules is multiplied by
−1, then the graphs are reflected through the x-axis.
For example, the graph of y = −x4 (at right)
is a reflection, through the x-axis, of the graph of y = x4.
Similarly, the graph of y = −x4 + x2 = −(x4 − x2) is a
reflection through the x-axis of the graph of y = x4 − x2.
y = –x 4
y
Note: The above graphs can be translated horizontally or
0
vertically but this is considered in chapter 2, Other graphs
and modelling.
To find the x-intercepts of a quartic function, let y = 0 and
y = –x 4 + x2
solve the equation for x.
Repeated factors touch the x-axis as they do for cubic and quadratic functions.
WORKED Example 27
Sketch the graph of y = x4 − x3 − 7x2 + 5x + 10, showing all intercepts.
THINK
1
2
3
Find the y-intercept.
Let y = P(x).
Find two linear factors of the quartic
expression, if possible, using the factor
theorem.
WRITE/DRAW
When x = 0, y = 10
The y-intercept is 10.
Let P(x) = x4 − x3 − 7x2 + 5x + 10
P(1) = (1)4 − (1)3 − 7(1)2 + 5(1) + 10
=8
≠0
P(−1) = (−1)4 − (−1)3 − 7(−1)2 + 5(−1) + 10
=0
(x + 1) is a factor.
P(2) = (2)4 − (2)3 − 7(2)2 + 5(2) + 10
=0
(x − 2) is a factor.
x
MQ12 Maths Methods / Final Pages / 18/11/05
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Chapter 1 Graphs and polynomials
THINK
Find the product of the two linear
factors.
Use long division to divide the quartic
by the quadratic factor x2 − x − 2 (or
use another method).
4
5
43
WRITE/DRAW
(x + 1)(x − 2) = x2 − x − 2
x2 − 5x − 55
x − x − 2 ) x − x − 7x2 + 5x + 10
−(x4 − x3 − 2x2)
0 − 5x2 + 5x + 10
−(−5x2 + 5x + 10)
0
2
y = (x + 1)(x − 2)(x − 5)
2
4
3
6
Express the quartic in factorised form.
7
Factorise the quadratic factor, x2 − 5,
using difference of perfect squares.
y = (x + 1)(x − 2)(x +
5 )(x −
5)
8
To find the x-intercepts, set y equal to
zero.
Let y = (x + 1)(x − 2)(x +
5 )(x −
5) = 0
9
Solve for x using the Null Factor Law.
10
State the x-intercepts.
Sketch the graph of the quartic.
11
x = −1, 2, ± 5
The x-intercepts are −1, 2, − 5 and
5.
y
(0, 10)
(–1, 0)
12
Check the graph using a graphics
calculator.
(2, 0)
(– 5, 0)
–3 –2
–1 0
1
( 5, 0)
x
2 3
WORKED Example 28
Sketch the graphs of each of the following equations, showing the coordinates of all
intercepts. Use a graphics calculator to find the coordinates of the turning points,
rounding to 2 decimal places as appropriate.
a y = x2 (x − 1)(x + 2)
b y = −(x + 3)2(x − 1)
THINK
WRITE/DRAW
a
a y = x2(x − 1)(x + 2)
When x = 0, y = 0
The y-intercept is 0
When y = 0,
x2(x − 1)(x + 2) = 0
x = −2, 0, 1
The graph touches the x-axis at x = 0.
The other x-intercepts are 2 and 1.
2
State the function.
Find the y-intercept.
3
Find the x-intercepts.
4
State the x-intercepts, noting where
the graph touches and where it cuts
the x-axis.
1
Continued over page
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44
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
THINK
5
State the coordinates of the turning
points.
6
Sketch the graph of the quartic,
using a graphics calculator to assist.
WRITE/DRAW
The minimum turning points are
(−1.44, −2.83) and (0.69, −0.40).
The maximum turning point is (0, 0).
y
Fig. 1.9 to go here
(0, 0)
0
(–2, 0)
(1, 0)
x
(0.69, –0.40)
(–1.44, –2.83)
b
b y = −(x + 3)2(x − 1)2
1
State the function.
