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Reg. No.: 2011/011959/07
Facilitator's Guide 2/2 − Grade 9
CAPS-aligned
Prof. C Vermeulen, Lead author
H Otto M Sherman
Patterns, functions and algebra (2)
THEME 13
FUNCTIONS AND RELATIONSHIPS
CAPS learning requirements
Learners should know and be able to apply the following:
• Determine input values, output values or rules for patterns and relationships using:
◦ flow diagrams
◦ tables
◦ formulae
◦ equations.
• Determine, interpret and justify equivalence of different descriptions of the same relationship or rule presented:
◦ verbally
◦ in flow diagrams
◦ in tables
◦ by formulae
◦ by equations
◦ by graphs on a Cartesian plane.
Term 3
Duration 1 week (5 hours)
Paper 1
Facilitator
tips
• This theme deals with content covered in Grades 7 and 8, as well as Theme 5 of this book. Progression should be in terms of complexity and context. Therefore, types of numbers used as input and output values should include natural numbers, integers and rational numbers.
SAMPLE
Weight November exam: 5 ± 2 marks out of 75.
• Learners must be able to convert between any form in which a function can be represented, and they must be able to identify equivalent forms in which a function is expressed. This has largely been covered in Themes 4 and 5 already.
In this theme we include graphs as a form in which functions can be represented.
• Embedded in this theme we also provide ample opportunities for learners to substitute values into formulae, which will feature again in Theme 16. This is an important skill, and learners must master this.
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.
• 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
◦ as a flow diagram
◦ as an equation or formula.
In subtheme 2 we will extend this list to include graphs.
• 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
The flow diagram represents a function:
a) Write the rule in words.
b) Give the rule as an equation.
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) 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.
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
So, x = 16
SAMPLE
Subtract 4 8 × 2 = 16
i) For i): replace x by 0:
ii) For ii): replace x by 6:
iii) For iii): replace y by 12: 12 = x 2 + 4
Multiply by 2
Check the answer by applying the flow diagram rule to
This is an equation that we must solve to find the value of x:
24 = x + 8
Multiply both sides by 2
∴ 16 = x Subtract 8 from both sides ∴ x = 16 d) x 8
e) Integers
Worked example 2: Input values, output values and rules
The table represents a function: x − 4 − 1 3 4 (iii)
y = (x + 1) 2 − 4 5 (i) (ii) 17 45
a) Use the rule [y = (x + 1) 2 − 4] 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).
The answers are the same.
For 2.4: y = − 4x + 4
For 2.5: y = (x − 1)( 4) = − 4x + 4
So, the two rules are identical.
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
Learners 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, they 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, they must be able to pick out equivalent forms of this function from a variety of flow diagrams, tables or verbal descriptions. They must also be able to justify (explain) their 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 them 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. (Learners also saw this graph in Theme 1):
a) Complete the following table so that it represents the same relationship as the graph:
c) Suggest a method other than used in question b) to determine if
could be equivalent forms.
b) Which of the following equations are equivalent forms of the graph?
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. (Learners also saw this graph in Theme 1):
1.1 Complete the following table so that it represents the same relationship as the graph: x 3 4 6 8 12 16 y 16 12 8 6 4 3
1.2 Which of the following equations are equivalent forms of the graph? y
Use the table in question 1.1 to justify your choice.
Therefore, y = 48 x is an equivalent form of the graph.
