Understanding how equations are represented on a graph

UNDERSTANDING HOW EQUATIONS ARE REPRESENTED ON A GRAPH ACTIVITY 1 To understand how equations can be represented on a graph [LO 2.2, 2.5, 2.6] Example: The equation y = 3x + 2 tells one how the values of x and y are connected – it shows the relationship between two variables, x and y. For example, if x is 5, then y can be calculated from 3 × 5 + 2, giving 17. So, we substitute the value 5 for the x, and complete the calculation.

The table shows some of the answers. The point of making a table is that it gives us coordinates, which we can plot on a set of axes. From these we can draw a graph, which is a picture of the relationship between x and y. 1 In a group of 4 or 5 learners, complete the tables from the equations below. Each one should do a different table, and then discuss the answers and copy the others’ tables into your book. Below each table is a set of axes on which to draw the graph. All of these graphs will be straight lines, so you may connect the points you plot from each table.

2 In your group, discuss what you see in your graphs. Compare the graph with the equation. Here are a few suggestions for you to investigate: 2.1 What does the coefficient of x in the equation do to the graph? What happens when the coefficient is negative? 2.2 In 1.6 the table has only two columns. Is it necessary to have more than two columns if you know that the graph will be a straight line? 2.3 Compare 1.1 and 1.5 to see if you can find out what the constant does to the graph. The table summarises how equations, tables and graphs agree with one another. You have to memorise this information – you can’t work correctly with graphs if you don’t know the words Equation: x y Equation: Independent variable Dependent variable Flow diagram:: Input value Output value Table: First row Second row Coordinates: 1st coordinate 2nd coordinate Graph: x– axis y– axis Graph: Horizontal axis Vertical axis

ACTIVITY 2 To understand and apply all the characteristics of the straight-line graph [LO 2.5] 1 In the previous activity you used equations to make tables from which graphs were drawn. One can also draw a graph from an equation without making a table. As we saw in graph 1.6 above, if we know that we have to draw a straight line, then two points on the graph paper are sufficient to enable us to draw the line. We don’t need a table if we have the equation of the straight line. In this section we will refer repeatedly to the six graphs in the previous exercise. First we have to examine the structure of the equation: y = mx + c is the standard form of the equation: y is on the left of the equal sign, and y has no coefficient (which is the same as saying that its coefficient is 1, which we don’t write) After the equal sign there may be one or two terms (compare equations 1.5 and 1.6 above). If there are two terms, then the term in x is written first – the x may have any number as a coefficient: positive, negative or fractional. In the standard form we write an m to stand for the coefficient. If the coefficient is 1, then again we don’t write it. The c stands for a constant – any number that is not a variable: it can also be negative, positive or fractional. If we write y = mx+ c then this is the general equation for all straight lines. But y = –3x+ 2 is the defining equation for a specific straight line. Writing equations in the standard form: First an example: 6x+ 2y – 1 = 0 Keep the term in y on the left; move the other two to the right. 2y = –6x + 1 Now make the coefficient of y = 1 by dividing all the terms by 2. y= –3x + ½ This the standard form. Here m = –3 and c = ½. Now you practise some – also write down what m and c are, as above. 1.1 2x + y = 3 1.2 3y – 9 = 6x 1.3 3x = 6y 1.4 2y – 8 = 0

2 Understanding the gradient. Previously we mentioned that the steepness of a graph can be calculated – this is very easy if the graph is a straight line, because it is equally steep everywhere – we say that the gradient of a straight-line graph remains constant. Study the values of m in the six graphs in the previous exercise If your work is correct, you will have noticed that the graphs slope up to the right where m is positive, and the graphs slope down to the right where m is negative. In other word, m tells us about the gradient. (What do you think happens in y = 4, the odd one out?) With m positive, the number of units the line goes up for every unit it goes right gives us m (the gradient). When m is negative, we count how many units the line goes down for every unit it goes right. Here are two examples. By completing right-angled triangles in a convenient position on the lines in the graph, we can easily calculate the two gradients, as follows: For the top line: m=−25 size 12{m= - { {2} over {5} } } {}, because the line goes down to the right, we know the gradient is negative; 2 is the height of the triangle and 5 is its length. For the bottom line: m=+69=23 size 12{m"=+" { {6} over {9} } = { {2} over {3} } } {}, with 6 the height of the triangle, and 9 its length. We don’t write the +, and we simplify the fraction. 2.1 Now go back to the previous six graphs and do the same so that you can confirm that the m in the equation agrees with the gradient you calculate from the graph itself. Also notice how the size of m tells you how steep the graph is. 3 Finding out where the graph cuts the y–axis (called the y–intercept): If you study the equations of the six graphs, you will notice that the constant term (c) in the standard form tells us exactly where the graph cuts the y–axis! For example, in y = 3x –4, the y–intercept is at –4 on the y–axis. Confirm that this is true for all six graphs. Now we have a method for drawing graphs from an equation in the standard form. We don’t have to make a table – we simply use the y–intercept (given to us by c), and the gradient (given by m). On the graph paper, mark the y–intercept. Now use the gradient in the form of a fraction; if it is a whole number, then write it with 1 as a denominator. From the y–intercept, count as many units to the right as the denominator. From there count as many units as the numerator up, if m is positive, or down, if m is negative. Here are two examples:

(a) y=23x−2 size 12{ size 11{y```=``` { { size 11{2}} over { size 11{3}} } `x`` - ``2}} {} The y-intercept is –2, marked on the y-axis with a circle. The gradient is 23 size 12{ { {2} over {3} } } {}, so we move from the circle three units to the right, and then 2 units up (not down – the gradient is positive). Another circle marks the spot we end up at. And now we draw the straight line through these two spots.