2
Find the y-intercept.
3
Find the x-intercepts.
4
State the points where the graph
touches the x-axis from the repeated
factors.
The graph touches the x-axis at x = −3 and
x = 1.
5
State the coordinates of the turning
points.
The maximum turning points are (−3, 0) and
(1, 0), and the minimum turning point is
(−1, −16).
6
Sketch the graph of the quartic,
using a graphics calculator to assist.
When x = 0,
y = −(3)2(−1)2
= −9
The y-intercept is −9.
When y = 0,
0 = −(x + 3)2(x − 1)2
x = −3, 1
y
(–3, 0)
(1, 0)
0
x
(0, –9)
(–1, –16)
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Chapter 1 Graphs and polynomials
WORKED Example 29
y
Determine the equation of the graph shown.
3
-3 –1 0 1 2
THINK
1
State the x-intercepts.
2
Write the equation using factor form
with a dilation factor of a.
State the y-intercept.
Substitute the coordinates of the point
where the graph crosses the y-axis into
the equation.
Solve the equation to find a.
3
4
5
Write the equation.
x
WRITE
The x-intercepts are −3, −1, 1, 2.
y = a(x + 3)(x + 1)(x − 1)(x − 2)
The y-intercept is 3.
(0, 3) ⇒ 3 = a(0 + 3)(0 + 1)(0 − 1)(0 − 2)
3=a×6
a=
6
1
--2
y = 1--- (x − 1)(x − 2)(x + 3)(x + 1)
2
WORKED Example 30
Sketch the graph of each of the following restricted functions, using the unrestricted
function as a guide. State the domain and the range in each case.
a y = (x + 1)3(x − 2), x ∈ (−∞, −1]
b y = −x4 − 2x2, x ∈ (−1, 1]
THINK
WRITE/DRAW
a
a y = (x + 1)3(x − 2), x ∈ (−∞, −1]
When x = 0,
y = (1)3(−2)
= −2
The y-intercept is −2.
When
y = 0,
3
(x + 1) (x − 2) = 0
x = −1 or 2
The x-intercepts are −1 and 2.
1
2
3
4
5
6
7
State the function.
Find the y-intercept.
State the y-intercept.
Find the x-intercept.
Solve for x.
State the x-intercepts.
Sketch the graph over the domain
(−∞, −1], using knowledge of basic
shapes or a graphics calculator to
assist. (The cubed factor indicates a
point of inflection.)
45
y
(2, 0)
0
(–1, 0)
(0, –2)
x
y = (x + 1)3 (x – 2)
Continued over page
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46
Maths Quest 12 Mathematical Methods
THINK
b
MQ12 Maths Methods / Final Pages / 18/11/05
WRITE/DRAW
8
State the domain, which is given
with the rule.
The domain is (−∞, −1].
9
From the graph, state the range.
The range is [0, ∞).
1
State the function.
2
Find the y-intercept.
When x = 0,
y = −(0)4 −2(0)2
=0
3
State the y-intercept.
The y-intercept is 0.
4
Find the x-intercepts.
When
y=0
4
−x − 2x4 = 0
5
Factorise the quartic expression.
6
Solve for x.
x = 0 is the only solution (as x2 + 2 ≠ 0).
7
State the x-intercepts.
The only x-intercept is 0.
8
Find y when x is one end point of the
domain.
When x = −1,
y = −(−1)4 − 2(−1)2
= −3
9
State the coordinates and whether it
is an open or closed point.
⇒ (−1, −3) is an open end point.
10
Find y when x is the other end point
of the domain.
When x = 1,
y = −(1)4 − 2(1)2
= −3
11
State the coordinates and whether it
is an open or closed point.
⇒ (1, −3) is a closed end point.
12
Sketch the graph of the quartic,
using knowledge of basic shapes or a
graphics calculator to assist, over the
domain.
b y = −x4 − 2x2, x ∈ (−1, 1]
−x2(x2 + 2) = 0
y
(0, 0)
0
(–1, –3)
x
(1, –3)
y = –x 4 – 2x2
13
State the domain, which is given
with the rule.
