A8 Linear and real-life graphs

A8 Linear and real-life graphs
KS4 Mathematics A8 Linear and real-life graphs

Contents A8 Linear and real-life graphs A A8.1 Linear graphs A A8.2 Gradients and intercepts A A8.3 Parallel and perpendicular lines A A8.4 Interpreting real-life graphs A A8.5 Distance-time graphs A A8.6 Speed-time graphs

(3, 5) (3, 5) (3, 5) Coordinate pairs
When we write a coordinate, for example, (3, 5) (3, 5) (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 the y-coordinate. Together, the x-coordinate and the y-coordinate are called a coordinate pair.

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. Explain 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. This line is called x = 2. x = 2

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 Stress that the graph of x = ‘a constant number’ will always be parallel to the y-axis. In other words, it will always be vertical. 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). x = –10 x = –3 x = 4 x = 9

Graphs parallel to the x-axis
What do these coordinate pairs have in common? (0, 1), (4, 1), (–2, 1), (2, 1), (1, 1) and (–3, 1)? The y-coordinate in each pair is equal to 1. Look at what happens when these points are plotted on a graph. x y All of the points lie on a straight line parallel to the x-axis. y = 1 Explain 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). Name five other points that will lie on this line. This line is called y = 1.

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 Stress that the graph of y = ‘a constant number’ will always be parallel to the x-axis. In other words, it will always be horizontal. 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). y = –2 y = –5

Plotting graphs of linear 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, –1), (4, 2), (–2, –4), (0, –2), (–1, –3) and (3.5, 1.5)? In each pair, the y-coordinate is 2 less 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 the the points will lie on a straight diagonal line. Stress 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). Ask pupils to suggest the coordinates of any other points that will lie on this line. Look for imaginative answers. y = x – 2 We can draw a graph of the function y = x – 2 by plotting points that obey this function.

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 = 2x + 5 We can use a table as follows: x y = 2x + 5 –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 extra 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 required function. –1 1 3 5 7 9 11 (–3, –1) (–2, 1) (–1, 3) (0, 5) (1, 7) (2, 9) (3, 11)

For example, y to draw a graph of y = 2x + 5: 1) Complete a table of values: x y = 2x + 5 –3 –2 –1 1 2 3 5 7 9 11 y = 2x + 5 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. x 4) Label the line. 5) Check that other points on the line fit the rule.

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.

A8.2 Gradients and intercepts

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 Positive gradient Zero gradient Negative gradient If a line is vertical, its gradient cannot be specified.

Calculating gradients
By looking at a variety of examples, establish that if we are given the coordinates of any two points on a line we can find its gradient by taking the y-coordinate of the first point subtracted from the y-coordinate of the second point and diving it by the x-coordinate of the first point subtracted from the x-coordinate of the second point. Demonstrate to pupils that when the line slopes downwards the change in the vertical distance is negative as we move from left to right and so the gradient is negative. Show that choosing different points on the same line will give the same gradient 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 often most useful to leave gradients as improper 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 vary the points to give lines of specified 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.

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 Explain how drawing a right-angled triangle on the line helps us calculate its gradient as shown in the previous activity. Explain too 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. the gradient = y2 – y1 x2 – x1

Investigating linear graphs
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). Establish that m is the gradient of the line and c is the intercept on the y-axis.

The general 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 having 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).

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)

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. Rearrange the equation by performing the same operations on both sides, 2y + x = 4 2y = –x + 4 subtract x from both sides: 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 having to draw up a table of values. y = –x + 4 2 divide both sides by 2: y = – x + 2 1 2

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. Explain that the equation of a (straight) line can always be arranged to be in the form y=mx + c. (This is not true for lines parallel to the y-axis.) 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 having to draw up a table of values. y = – x + 2 1 2 1 2 – So the gradient of the line is and the y-intercept is (0, 2).

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 subtract 5 from both sides: 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. divide both sides by 3: 2 = m m = 2 The equation of the line is therefore y = 2x + 5.

What is the equation of the line?