(b) y = –x + 3 The y-intercept is 3, marked by a circle. The gradient is –1, which we change to −11 size 12{ - { {1} over {1} } } {}. This tells us to move one unit (denominator) to the right and then one unit (numerator) down (not up). We end up at the point ( 1 ; 2 ), marked by a second circle. Draw the line through the two circled points. Draw the following graphs using the y-intercept/gradient method you have just studied. 3.1 y=−3x+1 size 12{ size 13{y```=`` - 3x``+``1}} {} 3.2 y=13x−52 size 12{ size 13{y```=``` { { size 13{1}} over { size 13{3}} } x``` - ``` { { size 13{5}} over { size 13{2}} } }} {} 3.3 y=−34x size 12{ size 13{y```=``` { { size 13{ - 3}} over { size 13{4}} } x}} {} 3.4 4x – 3y = 5 4 In problem 3.4 above, you should have written the equation in the standard form to be able to use m and cfor the y–intercept/gradient method. This is a lot of extra work. There is another way to find the two points needed to be able to draw a straight–line graph. If we can find out where the graph cuts the x–axis as well as the y–axis, then we can simply draw the line through the two intercepts!y–interceptx–intercepty = 3x – 4( 0 ; –4 )113;0 size 12{ left (1 { { size 8{1} } over { size 8{3} } } `;`0 right )} {}y = –4x + 3( 0 ; 3 )34;0 size 12{ left ( { { size 8{3} } over { size 8{4} } } `;`0 right )} {}y = ½x +1( 0 ; 1 )( –2 ; 0 )

Going back again to the previous six graph problems, the table shows the x– and y–intercepts for three of them in the form of coordinates. The important thing to notice is that the y–intercept always has a zero in the position of the x–coordinate, and the x–intercept always has a zero in the position of the y–coordinate. This means that if we take the equation as it is and make the x zero and simplify to find the value of y, it will give us the y–intercept. Making the y zero and finding x, gives us the x–intercept. Here is how to do it for the equation 9 – 6x = 3y (definitely not in the standard form):

Finding the y–intercept: Substitute 0 for x: 9 – 6(0) = 3y 9 – 0 = 3y 9 = 3y 3y = 9 y = 3 The y–intercept in coordinate form is ( 0 ; 3 ) Mark this point on the graph. Finding the x–intercept: Substitute 0 for y: 9 – 6x= 3(0) 9 – 6x = 0 9 = 6x 6x= 9  x=96=32 size 12{x= { { size 8{9} } over { size 8{6} } } = { { size 8{3} } over { size 8{2} } } } {} The x–intercept in coordinate form is 32;0 size 12{ left ( { { size 8{3} } over { size 8{2} } } `;``0 right )} {} Mark this point on the graph. Finally draw a line through the two points. Alongside is the sketch. This is a very easy and convenient method. If you work carefully and with concentration, it won’t easily go wrong. Practise the method on the following equations: 4.1 4y + 3x = 4 4.2 6y + 15 = 2x 4.3 3x + 4y = 0 4.4 3y + 5 = 4x 4.5 2y + 8 = 6x 4.6 4y – 2x – 4 = 0 Does this method remind you of something? 5 We still have to consider a few special cases. With the equation written in the standard form, we can deduce a great deal about the graph The standard form of the straight–line equation is y = mx + c, as we know. If c is zero, then the equation becomes y = mx; if m is zero, then the equation becomes y = c. 1 2 3 – 1 – 2– 3 0123– 1– 2– 3 y = mx + c, (where neither m nor c is zero), is the equation which produces lines which do not pass through the origin, nor are they horizontal or vertical. The first diagram shows some of these graphs. You can use either the y–intercept/gradient method or the two–intercept method for drawing these graphs. 1 2 3 – 1 – 2– 3 0123– 1– 2– 3y = mx (c is zero) produces lines which are neither horizontal nor vertical. They do pass through the origin, which is understandable as c is zero, meaning the y–intercept is zero. The second diagram shows a few of these. The y–intercept/gradient method is the simplest for drawing these graphs. y = c is the equation of a horizontal line, as you have already seen Draw them by drawing a horizontal line through the y–intercept (c). 1 2 3 – 1 – 2– 3 0123– 1– 2– 3If the equation of the line is x = k, where k is a constant, then this is a vertical line with k the x–intercept. Draw them by finding k on the x–axis and drawing a vertical line through that point.The third diagram shows some of the horizontal and vertical graphs. With all the advice above, you should be able to figure out the equations of these twelve graphs. If not, the next section will help.