The domain is (−1, 1].
14
From the graph, state the range.
The range is [−3, 0].
5_61_02561_MQV12MM - 01_tb Page 47 Friday, November 18, 2005 2:15 AM
MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
47
remember
remember
Quartic graphs
1. General equation is y = ax4 + bx3 + cx2 + dx + e
2. Basic shape of quartic graphs:
(a) If a > 0:
y
y = ax4
x
0
y
y = ax4 + cx2, c ≥ 0
x
0
y
y = ax2(x − b)(x − c)
b
c
0
x
y
y = a(x − b)2(x − c)2
0
b
x
c
y
y = a(x − b)(x − c)3
b
0
c
x
y
y = a(x − b)(x − c)(x − d)(x − e)
b
c
0 d
e x
(b) If a < 0, then the reflection through the x-axis of the types of graph in the
figures above is obtained.
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48
Maths Quest 12 Mathematical Methods
1G
WORKED
Example
d
hca
27
Mat
MQ12 Maths Methods / Final Pages / 18/11/05
Quartic graphs
— factor form
1 Sketch the graph of each of the following showing all intercepts. Verify the
shape of the graph using a graphics calculator.
a y = (x − 2)(x + 3)(x − 4)(x + 1)
b y = (x2 − 1)(x + 2)(x − 5)
y = 2x4 + 6x3 − 16x2 − 24x + 32
d y = x4 + 4x3 − 11x2 − 30x
c
Mat
d
hca
Quartic graphs
e y = x4 + 4x3 − 12x − 9
f
y = x4 − 4x2 + 4
g y = 30x − 37x2 + 15x3 − 2x4
h y = 6x4 + 11x3 − 37x2 − 36x + 36
Quartic
graphs
1.8
SkillS
HEET
WORKED
Example
Solving
quartic
equations
EXCE
a y = x2 (x − 2)(x − 3)
c y = (x − 1)2(x + 1)(x + 3)
b y = −(x + 1)2(x − 1)2
d y = (x + 2)3(1 − x)
et
reads
L Sp he
28
2 Sketch the graph of each of the following equations, showing the coordinates of all
intercepts. Use a graphics calculator to find the coordinates of the turning points,
rounding to 2 decimal places as appropriate.
Quartic
graphs —
factor form
3 multiple choice
Consider the function f (x) = x4 − 8x2 + 16.
a When factorised, f (x) is equal to:
A (x + 2)(x − 2)(x − 1)(x + 4)
B (x − 1)(x − 4)(x + 4)
C (x + 3)(x − 2)(x − 1)(x + 1)
D (x − 2)3(x + 2)
E (x − 2)2(x + 2)2
b The graph of f (x) is best represented by:
A
B
y
0
–2
2
x
–16
D
0
–2
E
y
16
C
y
16
x
2
y
4
–2
0
2
x
–2
0
2
x
y
16
–2
0
2
x
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49
Chapter 1 Graphs and polynomials
c
If the domain of f (x) is restricted to [−2, 2], then the range is:
A [0, 16]
B [0, 10]
+
E [0, ∞)
D R
C [−2, 12]
d If the range of f (x) is restricted to (0, 25) then the maximal domain is:
A [−2, 3)
B (−2, 3)
D (−3, 3)
E (−3, 4)
C (−3, 2)
e If the domain of f (x) is restricted to (−1, 0), then the range is:
f
WORKED
Example
29
A (0, 16)
B (0, 4)
D (9, 16)
E [9, ∞)
C (−1, 9)
If the domain of f (x) is restricted to [0, ∞), then the range is:
A R
B R+
D [0, 16)
E [2, ∞)
C [0, ∞)
4 Determine the equation of each of the following graphs.
a
b
y
y
12
6
–2
-1 0
1
3
x
–3
c
–10
d
y
2
4
1
3 4
x
y
8
24
–10
2
4
x
–2
0
x
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50
MQ12 Maths Methods / Final Pages / 18/11/05
Maths Quest 12 Mathematical Methods
WORKED
Example
30
5 Sketch the graph of each of the following restricted functions, using the unrestricted
function as a guide. State i the domain and ii the range in each case.