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: x 1 0 1 2 3 4 y 3 2 1 0 1 2
2.2 Find the rule that describes this function. Write it as an equation.
y = − x + 2
2.3 Which of the following equations will also describe this function? Use the table in question 2.1 to answer the question.
y = 2x − (x − 2)
− 1)
2.4 Solve the following equations using your table in question 2.3.
To solve, find the x-value that produces the given output value ( 1; 1; 0 and 3): x + 2 = − 1: x = 3 2x − (x − 2) = 1: x = − 1 x − 2(x − 1) = 0: x = 2 (x − 1) 2 − x(x − 1) + 1 = 3: x = − 1
Therefore, both
SAMPLE
2.5 Solve the following equations using the graph.
To solve, read from the graph the x-coordinate of the point with y-coordinate 1, 0 and 3: − x + 2 = − 1: x = 3 x − 2(x − 1) = 0: x = 2
(x − 1) 2 − x(x − 1) + 1 = 3: x = − 1
2.6 Now solve the following equations algebraically:
2.6.1 2x − (x − 2) = 1
∴ 2x − x + 2 = 1
∴ x = − 1
2.6.2 (x − 1) 2 − x(x − 1) + 1 = 3
∴ x 2 − 2x + 1 − x 2 + x + 1 = 3
∴ − x = 1
∴ x = − 1
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
y = (x + 2) 2
y = (x − 2)(x + 2)
Use the table in question 3.1 to justify your choice.
SAMPLE
Therefore, both y = x 2 − 4 and y = (x − 2)(x + 2) are equivalent forms of the graph, and therefore also represent this function.
3.3 Use the graph to find the input value(s) if the output value is 12. x = − 4 and x = 4
4. The function y = x 2 + 2x + 4 is represented by the graph.
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 y = − x 2 + 2x + 4: y = − (x 2 + 2x + 4) = − x 2 − 2x − 4
4.4 Use the completed table in question 4.1 to solve the following equations:
4.4.1 x 2 + 2x + 4 = − 4: x = − 2 or x = 4
4.4.2 x 2 + 2x = 1
∴ − x 2 + 2x + 4 = 1 + 4 Add 4 to each side to get the original expression
∴ − x 2 + 2x + 4 = 5
∴ x = 1 y = 5 when x = 1, at the point (1; 5)
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: x 0 1 1,5 3 3,75 y 2 3 3,828 9
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 : x = 3
2 x = 1
Self-evaluation
Learners use the following scale to determine how comfortable they are with each topic in the table that follows:
1. Alarm! Learners don’t feel comfortable with the topic at all and need help.
2. Help! Learners don’t feel comfortable, but just need more time to work through the topic again.
3. OK! Learners feel moderately comfortable with the topic, but still struggle sometimes.
4. Sharp! Learners feel comfortable with the topic.
5. Party time! Learners feel totally comfortable with the topic and can even answer more complicated questions about it.
Learners complete the following table.
Learners know the five different forms in which to express a function.
Learners can convert between any of these equivalent forms.
Learners can recognise equivalent forms.
Learners can justify equivalent forms.
Learners know what is meant by input values and output values.
Learners can calculate input values.
Learners can calculate output values.
Learners can identify the type of number used as input values.
Summary of theme
Now that we have completed this theme, learners 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
output values. Show your answers in a table.
2. In each case, use the given function rule and find the missing values.
3. Look at this flow diagram:
3.1 Determine the rule for this function.
3.2 Use the rule to find the values of (i), (ii) and (iii):
1 2,5 (iii) 56
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.
4.3 Which of the following equations will also describe this function? Use the table in question 4.1 to answer the question.
Therefore, both y = 1 2 (x − 5) and y = x − 1 2 (x + 1) − 2 describe the function.
4.4 Use simplification to justify your answer in question 4.3.
Therefore, both simplify to y =
4.5 Solve the following equations using:
4.5.1 your table in question 4.3. 1 2 x − 2 1 2 = − 1
question 4.2.
Find the x-value that corresponds with a y-value of 1. That is x = 3. 1 2 (x − 5) = 1
Find the x-value that corresponds with a y-value of 1. That is x = 7.
4.5.2 the graph. 1 2 x − 2 1 2 = − 1
Find the x-coordinate of a point with y-coordinate of 1. That is x = 3. 1 2 (x − 5) = 1
Find the x-coordinate of a point with y-coordinate of 1. That is x = 7.
