Chapter 8 Graphs and Functions:


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1 Chapter 8 Graphs and Functions: Cartesian axes, coordinates and points 8.1 Pictorially we plot points and graphs in a plane (flat space) using a set of Cartesian axes traditionally called the x and y axes to locate points on it. See diagram below The vertical axis is the y axis and the horizontal axis is the x axis. Any point in this plane has a set of numbers called an ordered pair, which is a set of coordinates that locate a point in the plane. In the diagram above the top right point has coordinates (x=5,y=7) i.e. 5 units on the x axis and 7 units on the y axis. It is in quadrant 1. However because we always write the x coordinate first we just write (5,7) instead of (x=5,y=7). The top right quadrant is quadrant 1, then going anticlockwise we get the other 3 quadrants. For example the bottom left point has coordinates (5,6) i.e. 5 units on the x axis and 6 units on the y axis. It is in quadrant 3. E8.1 Give the coordinates of the point in the (i) second quadrant (ii) fourth quadrant A8.1 The point in the (i) second quadrant is (5,4). (ii) fourth quadrant is (3,5). 8.2 The point (3,2) is located 3 units to the right and 2 units below the origin. It is in the 4 th (bottom right) quadrant. E8.2 Complete the Table below on the location of a point relative to the origin. The first row has been done for you. Location Coordinates Quadrant 3 to the right and 2 (3,2) 4 th below origin 87
2 (5, 2) At the origin Junction of all 4 (, 4) Boundary Between 1 st 4 to the left and 2 above. (2,0) Plot these points on the axes above. and 2 nd. (5, ) Boundary between 1 st and 4 th A8.2 Location Coordinates Quadrant 3 to the right and 2 (3,2) 4 th directly below origin 5 to the left, 2 below (5, 2) 2 nd At the origin (0,0) Junction of all 4 4 above the origin (0, 4) Boundary Between 1 st and 2 nd. 2 to the left (2,0) Between 2 nd And 3 rd. 4 to the left and 2 (4,2) 2 nd above. 5 to the right (5,0 ) Boundary between 1 st and 4 th Equations with infinite solutions 8.3 We know that x + 3 = 5 is an equation that we can solve by inspection. There is only one solution, x = 2. Now consider the equation, x + y =10. It has two unknowns and we can produce many solutions by inspection such as x = 5 and y = 5, x = 0 and y = 10, x = 1 and y = 9. Also solutions with negative values such as x = 4 and y = 14 etc. Why does the first equation have one solution and the second equation have an infinite number? The reason is that the first equation has just one unknown, x whereas the second equation has two unknowns x and y and yet we have just one piece of information. E8.3 How many solutions do the following equations have? 88
3 (i) 2x + 7 = 9. Identify it. (ii) 5x + 6y = 10 Can you identify some of them? A8.3 (i) 2x + 7 = 9. Has 1 solution, x = 1. (ii) 5x + 2y = 10. Has an infinite number of solutions e.g. (2,0), (0,5), (1,2.5) etc. Can you identify some of them? 8.4 Since 1 equation with two unknowns has an infinite number of solutions we can produce a graph by plotting the set of solutions on a set of Cartesian axes. Here are some solutions to the equation: y = 2x + 3 (0,3) because y = 3 when x = 0. (2,7) because y = 7 when x = 2. (3, 3) because when x = 3, y = 3. E8.4 Complete the following coordinates for some of the solutions of the equation: y = 2x + 3 (2, ) ; (3, ) ; (1, ) ; (1, ) ; (, 4) ; (, 3) ; (0, ) Plot the points on the axes given below A8.4 Complete the following coordinates for some of the solutions of the equation: y = 2x + 3 (2,1) ; (3,3) ; (1,1) ; (1,5) ; (0.5, 4) ; (3, 3) ; (0,3) 89
4 Linear graphs 8.5 What is the graph of the entire solution set? It is a straight line as shown below: Graph of y =2x + 3 E8.5 Examine the graph and identify the points where the graph crosses the y axis and the x axis. Graph crosses y axis at (0,3) and cross the x axis at (1.5,0) In graphing the equation, y = 2x + 3 let s see what the significance of the coefficient of 2 and and the constant term 3 is. Note the 2 is the coefficient of the x term (linear term) and the 3 is the constant term. First consider the constant term, 3. Notice that when x = 0 then y = 3. Hence (0,3) is a point on the graph. Which quadrant is it in? It is on the boundary between the 1 st and 2 nd quadrants i.e. it lies on the y axis. We call this point the y intercept. So we say either the y intercept is 3 or we say the y intercept is the point (0,3). We say that in the graph y = mx + c, c is the y intercept. E8.6 What is the y intercept of the following graphs? (i) y = 2x + 4 (ii) y = 2x 3 (iii) y = 2x? A8.6 The y intercept of (i) y = 2x + 4 is 4 i.e. the point (0,4). (ii) y = 2x 3 is 3 i.e. the point (0,3). (iii) y = 2x is 0 i.e. the point (0,0). 8.7 Now what is the significance of the coefficient of x of the RHS (2 in this case) in the graph of y = 2x + 3? 90