a y = x(x − 1)3, x ∈ R−
b y = (2 − x)(x2 − 4)(x + 3), x ∈ [2, 3]
c y = x4 − x2, x ∈ [1, ∞)
d y = 9x4 − 30x3 + 13x2 + 20x + 4, x ∈ (−2, −1]
e y = −(x − 2)2(x + 1)2, x ∈ (−∞, −2]
f y = x4 − 6x2 − 27, x ∈ [3, ∞)
g y = (x + 2)3(x − 3), x ∈ [−3, 0)
h y = 4x2 − x4, x ∈ [−3, −2]
Verify your answers using a graphics calculator.
6 The function f (x) = x4 + ax3 − 4x2 + bx + 6 has x-intercepts (2, 0) and (−3, 0). Find the
values of a and b.
7 The function f (x) = x4 + ax3 + bx2 − x + 6 has x-intercepts (1, 0) and (−3, 0). Find the
values of a and b.
8 The functions y = (a − 2b)x4 − 3x − 2 and y = x4 − x3 + (a + 5b)x2 − 5x + 7 both have
an x-intercept of 1. Find the value of a and b.
Quartics and beyond
Use a graphing program such as Graphmatica or one of the Maths Quest Mathcad
files to assist in answering the following questions.
1 Investigate graphs of functions of the form f(x) = x n for values of n from 4 to 9.
2 What do graphs of functions for which n is even have in common?
3 What does an odd value of n do to the graph?
4 Investigate graphs of functions of the form y = (x − a)n(x − b)m(x − c)p for
d
Mat
hca
various values of the pronumerals in the equation, for m, n and p ≤ 4. Write a
report on your findings.
Single
function
grapher
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Chapter 1 Graphs and polynomialhs
51
summary
Pascal’s triangle
1
1
1
1
1
1
1
2
3
4
5
1
3
6
10
1
4
10
1
5
1
Binomial theorem
(ax + b)n =  n (ax)nbo +  n (ax)n − 1b + . . . +  n  (ax)bn − 1 +  n (ax)obn
 0
 1
 n – 1
 n
Notes:
Indices add to n.
There are n + 1 terms in the expansion.
The (r + 1)th term is  n (ax)n − r(b)r.
 r
Polynomials
• If P(x) = a n x n + a n − 1 x n − 1 + . . . + a 2 x2 + a 1 x + a 0 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.
• Remainder theorem:
If P(x) is divided by (x − a), then the remainder is P(a). If P(x) is divided by
(ax + b) then the remainder is P  – b---
 a
• Factor theorem:
1. If P(a) = 0, then (x − a) is a factor of P(x) or if (ax + b) is a factor of P(x), then
b
P(− --- ) = 0
a
2. If (x − a) is a factor of P(x) then a must be a factor of the term independent of x.
Linear graphs
• Linear equations are polynomials of degree 1.
• General equation is
ax + by + c = 0
or
y = mx + c
where m = gradient
c = y-intercept
y2 – y1
• The gradient
m = --------------x2 – x1
• Equation if a point and the gradient is known:
y − y1 = m(x − x1)
• Parallel lines have the same gradient.
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52
Maths Quest 12 Mathematical Methods
• If m1 and m2 are the gradients of perpendicular lines, then:
m1 × m2 = −1
1
or
m1 = − -----m2
Quadratic graphs
• Quadratic equations are polynomials of degree 2.
• General equation is
y = ax2 + bx + c
−b ± b 2 – 4ac
x = -------------------------------------2a
• Quadratic formula is
• Discriminant = b2 − 4ac and
1. if b2 − 4ac > 0, there are 2 x-intercepts (and if b2 − 4ac is a perfect square, the
intercepts are rational)
2. if b2 − 4ac = 0, there is 1 x-intercept
3. if b2 − 4ac < 0, there are no x-intercepts.
• The turning point form of the quadratic is:
y = a(x − h)2 + k
and the turning point is (h, k).
b
• The equation of the axis of symmetry of a parabola is – ------ .
