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Patterns, functions and algebra (2)
THEME 13
FUNCTIONS AND RELATIONSHIPS
Introduction
In this theme we will revise and extend work covered in Theme 5 regarding:
• relationships between two variables (also known as functions)
• the different forms in which we can express a function
• finding input values and output values of functions
• equivalent forms of functions, including graphs.
1. INPUT VALUES, OUTPUT VALUES AND RULES
We briefly revise important concepts from Theme 5:
• A function is a relationship between two sets of values.
• These two sets of values are known as variables.
• There are input variables and output variables.
• The input variable is also known as the independent variable.
• The output variable is also known as the dependent variable, since its values depend on the values of the input variable.
◦ as a flow diagram
◦ as an equation or formula.
In subtheme 2 we will extend this list to include graphs.
SAMPLE
• A function has a rule that links or relates the input and output variables.
• Any function can be represented using a number of forms:
◦ verbally (in words)
◦ in a table
• An equation or formula is the most direct way to show the function rule. For example, y = 2x + 1 tells us that the output values (y) are produced by multiplying the input values (x) by 2 and then adding 1.
Worked example 1: Input values, output values and rules
flow diagram represents a function:
a) Write the rule in words.
b) Give the rule as an equation.
c) Find the missing values (i), (ii) and (iii) using the flow diagram.
d) Find the missing values (i), (ii) and (iii) using your equation in b).
e) Show all the input and output values in a table.
f) Which type of numbers are the input values?
Solutions
a) The input values (x) are divided by 2 and then 4 is added to give the output values (y).
b) y = 1 2 x + 4. We can also write this as y = x 2 + 4. c)
i) y = 0 ÷ 2 + 4 = 0 + 4 = 4
ii) y = 6 ÷ 2 + 4 = 3 + 4 = 7
iii) To find iii), start with the output value 12, and work backwards through the flow diagram, each time doing the inverse (opposite) operation:
12 − 4 = 8
8 × 2 = 16
Subtract 4
Multiply by 2
So, x = 16 Check the answer by applying the flow diagram rule to 16: 16 ÷ 2 + 4 = 8 + 4 = 12
i) For i): replace x by 0: y = 0 2 + 4 = 0 + 4 = 4
ii) For ii): replace x by 6: y = 6 2 + 4 = 3 + 4 = 7
iii) For iii): replace y by 12: 12 = x 2 + 4
This is an equation that we must solve to find the value of x:
24 = x + 8
∴ 16 = x
∴ x = 16 d)
x 8 0 16 6
y 0 4 12 7
e) Integers
Worked example 2: Input values, output values and rules
The table represents a function: x 4 1 3 4 (iii)
Multiply both sides by 2
Subtract 8 from both sides
SAMPLE
a) Use the rule [y = (x + 1) 2 4] to find the missing values.
∴ ± √ 49 = √ (x + 1) 2 Take the square root on both sides
∴ ± 7 = x + 1
∴ x + 1 = 7 OR x + 1 = − 7
∴ x = 6 OR x = − 8 Subtract 1 from each side
Check: If x = 6, then y = (6 + 1) 2 − 4 = (7) 2 − 4 = 49 − 4 = 45
If x = − 8, then y = ( − 8 + 1) 2 − 4 = ( 7) 2 − 4 = 49 − 4 = 45
b) Integers
Correct
Correct
Worked example 3: Input values, output values and rules
The flow diagram represents a function:
a) Show the input values and output values in a table.
b) Determine the rule for this function. Write the rule as an equation.
c) Use the rule to find the values of (i), (ii) and (iii):
a) Table:
For (iii): 24 = 3x − 6
x − 6 + 6
SAMPLE
b) 3 is added each time. The rule is y = 3x 6.
c) For (i): y = 3(− 1) − 6 = − 3 − 6 = − 9. For (ii): y = 3(2,5) − 6 = 7,5 − 6 = 1,5.
Add 6 on each side
Divide both sides by 3
Exercise 84: Input and output values
1. Answer the following questions:
1.1 If the input values are the integers from −2 to 2, find the output values using the rule y = − x 2 + 2x. Show your answers in a table.
1.2 If the input values are the rational numbers 1 3 ; − 5 6 ; 0 ; 2 3 and 1 6 , find the output values using the formula s = 6t − 3. Show your answers in a table.
1.3 If the input values are the integers from 2 to 2, find the output values using the equation y = 2 x−1 + 2x. Show your answers in a table.