Match the equations to the graphs

A8.3 Parallel and perpendicular lines

Investigating parallel lines

Parallel lines If two lines have the same gradient they are parallel.
Show that the lines 2y + 6x = 1 and y = –3x + 4 are parallel. We can show this by rearranging the first equation so that it is in the form y = mx + c. 2y + 6x = 1 2y = –6x + 1 subtract 6x from both sides: y = –6x + 1 2 divide both sides by 2: y = –3x + ½ The gradient m is –3 for both lines, so they are parallel.

Matching parallel lines
There are two levels of difficulty for this activity. Level 1 shows all the line in the form y = mx + c. In level 2 some of the lines are not in the form y =mx + c, and these must be rearranged to find the value of m.

Investigating perpendicular lines

Perpendicular lines If the gradients of two lines have a product of –1 then they are perpendicular. In general, if the gradient of a line is m, then the gradient of the line perpendicular to it is –1 m Write down the equation of the line that is perpendicular to y = –4x + 3 and passes through the point (0, –5). The gradient of the line y = –4x + 3 is –4. 1 4 . The gradient of the line perpendicular to it is therefore 1 4 The equation of the line with gradient and y-intercept –5 is, y = x – 5 1 4

Matching perpendicular lines
There are two levels of difficulty for this activity. Level 1 shows all the line in the form y = mx + c. In level 2 some of the lines are not in the form y = mx + c and these must be rearranged to find the value of m.

A8.4 Interpreting real-life graphs

Real-life graphs When we use graphs to illustrate real-life situations, instead of plotting y-values against x-values, we plot one physical quantity against another physical quantity. The resulting graph shows the rate that one quantity changes with another. For example, This graph shows the exchange rate from British pounds to American dollars. British pounds American dollars It is a straight line graph through the origin and so the equation of the line would be of the form y = mx. m represents the exchange rate (number of dollars to the pound). The value of m would be equal to the number of dollars in each pound. What would the value of m represent?

Real-life graphs This graph show the value of an investment as it gains interest cumulatively over time. The graph increases by increasing amounts. time investment value Each time interest is added it is calculated on an ever greater amount. This makes a small difference at first but as time goes on it makes a much greater difference. This is an example of an exponential increase.

Real-life graphs This graph show the mass of a newborn baby over the first month from birth. time mass The baby’s mass decreases slightly during the first week. Its mass then increases in decreasing amounts over the rest of the month.

Rates of change For each graph ask pupils whether y is large, small or becomes zero when x is large. Also ask whether y is large, small or becomes zero when x is small. Establish whether y increases as x increases or whether y decreases as x increases. Then decide whether it is increasing or decreasing by equal, increasing or decreasing amounts. Ask pupils to describe a real-life situation that could be described by each graph. For example, a graph that shows y increasing by increasing amounts could be showing acceleration if x is time and y is distance travelled.

Filling flasks Start by explaining that we are going to produce a graph of the depth of water in a flask as it fills with water. Note that the water flows out of the tap at a constant rate. As the first flask fills up, the graph of depth against time will be drawn. Ask pupils to predict the shape of the next graph before it is drawn, justifying their explanations. Establish that the wider the flask is at a given point the longer it will take for the water to increase in depth. Before filling a flask, ask a volunteer to use the pen tool to show the shape of the graph. Play the animation to see how close the pupil’s prediction was. Continue for each flask in turn. Ask pupils to explain why all the lines pass through the origin. Ask pupils to explain what would happen if the water from the tap did not flow out at a constant rate. For example, in real life the rate of the water coming out of the tap would speed up as the tap is turned on. How would this affect the shape of the graph? Ask pupils if we can use the graph to work out the capacity of each flask.

A8.5 Distance-time graphs

Formulae relating distance, time and speed
It is important to remember how distance, time and speed are related. Using a formula triangle can help, distance = speed × time DISTANCE SPEED TIME time = distance speed Recall the basic formulae relating distance, time and speed using the formula triangle. speed = distance time

Distance-time graphs In a distance-time graph the horizontal axis shows time and the vertical axis shows distance. For example, John takes his car to visit a friend. There are three parts to the journey: John drives at constant speed for 30 minutes until he reaches his friend’s house 20 miles away. time (mins) distance (miles) 15 30 45 60 75 90 105 120 5 10 20 He stays at his friend’s house for 45 minutes. Talk through the various stages of the journey. The distance refers to the distance from the starting point, in this case John’s house. During the first part of the journey John moves away from the starting point at a constant speed shown by the slope of the graph. While John stays at his friend’s house his distance from home does not change as time goes on and so the line is horizontal. In the last part of his journey the graph slopes downwards because as he returns home his distance from the starting point decreases. Ask pupils to calculate John’s speed for the first and last parts of the journey in miles per hour. Establish that the speed is given by the gradient of the graph. He then drives home at a constant speed and arrives home 45 minutes later.