2a
• The axis of symmetry is halfway between the x-intercepts.
Cubic graphs
• Cubic equations are polynomials of degree 3.
• General equation is y = ax3 + bx2 + cx + d
• Basic shapes of cubic graphs:
Positive cubic
Negative cubic
y
Basic form
y
y
y = a(x – b)3 + c
x
(b, c)
x
x
Factor form
Repeated factor
y
y
y = a(x – b)(x – c)(x – d)
where a > 0
b
c d
x
a
b
x
y = (x – a)2 (x – b)
If a < 0, then the reflections through the x-axis of the types of graph in the above
figures are obtained.
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Chapter 1 Graphs and polynomials
53
Quartic graphs
• Quartic equations are polynomials of degree 4.
• General equation is y = ax4 + bx3 + cx2 + dx + e
• Basic shapes of quartic graphs:
1. If a > 0:
y
y
0
b
x
0
y = ax
4
x
c
y = a(x − b)2(x − c)2
y
y
0
b
x
0
y = ax + cx , c ≥ 0
4
2
c
x
y = a(x − b)(x − c)3
y
y
b
b
0
c
c
0 d
e x
x
y = a(x − b)(x − c)(x − d)(x − e)
y = ax (x − b)(x − c)
2
2. If a < 0, then reflection through the x-axis of the types of graph above is
obtained.
Note: It is possible to translate the cubic and quartic graphs shown in the cubic graphs
and quartic graphs sections above.
Functions
• A function is fully defined if the rule and domain are given.
• The domain of a function is the set of values of x for which the function is defined.
• The range of a function is the set of values of y for which the function is fully
defined.
• Restricted domains can be represented by interval notation:
[a, b] = {x: a ≤ x ≤ b}
(a, b) = {x: a < x < b}
[a, b) = {x: a ≤ x < b}
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54
Maths Quest 12 Mathematical Methods
CHAPTER
review
Multiple choice
1A
1 When expanded, (1 − 2x)5 is equal to:
A 1 + 2x − 4x2 − 8x3 + 16x4 + 32x5
C 5 − 10x + 20x2 − 40x3 + 80x4 − 160x5
E 1 − 10x + 40x2 − 80x3 + 80x4 − 32x5
1A
1 8
2 The coefficient of x5 in the expansion of  4x – ----- is:

x 2
A 4096
B −131 072
B 1 − 2x + 4x2 − 8x3 + 16x4 − 32x5
D −1 + 2x − 4x2 + 8x3 − 16x4 + 32x5
C −4096
D −16 384
E 16 384
1A
1 10
3 Assuming descending powers of x, the fifth term of the expansion of  3x + --- is:

x
1B
81
81
C -----2D 729x2
E -----5x
x
4 Which of the following expressions is not a polynomial?
B x4 − 5x3 + 3x23 − 6x
A x3 + 3x − 1
--21
11
C x −x +x− 3
D x4 + 5x3 − 2x 2 + 5x − 3
E x6 − x5 + 2x4 − x3 + 4x − 2
1B
5 The value of P(−3) in the polynomial, P(x) = x5 − 4x3 − 3x2 + 10x + 1, is:
A −31
B −139
C −191
D 6
E 1
1B
6 The degree of the polynomial (5 − 6x + x3 + 7x6)(x2 − 3x4 + 2) when expanded is:
A 24
B 8
C 10
D 16
E 21
1C
7 The remainder when x5 + 2x4 + 4x3 − 5x + 3 is divided by (x + 3) is:
A −271
B 51
C −171
D 3
1C
8 For which one of the following polynomial expressions is (x − 2) not a factor?
B x4 − 2x3 − 6x2 − 8x + 2
A x3 + 3x2 − 4x − 12
4
3
2
C x + 2x − 7x − 8x + 12
D x3 + x2 − 10x + 8
3
2
E 2x + 3x − 9x − 10
1C
9 Which one of the following is a factor of 2x4 − 4x3 − 10x2 + 12x?
A (x − 2)
B (x + 3)
C (x + 1)
D (x − 4)
A 153 090x2
1D
B 243x4
10 The rule for the graph shown is:
A 2x + y + 4 = 0
C 2y − x − 4 = 0
E 4x + 2y = 0
E 108
E (x − 3)
y
B x − 2y − 4 = 0
D x + 2y − 4 = 0
2
0
4
x
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55
Chapter 1 Graphs and polynomials
Questions 11 and 12 refer to the graph at right, which
has a gradient of 2.