2. In each case, use the given rule to find the missing values:
3. Compare your answers in questions 2.4 and 2.5. What do you notice? Explain this algebraically (write the two flow diagrams as algebraic equations and simplify them if possible).
4. Look at this flow diagram:
4.1 Determine the rule for this function.
4.2 Use the rule to find the values of (i), (ii) and (iii):
SAMPLE
2. E QUIVALENT FORMS
You already know that any function can be represented in several ways, including words, in a table, with a flow diagram, and through an equation or formula. These are called equivalent forms. Forms are equivalent if they produce equal output values for the same input values. For any given function, they must be able to:
• convert between equivalent forms. For example, if a function is represented by a flow diagram, you must be able to draw up a table of values, as well as to give the rule in the form of an equation.
• recognise equivalent forms. For example, if a function is represented by an equation, you must be able to pick out equivalent forms of this function from a variety of flow diagrams, tables or verbal descriptions. You must also be able to justify (explain) your choice. Justification is based on the fact that equal output values are produced for the same input values.
The worked examples and exercises in subtheme 1, as well as in Theme 5, gave you plenty of opportunity to master these skills.
Now we are going to add another equivalent form, namely graphs. Let us review a few important facts about graphs:
• Graphs consist of a number of points that are plotted on the Cartesian plane.
• Each point has an x-coordinate and a y-coordinate.
• These coordinates lie opposite values on the x-axis and the y-axis.
• The x-coordinates of the points on a graph can be viewed as the input values, and the y-coordinates as the output values.
Worked example 4: Equivalent forms
The given graph shows the relationship between the number of bags of lemons bought and the cost. (You also saw this graph in Theme 1):
a) Complete the following table so that it represents the same relationship as the graph:
Use the table in question a) to justify your choice.
c) Suggest a method other than used in question b) to determine if y = 6(x + 2) + 6x OR y = 6(x + 2) + 6x − 12 could be equivalent forms.
b) Which of the following equations are equivalent forms of the graph?
y = 6x
y = 12x
y = x 2
y = 6(x + 2) + 6x
y = 6(x + 2) + 6x − 12
Therefore, both y = 12x and y = 6(x + 2) + 6x − 12 are equivalent forms of the graph as they produce equal output values for the same input values as the graph.
c) We have seen in question b) that y = 12x is an equivalent form of the graph. Any equation that simplifies to y = 12x will therefore also be an equivalent form.
So, simplify the expressions on the RHS of these equations:
y = 6(x + 2) + 6x
∴ y = 6x + 12 + 6x Distributive rule
∴ y = 12x + 12 Add like terms
This does not simplify to y = 12x, and is therefore not an equivalent form.
y = 6(x + 2) + 6x − 12
∴ y = 6x + 12 + 6x − 12 Distributive rule
∴ y = 12x Add and subtract like terms
This simplifies to y = 12x, and is therefore an equivalent form.
Exercise 85: Equivalent forms
1. The given graph shows the relationship between the number of workers and how long it takes them to complete a task. (You also saw this graph in Theme 1):
Theme 13: Functions and relationships
1.1 Complete the following table so that it represents the same relationship as the graph:
1.2 Which of the following equations are equivalent forms of the graph?
Use the table in question 1.1 to justify your choice.
2. The graph represents a function. The coordinates of the points are not shown. You need to read them using the values on the x- and y-axes.
2.1 Use the graph to complete the table:
2.2 Find the rule that describes this function. Write it as an equation.
2.3 Which of the following equations will also describe this function? Use the table in question 2.1 to answer the question. y = 2
2.5 Solve the following equations using the graph. x + 2 = − 1 x − 2(x − 1) = 0 (x − 1) 2 − x(x − 1) + 1 = 3
2.6 Now solve the following equations algebraically:
2.4 Solve the following equations using your table in question 2.3.
3. The graph represents a function. The coordinates of the points are not shown. You need to read them using the values on the x- and y-axes.
3.1 Use the graph to complete the table:
3.2 Which of the following equations are equivalent forms of the graph?
y = x 2 − 4 y = (x − 2) 2
Use the table in question 3.1 to justify your choice.
4.1 Use the equation to complete the following table:
4.2 Now plot the points from your table onto the graph.
4.3 Use simplification to find if any of the following equations will be an equivalent form to
4.4 Use the completed table in question 4.1 to solve the following equations:
3.3 Use the graph to find the input value(s) if the output value is 12. 4. The function y =
2 + 2x + 4 is represented by the graph.