5 Let us look at two random points on the graph, say point A at (1,5) and point B at (3,9) We can by looking at the x coordinates of the two points determine that B is 2 units to the right of A. We write x = x value of B x value of A. = 3 1 =2 We can by looking at the y coordinates of the two points determine that B is 4 units above A. We write y = y value of B y value of A = 9 5 = 4 So in going from A to B the graph is climbing 4 units up for every 2 units to the right. So we say the graph has a slope or gradient of 4 in 2 which we write as y 4 gradient 2. x 2 So the coefficient of the x term in the equation y = 2x + 3 is the gradient. Summarise: constant term = 3 is the y intercept coefficient of x term = the slope or gradient of the line given by y gradient x E8.7 Complete the Table below: The first row has been done for you. A8.7 Equation of line Slope Y intercept y = 2x y = 3x y = 4x 04 y = 5 2x y = Equation of line Slope Y intercept y = 2x y = 4x y = 3x y = 2x y = 4x 4 0 y =
6 y = 5 2x 2 5 y = y = 5x What is the equation of the line that passes through the points (0,3) and has a slope of 2? We know that in general the equation is of the form: y = mx + c, where m is the slope (coefficient of x term) and y intercept is c (the constant term). Since the slope is 2 (i.e. the coefficient of x is 2) y = mx + c becomes y = 2x + c. Since the graph passes through (0,3) we know the y intercept is 3; So c is 3 and y = 2x + c becomes y = 2x + 3. So the equation is y = 2x + 3 OR y = 3 2x. E8.8 What is the equation of the line that passes through the point i. (0, 1) and has a slope of 2 ii. (0, 2) and has a slope of 3 iii. (0,0) and has a slope of 1 A8.8 i. y = 2x 1 ii. y = 3x + 2 iii. y = x 8.9 What is the equation of the line that passes through the points (0,3) and (2,5)? We know that in general the equation is of the form: y = mx + c, where m is the slope (coefficient of x term) and y intercept is c (the constant term). Since the graph passes through (0,3), which is on the y axis, we know the y intercept is 3; so c is 3, so y = mx + c becomes y = mx + 3. The point (2,5) is 2 units to the right (2 0) and 2 units above (5 3) the point (0,3). y gradient 1. x So the slope is 2 in 2 i.e. m=1 92
7 so y = mx + 3 becomes y = x + 3. So the equation of the line is y = x + 3 E8.9 What is the equation of the line that passes through the points (0,1) and (1,4)? A8.9 (0,1) tells us that the y intercept is 1 so we can write y = mx 1. (0,1) and (1,4) tells us that the gradient is 5. So y = 5x Now consider the line that passes through the points (2,4) and (3,5). We can t determine the slope or the y intercept at sight easily: So we proceed as follows: gradient y x So m= 9/5 in the equation y = mx + c. 9x So write: y  c. 5 Now remember (3,5) is (x=3,y=5). So substitute in the equation above to get: 9(3) c 5 9(3) c x 9x 2 So our line y  c. becomes y E8.10 Find the equation of the line that passes through (3, 6) and (2,1) A8.10 gradient y x So m= 1 in the equation y = mx + c. 93
8 So write: y  x c. Now remember (2,1) is (x=1,y=1). So substitute in the equation above to get: 21 c c 3 So our line y  x c. becomes y 3 x Consider the line y = 2x + 3 again. It has a slope of 2 and the y intercept is 3. What is the x intercept? We know that at the x intercept y=0. If we substitute y=0 in y = 2x + 3 we get 0 = 2x We can solve this equation to get x So the x intercept is  and the point of interception on the x axis is 2 3 (  2, 0) Check by seeing the graph in box 8.5. E8.11 For the graph y = 4 3x what is i. the slope ii. y intercept iii. x intercept? A8.11 y = 4 3x i. slope = 3 ii. y intercept = 4 iii. 0 = 4 3x x intercept = 4/3 Graphs of Quadratic functions 8.12 A quadratic function is function of the form : y = ax 2 + bx + c where a,b,c are constants. There are three terms in the function the first is a term in x 2 (called the quadratic term), the second is a term in x (called a linear term) and the third is a constant term. This is a 2 nd degree function, whereas linear functions are first degree functions. The simplest quadratic is y = x 2 i.e. no linear term and no constant term. We can complete the ordered pairs using the x values given below: (5,25), (4,16), (3,9), (2,4), (1,1), (0,0), (1,1), (2,4), (3,9), (4,16), (5,25). Note the x values range from 5 to +5 but the y values are all positive 94
