The Graph of a Linear Equation

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1 4.1 The Graph of a Linear Equation 4.1 OBJECTIVES 1. Find three ordered pairs for an equation in two variables 2. Graph a line from three points 3. Graph a line b the intercept method 4. Graph a line that passes through the origin 5. Determine domain and range 6. Graph horizontal and vertical lines In previous algebra classes ou have solved equations in one variable such as Solving such an equation required finding the value of the variable, in this case, that made the equation a true statement. In this case, that value is 3, because 3( 3) 2 5( 3) 4 This is a true statement because each side of the equation is equal to 11; no other value for makes this statement true. The solution can be written in three different was. We can write 3, 3 which is read the set of all such that equals 3, or simpl 3, which is the set containing the number 3. What if we have an equation in two variables, such as 3 6? The solution set is defined in a similar manner. Definitions: Solution Set for an Equation in Two Variables The solution set for an equation in two variables is the set containing all ordered pairs of real numbers (, ) that will make the equation a true statement. The solution set for an equation in two variables is a set of ordered pairs. Tpicall, there will be an infinite number of ordered pairs that make an equation a true statement. We can find some of these ordered pairs b substituting a value for, then solving the remaining equation for. We will use that technique in Eample 1. Eample 1 Finding Ordered Pair Solutions Find three ordered pairs that are solutions for each equation. (a) 3 6 We will pick three values for, set up a table for ordered pairs, and then determine the related value for

2 THE GRAPH OF A LINEAR EQUATION SECTION NOTE To indicate the set of all solutions to the equation, we write {(, ) 3 6} Substituting 1 for, we get 3( 1) The ordered pair ( 1, 9) is a solution to the equation 3 6. Substituting 0 for, we get 3(0) The ordered pair (0, 6) is a solution to the equation 3 6. Substituting 1 for, we get 3(1) The ordered pair (1, 3) is a solution to the equation 3 6. Completing the table gives us the following: (b) 2 1 Let s tr a different set of values for. We will use the following table Substituting 5 for, we get 2( 5) The ordered pair ( 5, 11) is a solution to the equation 2 1. Substituting 0 for, we get 2(0)

3 192 CHAPTER 4 GRAPHS OF LINEAR EQUATIONS AND FUNCTIONS NOTE Again, the set of all solutions is {(, ) 2 1} The ordered pair (0, 1) is a solution to the equation 2 1. Substituting 5 for, we get 2(5) The ordered pair (5, 9) is a solution to the equation 2 1. Completing the table gives us the following: CHECK YOURSELF 1 Find three ordered pairs that are solutions for each equation. (a) 2 6 (b) 3 2 NOTE Wh can A and B not both be zero? First, recall that, although and are variables, A, B, and C are constants. With that in mind, look at the equation if A and B are both zero. (0) (0) C 0 0 C 0 C Because zero must be a constant, we are left with the statement 0 0 This would be a true statement regardless of the values of and. Its graph would be ever point in the plane. NOTE Because two points determine a line, technicall two points are all that are needed to graph the equation. You ma want to locate at least one other point as a check of our work. The graph of the solution set of an equation in two variables, usuall called the graph of the equation, is the set of all points with coordinates (, ) that satisf the equation. In this chapter, we are primaril interested in a particular kind of equation in and and the graph of that equation. The equations we refer to involve and to the first power, and the are called linear equations. Definitions: An equation of the form A B C Linear Equations in which A and B cannot both be zero, is called the standard form for a line. Its graph is alwas a line. Eample 2 Graphing b Plotting Points Graph the equation 5 This is a linear equation in two variables. To draw its graph, we can begin b assigning values to and finding the corresponding values for. For instance, if 1, we have Therefore, (1, 4) satisfies the equation and is on the graph of 5.

4 THE GRAPH OF A LINEAR EQUATION SECTION NOTE If ou first rewrite an equation so that is isolated on the left side, it can be easil entered and graphed with a graphing calculator. In this case, graph the equation 5 Similarl, (2, 3), (3, 2), and (4, 1) are in the graph. Often these results are recorded in a table of values, as shown below. We then plot the points determined and draw a line through those points (1, 4) (2, 3) (3, 2) (4, 1) Ever point on the graph of the equation 5 has coordinates that satisf the equation, and ever point with coordinates that satisf the equation lies on the line. CHECK YOURSELF 2 Graph the equation 2 6. NOTE An algorithm is a sequence of steps that solve a problem. The following algorithm summarizes our first approach to graphing a linear equation in two variables. Step b Step: To Graph a Linear Equation Step 1 Find at least three solutions for the equation, and write our results in a table of values. Step 2 Graph the points associated with the ordered pairs found in step 1. Step 3 Draw a line through the points plotted above to form the graph of the equation. Two particular points are often used in graphing an equation because the are ver eas to find. The intercept of a line is the point at which the line crosses the ais. If the intercept eists, it can be found b setting 0 in the equation and solving for. The intercept is the point at which the line crosses the ais. If the intercept eists, it is found b letting 0 and solving for. NOTE Solving for, we get To graph this result on our calculator, ou can enter Y 1 (1 2) 3 using the, T, u, n ke for. Eample 3 Graphing b the Intercept Method Use the intercepts to graph the equation 2 6 To find the intercept, let The intercept is (6, 0).

