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KS3 Mathematics A5 Functions and graphs

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1 KS3 Mathematics A5 Functions and graphs
The aim of this unit is to teach pupils to: Express functions and represent mappings Generate points and plot graphs of functions Material in this unit is linked to the Key Stage 3 Framework supplement of examples pp A5 Functions and graphs

2 A5 Functions and graphs Contents A A5.1 Function machines A
A5.2 Tables and mapping diagrams A A5.3 Finding functions A A5.4 Inverse functions A A5.5 Graphs of functions

3 Finding outputs given inputs
Start with one machine to demonstrate to pupils how the function machine works. Explain that a number is fed into the machine (this is the input), the machine performs an operation (or series of operations) on the number to produce the output. Change the number of machines and use the function editor to create a given chain of functions. Hide the output before pressing the orange button. Change the input and ask pupils to find the output. If required you can show the intermediate steps.

4 Introducing functions
A function is a rule which maps one number, sometimes called the input or x, onto another number, sometimes called the output or y. A function can be illustrated using a function diagram to show the operations performed on the input. x y × 3 + 2 A function can be written as an equation. For example, y = 3x + 2. Ask pupils if they know what a function is and if they know the difference between a formula and a function (these two words are sometimes confused). Formulae and functions can both be written as equations using an equals sign. Functions can also be written using a mapping arrow, for example, x  3x + 2. A function is a rule which maps one number, sometimes called the input, onto another number, sometimes called the output, using a sequence of operations. A formula, on the other hand, can link more than two quantities, for example A = lw. There does not need to be a unique output for a given input. Also, formulae relate to real-life or practical contexts while functions usually relate to abstract inputs, outputs and operations. Discuss the function y = 3x + 2 in terms of inputs and outputs. The function y = 3x + 2 means that the input, x, is multiplied by 3 and then 2 is added on to get the output, y. Ask pupils, to find y for different values of x. Then ask: If y is 38, what is x? Explain that if we are given the output and asked to find the input then we have to use inverse operations. Ask: What is the inverse of times 3 and add 2? Tell pupils that we must not only invert the operations but we must also reverse the order. Establish that the inverse of times 3 and add 2 is, subtract 2 and divide by 3. Performing these inverse operation in reverse order, we have x = 12 when y = 38 (because (38 – 2) ÷ 3 = 12). A function can can also be be written with a mapping arrow. For example, x  3x + 2.

5 Writing functions using algebra
Explain that for this function machine we input x. This will give us the output in terms of x. We can then write the function as an equation or with a mapping arrow. Explain the difference between these two forms of writing a function. Hide the output and ask pupils how given functions can be written algebraically. Show the intermediate steps if required.

6 Ordering machines Is there any difference between x y × 2 + 1 and x y
? Show that these machines are different by using the same input in both machines and comparing the outputs. The first function can be written as y = 2x + 1. The second function can be written as y = 2(x + 1) or 2x + 2.

7 Equivalent functions Explain why x y + 1 × 2 is equivalent to x y × 2
+ 2 When an addition is followed by a multiplication; the number that is added is also multiplied. Ask pupils to explain in their own words why these two functions are equivalent. This is also true when a subtraction is followed by a multiplication.

8 Ordering machines Is there any difference between x y ÷ 2 + 4 and x y
? The first function can be written as y = x 2 Show that these machines are different by using the same input in both machines and comparing the outputs. The second function can be written as y = or y = x 2 x + 4

9 Equivalent functions Explain why x y + 4 ÷ 2 is equivalent to x y ÷ 2
+ 2 Ask pupils to explain in their own words why these two functions are equivalent. When an addition is followed by a division then the number that is added is also divided. This is also true when a subtraction is followed by a division.

10 Equivalent function match
Use this activity, to match equivalent functions. Level 1, matches functions with two operations to their single operation equivalent. Level 2 is much more challenging and matches functions that add (or subtract) first and multiply (and divide) second with the equivalent function that multiplies (or divides) first and adds (or subtracts) second. After some practice pupils will notice that the operation that multiplies or divides will remain unchanged in the equivalent function. The addition or subtraction, when it appears before the multiplication or division, will be multiplied or divided in the equivalent function. Show that two functions are equivalent by using the same input in both functions and observing that the same output is produced. These equivalent functions can be found by writing the function algebraically. For example, 2(x + 1) is equivalent to 2x + 2 (by multiplying out the bracket). This means that ‘add 1 and multiply by 2’ is equivalent to ‘multiply by 2 and add 2’. 2x + 2 can be shown to be equivalent to 2(x + 1) by factorizing. Also, x + 6/2 is equivalent to x/2 + 3 (by dividing both terms in the numerator by the denominator). This means that ‘add 6 and divide by 2’ is equivalent to ‘divide by 2 and add 3’. x/2 + 3 can be shown to be equivalent to x + 6/2 by writing both terms over a common denominator, 2.