5. The function y = 2 x + 1 is represented by the graph.
5.1 Use the equation to complete the following table. Use a calculator where necessary:
5.2 Now plot the points from your table onto the graph.
5.3 Use the graph to solve the following equations:
5.3.1 2 x + 1 = 9
5.3.2 2 x = 1
Self-evaluation
Use the following scale to determine how comfortable you are with each topic in the table that follows:
1. Alarm! I don’t feel comfortable with the topic at all and need help.
2. Help! I don’t feel comfortable, but just need more time to work through the topic again.
3. OK! I feel moderately comfortable with the topic, but still struggle sometimes.
4. Sharp! I feel comfortable with the topic.
5. Party time! I feel totally comfortable with the topic and can even answer more complicated questions about it.
Complete the following table. Topic 1
I know the five different forms in which to express a function.
I can convert between any of these equivalent forms.
I can recognise equivalent forms.
I can justify equivalent forms.
I know what is meant by input values and output values.
I can calculate input values.
I can calculate output values.
I can identify the type of number used as input values.
Summary of theme
Now that you have completed this theme, you should know the following:
1. A function is a relationship between two variables.
2. One variable is the input variable (usually x) and the other variable is the output variable (usually y).
3. A function can be represented in a number of forms:
• verbally (in words)
• in a table
• as a flow diagram
• as an equation or formula
• using a graph.
4. If these forms produce equal output values for the same input values, they are called equivalent forms.
5. How to convert between equivalent forms.
6. How to recognise and justify equivalent forms.
End of theme exercise
1. If the input values of the function y = 1 2 x + 4 are − 6, − 1 and 1 2 , find the output values. Show your answers in a table.
2. In each case, use the given function rule and find the missing values.
3.2 Use the rule to find the values of (i), (ii) and (iii):
x − 1 2,5 (iii)
3. Look at this flow diagram:
SAMPLE
y (i) (ii) 24
4. The graph represents a function.
4.1 Use the graph to complete the table:
4.2 Find the rule that describes this function. Write it as an equation.
3.1 Determine the rule for this function.
4.3 Which of the following equations will also describe this function? Use the table in question 4.1 to answer the question. y = 1 2 (x − 2) y = 1 2 (x − 5) y
4.4 Use simplification to justify your answer in question 4.3.
4.5 Solve the following equations using:
4.5.1 your table in question 4.3.
4.5.2 the graph.
4.6 Now solve the equations in question 4.5 algebraically.
Patterns, functions and algebra (2)
THEME 14
ALGEBRAIC EXPRESSIONS
Introduction
In Theme 6 in the first term we revised and extended learners’ Grade 8 knowledge and skills about algebraic expressions. In this theme we will:
• revise the work covered in Theme 6 about expanding and simplifying algebraic expressions
• introduce factorising of algebraic expressions. This is the inverse (opposite) process to expanding algebraic expressions.
• use factorising to simplify calculations and algebraic fractions.
1. EXPANDING AND SIMPLIFYING ALGEBRAIC EXPRESSIONS
1.1 A dd and subtract algebraic expressions
Worked example 1: Add and subtract algebraic expressions
Simplify each of the following expressions:
SAMPLE
b) (12 x 2 − 0,5x + 2,3) + (3 x 2 + x − 3,7)
or subtract like terms
= 12 x 2 − 0,5x + 2,3 + 3 x 2 + x − 3,7 Remove brackets
= 15 x 2 + 0,5x − 1,4
c) (5 p 2 0,8p + 6) (12 p 2 + 1,2p 3)
Add or subtract like terms
= 5 p 2 − 0,8p + 6 − 12 p 2 − 1,2p + 3 Remove brackets = − 7 p 2 − 2p + 9
Add or subtract like terms
d) (18 a 3 12 a 2 + 5) + (7 a 2 13a + 8) (15 a 3 + 11 a 2 + 4a 15)
= 18 a 3 − 12 a 2 + 5 + 7 a 2 − 13a + 8 − 15 a 3 − 11 a 2 − 4a + 15
Remove brackets
= 3 a 3 16 a 2 17a + 28
REMEMBER
Add or subtract like terms
If the operation in front of brackets is addition (+), the signs of the terms inside the brackets remain the same when the brackets are removed.
If the operation in front of brackets is subtraction (−), the signs of the terms inside the brackets change when the brackets are removed.
Exercise 86: Add and subtract algebraic expressions
Simplify:
1.2 Mult iply and divide algebraic expressions
Simplify:
Worked example 2: Multiply and divide algebraic expressions
REMEMBER
SAMPLE
• When we multiply algebraic expressions, we
multiply the coefficients, and
add the exponents of identical bases.