4.6 Now solve the equations in question 4.5 algebraically.
2 x − 2 1 2 = − 1
1 2 x − 5 2 = − 1 Multiply by 2 ∴ x − 5 = − 2 ∴ x = 3
SAMPLE
1 2 (x − 5) = 1 ∴ x − 5 = 2 Multiply by 2 ∴ x = 7
Patterns, functions and algebra (2)
THEME 14
ALGEBRAIC EXPRESSIONS
CAPS learning requirements
Learners should know and be able to apply the following:
1. Algebraic language
As for Theme 6.
2. Expand and simplify algebraic expressions
As for Theme 6.
3. Factorise algebraic expressions that involve:
• common factors
• difference of two squares
• trinomials of the form:
◦ x 2 + bx + c
◦ a x 2 + bx + c, where a is a common factor.
• Simplify algebraic expressions that involve the above factorisation processes.
• Simplify algebraic fractions using factorisation.
Term 3
Duration 2 weeks (9 hours)
Paper 1
SAMPLE
Weight November exam: 5 ± 2 marks out of 75.
Facilitator tips
• The first part of this theme revises the content covered in Theme 6. As in Theme 6, this is fundamentally important work, and sufficient time should be spent on this.
• Factorising algebraic expressions is introduced in this theme as new content. Learners need to thoroughly master the skill of factorisation, as it has many applications, including solving equations in Theme 15.
• It is important that learners realise that factorisation and expansion (removing brackets through multiplication) are two inverse (opposite) processes.
• Encourage learners to always check their factorisation answers by multiplying out the factors to ensure that the original expression (the product) is obtained.
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:
a)
b) (12 x 2 0,5x + 2,3) + (3 x 2 + x 3,7)
c) (5 p 2 − 0,8p + 6) − (12 p 2 + 1,2p − 3)
d) (18 a 3 12 a 2 + 5) + (7 a 2 13a + 8) (15 a 3 + 11 a 2 + 4a 15)
Solutions
a)
= 12 a 2 − 7
= 5 a 2 + 9a b 2 + 8 a 2 b − 2b
b) (12 x 2 0,5x + 2,3) + (3 x 2 + x 3,7)
Add 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)
= 5 p 2 − 0,8p + 6 − 12 p 2 − 1,2p + 3
= − 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
SAMPLE
Remove brackets
Add or subtract like terms
REMEMBER
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.
Simplify:
Exercise 86: Add and subtract algebraic expressions
+ 3) = 7 a 3 + 6 a 2 − 3a − 5 + 3 a 3 − 5 a 2 − 4a + 3 = 10 a 3 + a 2 − 7a − 2 4. (4 b 3 3 b 2 8b 3) (7 b 3 8 b 2 + 9b 5) = 4 b 3 − 3 b 2 − 8b − 3 − 7 b 3 + 8 b 2 − 9b + 5
= − 3 b 3 + 5 b 2 − 17b + 2
5. (x 2 + 2x 3) (3 x 2 x 4) + (2 x 2 6x 1)
= x 2 + 2x − 3 − 3 x 2 + x + 4 + 2 x 2 − 6x − 1
= − 3x
1.2 Mult iply and divide algebraic expressions
Worked example 2: Multiply and divide algebraic expressions
Simplify: a)
3 a 2 b c 3 × 0,5 a 3 b × ( − 2a b 3 c 4)
b) 2 x 2 y z 3(0,3 x 2 yz 1 4 x y 5 z 4)
c) 3a b 2(2ab 4a b 3 + a 3 b 5) d)
a 4 b 8
a 2 b 4
x 7 y 8 − 9 x 8 y 5+ 3 x 4 y 5 3 x 4 y 5
2 (3x − 1) 2 − (2 − 3x)
x(x − 4)(2x + 3) + x (2x − 3) 2 4x
REMEMBER
• When we multiply algebraic expressions, we
SAMPLE
◦ 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
2) × a 2 + 3 + 1 × b 1 + 1 + 3 × c 3 + 4
3 a 6 b 5 c 7 b) 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 Distributive rule