11 A5.2 Tables and mapping diagrams
Contents A5 Functions and graphs A A5.1 Function machines A A5.2 Tables and mapping diagrams A A5.3 Finding functions A A5.4 Inverse functions A A5.5 Graphs of functions

12 Using a table We can use a table to record the inputs and outputs of a function. We can show the function y = 2x + 5 as x y × 2 + 5 3 3, 1, 6, 4, 1.5 3, 1, 6, 4 3, 1, 6 3, 1 11, 7, 17, 13 11, 7, 17 11, 7, 17, 13, 8 11 11, 7 and the corresponding table as: x y 3 3 1 1 6 6 4 4 1.5 1.5 11 11 7 7 17 17 13 13 8

13 Using a table with ordered values
It is often useful to enter inputs into a table in numerical order. We can show the function y = 3(x + 1) as x y + 1 × 3 1 1, 2, 3, 4, 5 1, 2, 3, 4 1, 2, 3 1, 2 6, 9 6, 9, 12, 15 6 6, 9, 12 6, 9, 12, 15, 18 When the inputs are ordered and the corresponding table as: Emphasize that when the inputs of a function are entered in numerical order, the outputs form a sequence. Point out that the nth term of this sequence is 3(n + 1) (or 3n + 3). This can be useful when we are given inputs and outputs and need to find the function. Although there are similarities between sequences and functions, there is a crucial difference between n, the term number in a sequence and x, the input of a function. That is that n can only be a whole positive number, whereas x can be any number, positive, negative or fractional. x y 1 1 2 2 3 3 4 4 5 5 the outputs form a sequence. 6 6 9 9 12 12 15 15 18

14 Recording inputs and outputs in a table
Use this activity to demonstrate how functions can be entered into a table. Ask pupils: How could we show this function as a graph? Establish that each value we have chosen for x and the corresponding value for y, form a coordinate pair. Plotting these coordinates would give us a graph of the function.

15 Mapping diagrams We can show functions using mapping diagrams.
For example, we can draw a mapping diagram of x  2x + 1. Inputs along the top can be mapped to outputs along the bottom. 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1

16 Mapping diagrams of x x + c
What happens when we draw the mapping diagram for a function of the form x  x + c, such as x  x + 1, x  x + 2 or x  x + 3? x  x + 2 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 The lines are parallel.

17 Mapping diagrams of x mx
What happens when we draw the mapping diagram for a function of the form x  mx, such as x  2x, x  3x or x  4x, and we project the mapping arrows backwards? For example: x  2x 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 The lines meet at a point on the zero line.

18 The identity function The function x  x is called the identity function. The identity function maps any given number onto itself. We can show this in a mapping diagram. x  x 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 Every number is mapped onto itself.

19 A5 Functions and graphs Contents A A5.1 Function machines A
A5.2 Tables and mapping diagrams A A5.3 Finding functions A A5.4 Inverse functions A A5.5 Graphs of functions

20 Finding functions given inputs and outputs
Hide the functions then adjust them. By entering different inputs and recording the corresponding output, challenge pupils to find the function. Show the operation in one of the machines and challenge pupils to find the other. Notice that by entering the inputs in numerical order, starting with one, the outputs form a linear sequence. Remind pupils how to find the nth term of a linear sequence and use the rule for the nth term to find the function. Another way to find the function would be to plot a graph of the points in the table and use the value of the gradient, m, and the y-intercept, c, to write the function in the form y = mx + c. Point out that the same function can often be expressed in different ways. For example, 2(x – 2) is equivalent to 2x – 4. This means that ‘subtract 2 and multiply by 2’ is equivalent to ‘multiply by 2 and subtract 4’. Also x + 6/3 is equivalent to x/ This means that ‘add 6 and divide by 3’ is equivalent to ‘divide by 3 and add 2’. Link: A4 Sequences – finding the nth term.

21 A5 Functions and graphs Contents A A5.1 Function machines A
A5.2 Tables and mapping diagrams A A5.3 Finding functions A A5.4 Inverse functions A A5.5 Graphs of functions

22 Think of a number This activity shows how ‘Think of a number’ problems can be solved using inverse operations in reverse order.