• When we multiply a monomial (one term) and a binomial (two terms) or a trinomial (three terms), we apply the distributive rule a(b + c) = ab + ac
• When we multiply a binomial with another binomial, we multiply each term in the first set of brackets with each term in the second set of brackets. Remember this by using FOIL (First, outer, inner, last).
• When we square a binomial, the final answer can be obtained in a short way as follows:
◦ Square the first term [a 2]
◦ Multiply the first term and second term, and multiply the answer by 2 [a × b × 2 = 2ab]
◦ Square the second term [b 2]
• When we divide a monomial (one term) by another monomial, we:
◦ divide the coefficients, and
◦ subtract the exponents of identical bases.
• When we divide a binomial or a trinomial by a monomial, we divide each term of the numerator (above the line) by the denominator (below the line).
Solutions
b)
c)
2 x 2 y z 3(0,3 x 2 yz 1 4 x y 5 z 4)
= 2 x 2 y z 3 × 0,3 x 2 yz − 2 x 2 y z 3 × 1 4 x y 5 z 4
= 0,6 x 4 y 2 z 4 − 1 2 x 3 y 6 z 7
3a b 2(2ab 4a b 3 + a 3 b 5)
= 3a b 2 × 2ab − 3a b 2 × 4a b 3+ 3a b 2 × a 3 b 5
64 a 4 b 8 8 a 2 b 4 = 64 ÷ ( 8) × a 4−2 b 8−4 = 8 a 2 b 4 Divide coefficients, subtract exponents
27 x 7 y 8 9 x 8 y 5+ 3 x 4 y 5 3 x 4 y 5 = 27 x 7 y 8 3 x 4 y 5 9 x 8 y 5 3
Divide each term in numerator = 9 x 3 y 3 − 3 x 4 × 1 + 1 y 5 y 5 = 1 and 3 x 4 y 5 3 x 4 y 5 = 1 = 9 x 3 y 3 − 3 x 4 + 1 Simplify h) 2 (3x 1) 2 (2 3x) = 2(9 x 2 6x + 1) (2 3x) = 18 x 2 − 12x + 2 − 2 + 3x Remove brackets = 18 x 2 − 9x Add or subtract like terms
4(3x + 1)(3x 1) (x 2) 2 + 4x(2 3x) = 4(9 x 2 − 3x + 3x − 1) − (x 2 − 4x + 4) + 4x(2 − 3x) FOIL and square binomial = 36 x 2 − 12x + 12x − 4 − x 2 + 4x − 4 + 8x − 12 x 2 Distributive rule = 23 x 2 + 12x − 8 Add or subtract like terms
SAMPLE
Exercise 87: Multiply and divide algebraic expressions
Exercise 89: Substitution into algebraic expressions
SAMPLE
1. Find through substitution the value of the following expressions for the given values of the variables:
1.1 (2x − 1) 2 if x = − 1
= 3 and y = − 2
3 √ x 2 − 11y if x = 7 and y = 2 1.7 If x = 0 and y = 4, determine 2xy − 3 y 2 1.8 If x = 1 2 and y = 2
2. FACTORISING ALGEBRAIC EXPRESSIONS
What is a factor?
The factors of a number are numbers we multiply to get the number. For example, 2 and 3 are factors of 6, because 2 × 3 = 6.
We can also think of a factor of a number as a number that we can divide into that number without leaving a remainder. For example, 2 is a factor of 6, because 6 ÷ 2 = 3.
What is a product?
This is the answer when two or more factors are multiplied. For example, in 2 × 3 = 6, the product is 6.
Algebraic expressions can also have factors. For example:
• In a × a = a 2, the factors are a and a and the product is a 2 .
• In 2(3x + 1) = 6x + 2, the factors are 2 and 3x + 1 and the product is 6x + 2.
NOTE
• Factorising a number or an algebraic expression means finding the factors of that number or expression.
• Factorising is the inverse (or opposite) process of multiplication.
• There are various ways of factorising algebraic expressions (which will be covered in Grade 9):
◦ common factor
◦ difference of squares
◦ trinomial.
2.1 C ommon factor
A common factor is a number (or expression) that can be divided into two or more different numbers (or expressions) without leaving a remainder. Numbers (and expressions) often share more than one common factor.
For example, 2 is a common factor of 4 and 12.
However, 4 is also a common factor of 4 and 12.
In this case 4 is the highest common factor (HCF).
x is a common factor of x 2 and x 3 .
However, x 2 is also a common factor of x 2 and x 3 .
In this case x 2 is the highest common factor.