9 because the square of a number is always +ve. E8.12 Complete the following points on this quadratic: (,25), (,0), (,25) so that the points are distinct (i.e. separate) A8.12 ( 5,25), (0,0), (5,25) 8.13 The graph is plotted below E8.13 What is the y intercept? What is the y intercept? the x intercept? What is the axis of symmetry? What is the minimum point? What is the minimum y value? A8.13 y intercept is 0, the x intercept is 0, the axis of symmetry is x = 0, the minimum point is (0,0), the minimum value is Now suppose we want to sketch (rather than draw) the quadratic, y = x 2 2x 15 We can find the y intercept by putting x = 0 in y = x 2 2x 15 to get y = 15. So the graph crosses the y axis at (0,15). Now to find the x intercepts we put y = 0 to get: 0 = x 2 2x 15. Write this as x 2 2x 15 = 0 and solve viz. (x 5)(x + 3) = 0 So x = 3 and x = 5. The graph crosses the x axis at (3,0) and (5,0). The axis of symmetry is a vertical line bisecting the line joining (3,0) and (5,0). 95
10 How do we find it? The middle value of any two numbers is its arithmetic average. (3 5) x The average of x = 3 and x = 5 is thus 2 x 1 The axis of symmetry is x = 1. Now we get the minimum y value using the fact that minimum y value is on the axis of symmetry, x = 1. So putting x=1 in y = x 2 2x 15 we get y min = 1 2 2(1) 15 = 17. So the minimum point of the graph is (1,17). We can now sketch our graph by putting all this information together: Point where graph crosses y axis. (0,15) Two axis points where graph (3,0) and (5,0) crosses x axis. Minimum point on the axis of (1,17) symmetry The sketch is confirmed by this drawing E8.14 Sketch the graph of the quadratic function y = x 2 2x 8 1. Find the y intercept by putting x = 0 in y = x 2 2x 8 to get y =. 2. Find the x intercepts by putting put y = 0 to get:. = x 2 2x 8 and solve viz. (x )(x ) = 0 So x =.. and x =.. 96
11 The two points of interception on the x axis are (,.) and (, ) The axis of symmetry is a vertical line bisecting the line joining (., ) and (, ) The middle x value is its arithmetic average. The average of x = and x = is thus x =.. The axis of symmetry is x =... The minimum y value is on the axis of symmetry, x =.. So putting x=. in y = x 2 2x 8 we get y min =. So the minimum point of the graph is (.,.). We can now sketch our graph by putting all this information together: Point of interception on y axis (, ) Two points of interception on x (, ) and (, ) axis Minimum point on the axis of (, ) symmetry Sketch: A8.14 Sketch the graph of the quadratic function y = x 2 2x 8 1. Find the y intercept by putting x = 0 in y = x 2 2x 8 to get y = 8. So the y intercept is the point (0,8). 2. Find the x intercepts by putting put y = 0 to get: 0 = x 2 2x 8 and solve viz. (x 4)(x + 2) = 0 So x = 4 and x = 2 are the x intercepts. 97
12 The two points of interception on the x axis are (4,0) and (2,0). The axis of symmetry is a vertical line bisecting the line joining (4,0) and (2,0). The middle x value is its arithmetic average. The average of x = 4 and x = 2 is thus x = (4 2)/2 = 1 i.e. the axis of symmetry is the vertical line, x = 1 The minimum y value is on the axis of symmetry, x = 1. So putting x = 1 in y = x 2 2x 8 we get y min = (1) 2 2(1) 8 = = 9. So the minimum point of the graph is (1,9). We can now sketch our graph by putting all this information together: Point of interception on y axis (0,8) Two points of interception on x (2,0) and (0,4) axis Minimum point on the axis of (1,9) symmetry Sketch: 98
13 99
14 Exercise 8 1. Find the equation of the straight line with a gradient of 2 passing thru the point ( 1,1). 2. A line passes thru the points (0,3)and (3,0). Find Deltax, Deltay, the slope, and the equation of the line. 3. A variable P changes from 20 to 30. This causes a dependent variable Q to change from 100 to 25. Find DeltaP, deltaq, the slope, the proportional change in P, proportional change in Q and the ratio of the proportional change in P to the proportional change in Q. 4. (i) (ii) (iii) (iv) Sketch the graph of the straight line y = 2 3x what is the gradient? what is the y intercept? what is the x intercept? 5 Sketch the quadratic y = x 2 2x 8 100
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