23 Finding inputs given outputs
Suppose x 1 + 3 ÷ 8 How can we find the value of x? To find the value of x we start with the output and we perform the inverse operations in reverse order. – 3 × 8 5 1 x = 5

24 Finding inputs given outputs
Find the value of x for the following: x – 1 × 3 – 7 ÷ 3 + 7 2 – 1 x = 2 x 4 – 2 ÷ 5 + 6 Encourage pupils to check the solutions by feeding them back into the original function. The numbers used in these examples can be edited to make the problems more challenging. + 2 × 5 – 6 –8 4 x = –8

25 Finding inputs given outputs
Find the value of x for the following: x 24 × 5 – 11 ÷ 5 + 11 7 24 x = 7 x 4 – 6 × 4 + 9 Again, check the solutions by feeding them back into the original function. + 6 ÷ 4 – 9 4.75 4 x = 4.75

26 Finding the inverse function
We can write x  3x + 5 as x 3x + 5 × 3 + 5 To find the inverse of x  3x + 5 we start with x and we perform the inverse operations in reverse order. x – 5 3 ÷ 3 – 5 x x – 5 3 The inverse of x  3x is x 

27 Finding the inverse function
We can write x  x/4 + 1 as x + 1 ÷ 4 4 and we perform the inverse operations in reverse order. To find the inverse of x  x/4 + 1 we start with x × 4 – 1 4(x – 1) x The inverse of x  is x  4(x – 1) + 1 x 4

28 Finding the inverse function
We can write x  3 – 2x as x –2x + 3 × –2 + 3 (= 3 – 2x) To find the inverse of x  3 – 2x we start with x and we perform the inverse operations in reverse order. 3 – x 2 x – 3 –2 ÷ –2 – 3 = x Point out that although 3 – 2x can be written as –2x + 3, it is usually preferable to rearrange expressions so that they do not start with a – sign. Emphasize that we can rewrite x – 3/– 2 as 3 – x/2 by multiplying the numerator and the denominator by –1. x – 3/– 2 is equivalent to 3 – x/2 but the latter is preferable because it does not have a negative integer as a denominator. 3 – x 2 The inverse of x  3 – 2x is x 

29 Functions and inverses
Use this activity to practise matching functions with their inverses.

30 A5 Functions and graphs Contents A A5.1 Function machines A
A5.2 Tables and mapping diagrams A A5.3 Finding functions A A5.4 Inverse functions A A5.5 Graphs of functions

31 (3, 5) (6, 2) (3, 5) Coordinate pairs
When we write a coordinate, for example, (3, 5) (6, 2) (3, 5) x-coordinate y-coordinate the first number is called the x-coordinate and the second number is called the y-coordinate. the first number is called the x-coordinate and the second number is called the y-coordinate. Link: S4 Coordinates and transformations 1 – coordinates. Together, the x-coordinate and the y-coordinate are called a coordinate pair.

32 Graphs parallel to the y-axis
What do these coordinate pairs have in common? (2, 3), (2, 1), (2, –2), (2, 4), (2, 0) and (2, –3)? The x-coordinate in each pair is equal to 2. Look what happens when these points are plotted on a graph. x y All of the points lie on a straight line parallel to the y-axis. Emphasize that as long as the x-coordinate is 2 the y-coordinate can be any number: positive, negative or decimal. Encourage pupils to be imaginative in their choice of points that lie on this line. For example, (2, ) (2, 43/78) or (2, – ). Name five other points that will lie on this line. O This line is called x = 2. x = 2

33 Graphs parallel to the y-axis
All graphs of the form x = c, where c is any number, will be parallel to the y-axis and will cut the x-axis at the point (c, 0). x y Conclude that the graph of x = ‘something’ will always be parallel to the y-axis. In other words, it will always be vertical (not horizontal like the x-axis). For each graph shown in the example, ask pupils to tell you the coordinate of the point where the line cuts the x-axis. Ask pupils to tell you the equation of the line that coincides with the y-axis (x = 0). O x = –10 x = –3 x = 4 x = 9

34 Graphs parallel to the x-axis
What do these coordinate pairs have in common? (0, 1), (3, 1), (–2, 1), (2, 1), (1, 1) and (–3, 1)? The y-coordinate in each pair is equal to 1. Look what happens when these points are plotted on a graph. All of the points lie on a straight line parallel to the x-axis. x y Emphasize that as long as the y-coordinate is 1 the x-coordinate can be any number: positive negative or decimal. Encourage pupils to be imaginative in their choice of points that lie on this line. For example, ( , 1) (56/87,,1) or (– , 1). y = 1 Name five other points that will lie on this line. O This line is called y = 1.