To factorise using a common factor of an algebraic expression such as 4 x 2 + 12 x 3 :
• Determine the HCF of the terms (4 x 2 in this case).
SAMPLE
• The second factor is obtained by dividing the common factor into each term of the original algebraic expression. This gives 1 + 3x . Therefore, 4 x 2 + 12 x 3 is factorised as 4 x 2(1 + 3x). Note that the second factor is written in brackets.
• So, the factors of 4 x 2 + 12 x 3 are 4 x 2 and 1 + 3x. Therefore, if we multiply 4 x 2 and 1 + 3x, we will obtain the original expression, 4 x 2 + 12 x 3 .
Worked example 5: Common factor
Factorise: a) 10a 5b b) 3 x 3 + 6xy 12 x 2
Solutions a) 10a 5b = 2.5 . a − 5 . b Determine HCF of the terms (5) = 5(2a − b) Write common factor outside brackets, remaining factor inside brackets b) 3 x 3 + 6xy 12 x 2 = 3 . x . x . x + 2.3 . x . y − 3.4 . x . x Determine HCF of the terms (3x) = 3 . x(x . x + 2 . y − 4 . x) Write common factor outside brackets, remaining factor inside brackets = 3x( x 2 + 2y − 4x) Final, simplified form
REMEMBER
Always check your answer by multiplying the factors to see if you get the original expression, that is the given product. This should become a habit.
3x(x 2 + 2y − 4x) = 3 x 3 + 6xy − 12 x 2 original expression
Factorise:
a)
b)
Worked example 6: More difficult common factors
3x(a b) + 5(a b)
3x(a b) + 5(b a)
c) ax − bx + 2a − 2b
Solutions
a) There are two terms in the expression, namely 3x(a b) and 5(a b).
The common factor is (a − b).
So, 3x(a − b) + 5(a − b)
= (a − b)(3x + 5) Remove common factor (a b) and write other factor (3x + 5) inside brackets
b) a b and b a are not the same. However, we can rewrite b a as (a b) (check this for yourself).
So, 3x(a − b) + 5(b − a) = 3x(a − b) + 5[− (a − b)] Rewrite b a as (a b) = 3x(a − b) − 5(a − b) +
= (a − b)(3x − 5)
SAMPLE
Remove common factor (a b) and write other factor (3x + 5) inside brackets
c) ax bx + 2a 2b = (ax − bx) + (2a − 2b) Group terms in pairs, each of which has a common factor = x(a − b) + 2(a − b) Remove common factor from each pair = (a − b)(x + 2) Remove common factor (a b)
Exercise 90: Common factor
1. Copy and complete: 1.1 3x + 12 = 3(.........) 1.2 2ab − 7a = a(.........) 1.3 8 p 2 + 4p = 4p(.........) 1.4 20 x 4 y + 15 x 3 y 2 = .........(4x + 3y)
2. Factorise: 2.1 3x + 6y 2.2 ax + ay 2.3 2ax + 4bx 2.4 8 x 2 − 12 x 3 + 4x
3. Factorise:
3.1
4. If a = 7 and b = 3, use the easiest way to find the value of:
• An expression such as a 2 b 2 is known as the difference of two squares.
Calculate the following, using the easiest method:
Dif ference of two squares Look at the expansion of (
SAMPLE
• The factors of the difference of two squares are two binomials, such as a b and a + b.
• To find the factors of the difference of squares:
◦ write two sets of brackets with opposite signs: (… – …)(… + ...)
◦ write inside the brackets:
Always multiply out the two factors to check if you get the original expression.
Worked example 7: Difference of two squares
Factorise:
a) a 2 − 16 b) 25 a 2 9 b 6
c) 16 (x y) 2 1
d) (2x − 3y) 2 − (5x + y) 2
Solutions
a) a 2 − 16
Recognise difference of two squares
= (a − 4)(a + 4) Factors are two nearly identical binomials; only signs differ
b) 25 a 2 9 b 6
Recognise difference of two squares
= (5a − 3 b 3)(5a + 3 b 3) Factors are two nearly identical binomials; only signs differ
c) 16 (x y) 2 1
Recognise difference of two squares
= [4(x − y) − 1][4(x − y) + 1] Factors are two nearly identical binomials
= (4x − 4y − 1)(4x − 4y + 1)
d)
(2x 3y) 2 (5x + y) 2
Remove inner brackets
Recognise difference of two squares
= [(2x − 3y) − (5x + y)][(2x − 3y) + (5x + y)] Two factors; note use of brackets