y
(2, b)
0
11 The value of b must be:
A 5
D −1
x
2
(–3, –5)
B 3
E 4
C 1
1D
12 The y-intercept is:
A (0, 3)
C (0, 1--- )
B (0, 2)
E (− 1--- , 0)
D (0, 1)
2
2
13 If 3x + 4x − 5 = 0, then the value of the discriminant is:
2
A
B −44
76
1D
C
D − --2-
– 44
1E
E 76
3
Questions 14 and 15 refer to the function with the rule: y = 2x2 + 8x − 10 where x ∈ (−6, 2).
1E
14 Which one of the following graphs could represent this function?
A
y
(–6, 14)
(2, 14)
0
–6 –5
2
B
y
(–6, 14)
x
–10
D
E
y
(–6, 22)
(2, 14)
y
(–2, 14)
x
0 12
–10
–6 –5
C
y
(6, 14)
–2 –1 0
–10
56
x
(2, 10)
(–6, 6)
–1 0 2
–2
–6
x
0 12
–6 –5
–10
x
(2, –3.6)
1E
15 The range of this function is:
A (−18, 14)
B (−10, 14)
C [−18, 14)
D [−18, 14]
E (−14, 10)
16 The graph of y = −3x3 could be:
A
B
y
–1
0
1
x
C
y
0
x
D
y
0
E
y
x
0
1
x
1F
y
–1 0
1
17 Which of the following intercepts does the graph of f (x) = −6 + 11x + 3x2 − 2x3 have?
A ( 1--- , 0), (−2, 0), (3, 0) and (0, −6)
2
C (− 1--- , 0), (−2, 0), (3, 0) and (0, 6)
2
E ( 1--- , 0), (−3, 0), (2, 0) and (0, −6)
2
B (−2, 0), (2, 0), (3, 0) and (0, −6)
D (−2, 0), (−1, 0), (3, 0) and (0, −6)
x
1F
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56
1F
1G
1G
Maths Quest 12 Mathematical Methods
18 The rule for the graph shown at right could be:
A f (x) = (x − 1) (x + 3)
C f (x) = (x + 1)2(3 − x)
E f (x) = (x − 3)(x + 1)2
2
y
B f (x) = (x + 1)(x − 3)
D f (x) = (x2 − 1)(x + 3)
f(x)
2
–1 0
x
3
–3
19 The rule for the graph shown at right could be:
y
f(x)
B f (x) = −x(x − 2)
D f (x) = x(x − 2)3
A f (x) = x(x + 2)
C f (x) = x2(x − 2)2
E f (x) = x(2 − x)2
3
2
0
x
2
20 The graph of y = (x + 3)2(x − 1)(x − 3) is best represented by:
A
B
y
–3
–3
0
D
1
3
0
3
1
3
x
y
x
–3
E
1
0
x
y
–3
C
y
0
1
3
x
y
–3
0 1
3
x
Short answer
1A
1 Expand each of the following:
a (2y − 3x)5
x 2 8
b  --- – ---
 2 x
1B
1C
2 If (x2 − 1) is a factor of P(x) = −7 + ax + 5x2 + 15x3 + bx4, then find the values of a and b.
3 Factorise each of the following expressions:
a x3 − 12x2 + 17x + 90
b 2x4 + 7x3 − 31x2 + 36
1D
4 Find the equation of each of the straight lines described below.
a The line which passes through the points (−5, 6) and (1, −1).
b The line which is perpendicular to the line with equation 2x − y + 10 = 0 and passes
through the point (3, 3).