35 Graphs parallel to the x-axis
All graphs of the form y = c, where c is any number, will be parallel to the x-axis and will cut the y-axis at the point (0, c). x y y = 5 y = 3 Conclude that the graph of y = ‘something’ will always be parallel to the x-axis. In other words, it will always be horizontal (not vertical like the y-axis). For each graph shown in the example, ask pupils to tell you the coordinate of the point where the line cuts the y-axis. Ask pupils to tell you the equation of the line that coincides with the x-axis (y = 0). O y = –2 y = –5

36 Drawing graphs of functions
The x-coordinate and the y-coordinate in a coordinate pair can be linked by a function. What do these coordinate pairs have in common? (1, 3), (4, 6), (–2, 0), (0, 2), (–1, 1) and (3.5, 5.5)? In each pair, the y-coordinate is 2 more than the x-coordinate. These coordinates are linked by the function: Ask pupils if they can visualize the shape that the graph will have. This might be easier if they consider the points (0, 2), (1, 3), (2, 4) (3, 5) etc. Establish that the points will lie on a straight diagonal line. Point out that the graphs of all linear functions are straight lines. A function is linear if the variables are not raised to any power (other than 1). The word linear itself contains the word line which may help pupils remember. Ask pupils to suggest the coordinates of any other points that will lie on this line. Praise the most imaginative answers. y = x + 2 We can draw a graph of the function y = x + 2 by plotting points that obey this function.

37 Drawing graphs of functions
Given a function, we can find coordinate points that obey the function by constructing a table of values. Suppose we want to plot points that obey the function y = x + 3 We can use a table as follows: x y = x + 3 –3 –2 –1 1 2 3 Explain that when we construct a table of values, the value of y depends on the value of x. That means that we choose the values for x and substitute them into the equation to get the corresponding value for y. The minimum number of points needed to draw a straight line is two, however, it is best to plot several points to ensure that no mistakes have been made. The points given by the table can then be plotted to give the graph of the given function. 1 2 3 4 5 6 (–3, 0) (–2, 1) (–1, 2) (0, 3) (1, 4) (2, 5) (3, 6)

38 Drawing graphs of functions
To draw a graph of y = x – 2: y x 1) Complete a table of values: y = x – 2 x y = x – 2 –3 –2 –1 1 2 3 –5 –4 –3 –2 –1 1 O 2) Plot the points on a coordinate grid. 3) Draw a line through the points. This slide summarizes the steps required to plot a graph using a table of values. 4) Label the line. 5) Check that other points on the line fit the rule.

39 Drawing graphs of functions
Start by choosing a simple function. Remind pupils that we can draw graphs of functions by plotting inputs along the x-axis against outputs along the y-axis. Talk through the substitution of each value of x in the table and click to reveal the corresponding value of y below it. Start with the positive x-values, if required and work backwards along the table to include the negative values. Explain that each pair of values for x and y corresponds to a coordinate that we can plot on the coordinate axis. For example, for the equation y = 2x, when x = 1, y = 2. This corresponds to the coordinate (1, 2). Click to plot each coordinate from the table of values onto the graph. Remind pupils that we always move along the x-axis and then up (or down) the y-axis when plotting coordinate points. A common mnemonic for this is ‘along the corridor and up the stairs’. Once all the points have been plotted ask pupils what they notice? (That is that all the points lie in a straight line.) Click ‘show line’ to draw a line through the points. Draw pupils attention to the fact that the line extends beyond either end of the points plotted on the graph. Use the crosshair button to find the coordinates of other points on the line. Verify that all of these points satisfy the equation. Ask pupils to suggest other coordinates that would lie on this line. Establish that the line could be infinitely long and praise the most imaginative correct answers.

40 The equation of a straight line
The general equation of a straight line can be written as: y = mx + c The value of m tells us the gradient of the line. The value of c tells us where the line crosses the y-axis. Explain that the equation of a line can always be arranged to be in the form y = mx + c. It is often useful to have the equation of a line in this form because it tells us the gradient of the line and where it cuts the x-axis. These two facts alone can enable us to draw the line without have to set up a table of values. Ask pupils: What they can deduce about two graphs that have the same value for m? Establish that if they have the same value for m, they will have the same gradient and will therefore be parallel. This is called the y-intercept and it has the coordinate (0, c). For example, the line y = 3x + 4 has a gradient of 3 and crosses the y-axis at the point (0, 4).