Finding speed from distance-time graphs
How do we calculate speed? Speed is calculated by dividing distance by time. In a distance-time graph this is given by the gradient of the graph. gradient = change in distance change in time time distance change in distance = speed The steeper the line, the faster the object is moving. change in time A zero gradient means that the object is not moving.

Interpreting distance-time graphs
Move the points to change the graph and ask pupils to interpret it. For example, ask pupils at what speed the object is traveling for different stage in the journey. Ask pupils to make up a story for a given graph.

Distance-time graphs When a distance-time graph is linear, the objects involved are moving at a constant speed. Most real-life objects do not always move at constant speed, however. It is more likely that they will speed up and slow down during the journey. Increase in speed over time is called acceleration. acceleration = change in speed time It is measured in metres per second per second or m/s2. When speed decreases over time is often is called deceleration.

Distance-time graphs Distance-time graphs that show acceleration or deceleration are curved. For example, This distance-time graph shows an object accelerating from rest before continuing at a constant speed. time distance This distance-time graph shows an object decelerating from constant speed before coming to rest. time distance Point out that the first graph gets steeper as time goes on. If the gradient is increasing then the object is speeding up. The second graph gets shallower as time goes on. If the gradient is decreasing then the object is slowing down.

Speed-time graphs Travel graphs can also be used to show change in speed over time. For example, this graph shows a car accelerating steadily from rest to a speed of 20 m/s. It then continues at a constant speed for 15 seconds. time (s) speed (m/s) 5 10 15 20 25 30 35 40 The brakes are then applied and it decelerates steadily to a stop. Talk through the various stages in the graph. Stress that when the graph slopes downwards a decrease in speed is indicated. Unlike in a distance-time graph it is still moving in the same direction. The car is moving in the same direction throughout.

Finding acceleration from speed-time graphs
Acceleration is calculated by dividing speed by time. In a speed-time graph this is given by the gradient of the graph. gradient = change in speed change in time time speed change in speed = acceleration The steeper the line, the greater the acceleration. Remind pupils that acceleration is measured in metres per second per second or m/s2. change in time A zero gradient means that the object is moving at a constant speed. A negative gradient means that the object is decelerating.

Finding distance from speed-time graphs
The following speed-time graph shows a car driving at a constant speed of 30 m/s for 2 minutes. time (s) speed (m/s) 15 30 45 60 75 90 105 120 5 10 20 What is the area under the graph? The area under the graph is rectangular and so we can find its area by multiplying its length by its height. Remind pupils that the distance travelled can be found by multiplying the average speed by the time taken. Use this to establish that the area under the graph corresponds to the distance travelled in metres. Area under graph = 20 × 120 = 240 What does this amount correspond to?

The area under a speed-time graph
This area under a speed-time graph corresponds to the distance travelled. time (s) speed (m/s) 5 10 15 20 25 30 35 40 For example, to find the distance travelled for the journey shown in this graph we find the area under it. The shape under the graph is a trapezium so, 15 Area = ½( ) × 20 Point out that we could also find the area under the graph by dividing it into a rectangle with two triangles either side. Demonstrate, if necessary, that this leads to the same answer. The units in the answer depend on the units used in the graph. Here we have m/s multiplied by s to give the answer in metres. = ½ × 55 × 20 = 550 So, distance travelled = 550 m

Interpreting speed-time graphs
Move the points to change the graph and ask pupils to interpret it. For example, ask pupils to calculate the acceleration and distance travelled over various stages in the journey. Ask pupils to find the total distance travelled by finding the area under the graph. Ask pupils to make up a story for a given graph.

A8 Linear and real-life graphs

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