1E
5 Sketch the graph of y = 8 − 2x − x2, by labelling the turning point and all intercepts. State its
domain and range.
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MQ12 Maths Methods / Final Pages / 18/11/05
Chapter 1 Graphs and polynomials
57
6 Sketch the graph of y = 3x2 + 8x − 3, x ∈ [−3, 0). State the range of this function.
7 a If (x + 3) is a factor of f (x) = −x3 + bx2 + ax − 18 and g(x) = ax2 + bx − 75, then find the
values of a and b.
b Sketch the graph of f (x) by labelling all intercepts.
8 Sketch the graph of y = 4x3 − 7x2 − 5x + 6, x ∈ [−2, −1). State the range of this function.
9 Solve the equation 6 + 7x − 27x2 + 17x3 − 3x4 = 0
10 Sketch the graph of f (x) = x4 − 7x3 + 12x2 + 4x − 16.
Analysis
1 An empty parfait glass has been left on a table with
the rim just touching a wall. Ants are marching in a
line down into the parfait glass and then up the other
side, following the path of a parabola. They begin
their journey where the glass touches the wall, 18 cm
above the table.
a The stem of the glass is 4 cm long and the
diameter of the top of the glass is 5 cm. Find the
rule for the quadratic function that describes the
shape of the glass.
b State the domain and range of the function.
c If there is fruit juice in the bottom of the glass to
a depth of 1 cm, find the coordinates of the point
where the ants first touch the juice. Round
answers to the nearest whole number.
d Using function notation, write the rule for the
surface of the cross-section of the juice in the
glass.
2 A ‘rogue satellite’ has its distance from Earth, d thousand kilometres, modelled by a cubic
function of time, t days after launch. After 1 day it reaches a maximum distance from Earth of
4000 kilometres, then after 2 days it is 2000 kilometres away. It effectively returns to Earth
after 3 days, then moves further and further away.
a What is the satellite’s initial distance
from Earth?
b Sketch the graph of d versus t for the
first 6 days of travel.
c Express d as a function of t.
The moon is approximately 240 000
kilometres from Earth.
d Which is closer to Earth after 8 days,
the satellite or the moon? By how
far?
The satellite is programmed to selfdestruct. This happens when it is
490 000 kilometres from Earth.
e What is the ‘life span’ of the satellite?
f State the domain and range of d(t).
1E
1F
1F
1G
1G
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58
Maths Quest 12 Mathematical Methods
3 The diagram shows a main road passing
N
River
W
E
through O, A, C and E. The road crosses
S
a river at point O and 3 kilometres further
along the road at point C. Between O and C,
O A
C Main road
the furthest the river is from the road is
E
8.54 kilometres, at a point D, 2.25 kilometres east of a north–
B
south line through O. Point A is 1 kilometre east of point O. If
D
point O is taken as the origin and the road as the x-axis, then
the path of the river can be modelled by a quartic function, as
shown in blue.
a Give the coordinates of C and D.
b Find the rule for the quartic function, f (x).
c How far is the river from the main road along the track
AB?
d A canoeing race, of at least 17 kilometres in length, along the river is being organised. It
is suggested that the race could start at O and finish at C. Is this course satisfactory? Why?
4 Willie Wonkie, of Willie Wonkie’s Construction Company, makes a sketch of the symmetrical
W for a large neon sign as shown below. The x- and y-axes represent the supporting crosspieces. The width of the W along the x-axis is 6 metres and the point on the vertical support
is 2 1--- metres above the horizontal support. The W can be modelled by a quartic function, with
4
all x-intercepts exactly evenly spaced.
y
x
a
b
c
d
CHAPTER
test
yourself
1
Find the rule for the letter W.
If the top of the W is 8 metres wide, find the coordinates of the highest points of the letter.
State the domain of the function.
Use a graphics calculator to find the coordinates of the lowest points of the W, giving the
answer to 3 decimal places. Hence find the range of the function.
e In order to test the strength of his design, Willie Wonkie moves the horizontal crosspiece
so that it just touches the lowest points of the W. Find the new rule that describes the W
now.
f State the domain and range of the new function.