41 Linear graphs with positive gradients
Use this activity to explore the effect of changing the value of the coefficient of x (m) and the value of the number that is added on (c). Start by keeping c set to 0 to explore graphs of the form y = mx. Establish that these graphs of this form: Are straight lines. Pass through the origin. Vary in steepness depending on the value of m. Ask pupils if they can explain why graphs of the form y = mx pass through the origin. Establish that at the origin the x-coordinate is equal to 0. Whatever we multiply 0 by, the answer is always 0. Therefore if x is 0, y is 0, regardless of the value of m, and the point (0, 0) is the origin. Next, explore graphs of the form y = x + c by keeping m set to 1 and changing the value of c. Establish that graphs of the form y = x + c: Have the same slope as y = x and so are parallel. Cut (or intersect) the y-axis at c, that is the point (0, c).

42 Investigating straight-line graphs
Remind pupils that the the equation of a graph tells us the relationship between the x-coordinate and the y-coordinate of every point on that line. If the x- and y-coordinates of a point do not satisfy the equation of the line then the point cannot lie on the line. Also, the equation of a line represents an infinite set of points. A function is linear if the variables are not raised to any power other than 1. Investigating the gradient Ask a volunteer to drag the line on the board so that it has a gradient of 2. Point out that this means that for every square we move along the line we move two squares up. Add another line and ask another volunteer to drag the line to a different position and drag the end points to again produce a line with a gradient of 2. Ask pupils what they notice about the lines. Establish that lines with the same gradient are parallel. Look at the equation of each line. Ask pupils what they notice about the equations. Establish that the multiplier (the coefficient) of x is 2 for each line. The constant value that is added on in each case varies depending on where the line meets the vertical axis (the y-axis). All of the lines are of the form y = 2x + c, where c indicates where the line cuts the y-axis. Clear the lines from the graph. Drag the lines up and down to different positions to show that two parallel lines will always have the same coefficient for x. Repeat for different gradients. Investigating the y-intercept Define the y-intercept as the point where a line crosses the y-axis. Ask a volunteer to drag the line on the board so that it crosses the y-axis at the point (0, 2). State that the y-intercept of this line is 2. Add another line and ask another volunteer to drag the line and vary the gradient to produce another line with a y-intercept of 2. Reveal the equations of each line. Drag the end points of the lines to change the gradient, while keeping the y-intercept fixed. Ask pupils what they notice about the equations of the lines, that is that all of the lines are of the form y = mx + 2. Conclude that the value of c tells us where the line cuts the y-axis. Clear the board and repeat, if required, with a different intercept, for example, –3. Note that, again, the multiplier (the coefficient) of x changes and the number that is added on the stays the same.

43 The gradient and the y-intercept
Complete this table: equation gradient y-intercept y = 3x + 4 y = – 5 y = 2 – 3x 1 –2 3 (0, 4) x 2 1 2 (0, –5) (0, 2) –3 Complete this activity as a class exercise. (0, 0) y = x y = –2x – 7 (0, –7)

44 Rearranging equations into the form y = mx + c
Sometimes the equation of a straight line graph is not given in the form y = mx + c. The equation of a straight line is 2y + x = 4. Find the gradient and the y-intercept of the line. We can rearrange the equation by transforming both sides in the same way: 2y + x = 4 Explain that if the equation of a line is linear (that is if x and y are not raised to any power except 1), it can be arranged to be in the form y = mx + c. It is often useful to have the equation of a line in this form because it tells us the gradient of the line and where it cuts the y-axis. These two facts alone can enable us to draw the line without have to draw up a table of values. 2y = –x + 4 y = –x + 4 2 y = – x + 2 1 2

45 Rearranging equations into the form y = mx + c
Sometimes the equation of a straight line graph is not given in the form y = mx + c. The equation of a straight line is 2y + x = 4. Find the gradient and the y-intercept of the line. Once the equation is in the form y = mx + c we can determine the value of the gradient and the y-intercept. y = – x + 2 1 2 1 2 So the gradient of the line is and the y-intercept is 2.

46 What is the equation? What is the equation of the line passing through the points Look at this diagram: y C A B E G H F D 5 10 -5 a) A and E? x = 2 b) A and F? y = x + 6 c) B and E? y = x – 2 d) C and D? y = 2 x Ask pupils to give you the equations of the lines passing through the required points by considering the gradient and the y-intercept of each line. Ask pupils to tell you which lines are parallel. Ask how we can use the equations of the lines to find out which ones are parallel. Establish, that parallel lines have the same gradient and therefore, the x’s have the same coefficient. e) E and G? y = 2 – x f) A and C? y = 10 – x

47 Substituting values into equations
A line with the equation y = mx + 5 passes through the point (3, 11). What is the value of m? To solve this problem we can substitute x = 3 and y = 11 into the equation y = mx + 5. This gives us: = 3m + 5 Subtracting 5: 6 = 3m Discuss ways to solve the problem. Some pupils may suggest plotting the point (3, 11) and drawing a straight line through this and the point (0, 5). The gradient of the resulting line will give the value for m. Ask pupils if they can suggest a method that does not involve drawing a graph. Establish that if the line passes through the point (3, 11) then we can substitute x = 3 and y = 11 into the equation y = mx + 5. Reveal the equation 11 = 3m + 5 on the board and talk through the steps leading to the solution of this equation. Dividing by 3: 2 = m m = 2 The equation of the line is therefore y = 2x + 5.

48 Pairs Pupils take turns to choose two cards with the object of matching an equation to its corresponding graph. Each time a card showing a graph is revealed discuss possible equations for that graph. Each time an equation is revealed discuss the properties of the corresponding graph. The game can be played with the cards face up or face down.

49 Matching statements Briefly discuss the graphs that would be produced by each of the equations on the left. Ask pupils what the gradient of each of the graphs would be and the y-intercept. Note: one of the graphs is not linear because it contains x–1 and does not, therefore, have a constant gradient. Discuss the fact that this graph never crosses the y-axis (x = 0) because when x = 0 we would be dividing by 0 which is undefined. Ask pupils to match each equation with the most appropriate statement by dragging matching pairs to fit together. Ask pupils to justify their choices with reference to the equation given.

50 Exploring gradients Drag the end points of the line to demonstrate lines of varying gradient. (Notice that the gradient of a vertical line will appear as ‘undefined’ and the gradient of a horizontal line as 0). Show that the steeper the line is the larger the gradient. We can find the gradient by asking, for every unit we move horizontally (from left to right) how many squares do we move vertically (from bottom to top)? Demonstrate to pupils that when the line slopes downwards the change in the vertical distance is negative and so the gradient is negative. Define the gradient as ‘the vertical distance between the end points (from bottom to top) over the horizontal distance between the end points (from left to right) ‘. Show how the length of the line can be changed without changing the gradient by using equivalent fractions. For example, if the vertical distance between the end points is 6 and the horizontal distance between the end points is 4, the gradient of the line is 3/2. If we change the vertical distance to 9 and the horizontal distance to 6, the gradient of the line is still 3/2. Point out that it is most useful to write gradients as vulgar fractions. For example, the gradient 3/2 tells us that for every 2 squares we move along we move 3 up. Hide the value of the gradient and ask pupils to tell you the gradients of given lines. Include gradients that need to be cancelled down and negative gradients. Ask volunteers to come to the board and drag the end points of the line to give required gradients. For example, if you ask the volunteer to make a line with a gradient of –2 they could have the horizontal distance between the end points as 5 and the vertical distance between the end points as –10. Challenge another pupil to make the line longer (or shorter) without changing the gradient. Ask pupils to give you the coordinates of any two points on the line. Establish that the gradient can by found by dividing the difference between the y-coordinates by the difference between the x-coordinates.

51 Gradients of straight-line graphs
The gradient of a line is a measure of how steep the line is. The gradient of a line can be positive, negative or zero if, moving from left to right, we have: y x an upwards slope y x a horizontal line y x a downwards slope O O O Positive gradient Zero gradient Negative gradient If a line is vertical its gradient cannot be specified.

52 Finding the gradient from two given points
If we are given any two points (x1, y1) and (x2, y2) on a line we can calculate the gradient of the line as follows: y the gradient = change in y change in x x (x2, y2) y2 – y1 (x1, y1) Draw a right-angled triangle between the two points on the line as follows: x2 – x1 Formally define the gradient as the change in y/the change in x. Explain that since, for a straight line, the change in y is proportional to the corresponding change in x, the gradient will be the same no matter which two points we choose on a line. Explain how drawing a right-angled triangle on the line help us calculate its gradient. O the gradient = y2 – y1 x2 – x1


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