5 194 CHAPTER 4 GRAPHS OF LINEAR EQUATIONS AND FUNCTIONS To find the intercept, let The intercept is (0, 3). Graphing the intercepts and drawing the line through those intercepts, we have the desired graph. 2 6 (6, 0) (0, 3) CHECK YOURSELF 3 Graph, using the intercept method The following algorithm summarizes the steps of graphing a line b the intercept method. Step b Step: Graphing b the Intercept Method Step 1 Find the intercept. Let 0, and solve for. Step 2 Find the intercept. Let 0, and solve for. Step 3 Plot the two intercepts determined in steps 1 and 2. Step 4 Draw a line through the intercepts. intercept ( 0) intercept ( 0) When can the intercept method not be used? Some lines have onl one intercept. For instance, the graph of 2 0 passes through the origin. In this case, other points must be used to graph the equation.

6 THE GRAPH OF A LINEAR EQUATION SECTION Eample 4 Graphing a Line That Passes Through the Origin NOTE Graph the equation 1 2 Note that the line passes through the origin. Graph 2 0. Letting 0 gives Thus (0, 0) is a solution, and the line has onl one intercept. We continue b choosing an other convenient values for. If 2: So (2, 1) is a solution. You can easil verif that (4, 2) is also a solution. Again, plotting the points and drawing the line through those points, we have the desired graph. 2 0 (0, 0) (2, 1) (4, 2) CHECK YOURSELF 4 Graph the equation 3 0. In Section 3.1, we defined the terms domain and range. Recall that the domain of a relation is the set of all the first elements in the ordered pairs. The range is the set of all the second elements. Recall that a line is the graph of a set of ordered pairs. In Eample 5, we will eamine the domain and range for the graph of a line. Eample 5 Finding the Domain and Range Find the domain and range for the relation described b the equation 5 We can analze the domain and range either graphicall or algebraicall. First, we will look at a graphical analsis. From Eample 2, let s look at the graph of the equation. (1, 4) (2, 3) (3, 2) (4, 1)

7 196 CHAPTER 4 GRAPHS OF LINEAR EQUATIONS AND FUNCTIONS The graph continues forever at both ends. For ever value of, there is an associated point on the line. Therefore, the domain (D) is the set of all real numbers. In set notation, we write D R This is read, The domain is the set of ever that is a real number. To find the range (R), we look at the graph to see what values are associated with. Note that ever is associated with some point. The range is written as R R This is read, The range is the set of ever that is a real number. Let s find the domain and range for the same relation b using an algebraic analsis. Look at the following equation. 5 To determine the domain, we need to find ever value of that allows us to solve for. That combination will result in an ordered pair (, ). The set of all those values is the domain of the relation. We can find a value for for an real value of. For eample, if 5, The ordered pair ( 5, 10) is part of the relation. As in our graphical analsis, the domain is D R B a similar argument, we can substitute an value for and solve the equation for. The range is R R CHECK YOURSELF 5 Find the domain and range for the relation described b the following equation. 4 Two tpes of linear equations are worth of special attention. Their graphs are lines that are parallel to the or ais, and the equations are special cases of the general form A B C in which either A 0 or B 0. Rules and Properties: 1. A line with an equation of the form k is horizontal (parallel to the ais). 2. A line with an equation of the form h is vertical (parallel to the ais). Vertical or Horizontal Lines

8 THE GRAPH OF A LINEAR EQUATION SECTION Eample 6 illustrates both cases. Eample 6 Graphing Horizontal and Vertical Lines NOTE Because part (a) is a function, it can be graphed on our calculator. Part (b) is not a function and cannot be graphed on our calculator. (a) Graph the line with equation 3 You can think of the equation in the equivalent form 0 3 Note that an ordered pair of the form (, 3) will satisf the equation. Because is multiplied b 0, will alwas be equal to 3. For instance, ( 2, 3) and (3, 3) are on the graph. The graph, a horizontal line, is shown below. 3 NOTE Notice that D 2 and R R The domain for a horizontal line is ever real number. The range is a single value. We write D R and R 3 (b) Graph the line with equation 2 In this case, ou can think of the equation in the equivalent form 0 2 Now an ordered pair of the form ( 2, ) will satisf the equation. Eamples are ( 2, 1) and ( 2, 3). The graph, a vertical line, is shown below. 2 CHECK YOURSELF 6 Graph each equation and state the domain and range. (a) 3 (b) 5

9 198 CHAPTER 4 GRAPHS OF LINEAR EQUATIONS AND FUNCTIONS CHECK YOURSELF ANSWERS 1. (a) Answers will var, but could include (0, 6); (b) Answers will var, but could include (0, 2) (2, 2) (1, 4) (0, 6) D R and 3 0 R R (0, 4) (3, 0) (0, 0) (3, 1) (6, 2) 6. (a) (b) 5 (5, 0) (0, 3) 3 D R R 3 D 5 R R

10 Name 4.1 Eercises Section Date In eercises 1 to 8, find three ordered pairs that are solutions to the given equations In eercises 9 to 26, graph each of the equations ANSWERS

11 ANSWERS

12 ANSWERS In eercises 27 to 38, find the and intercepts and then graph each equation

13 ANSWERS In eercises 39 to 46, find the domain and range of each of the relations For eercises 47 to 54, select a window that allows ou to see both the and intercepts on our calculator. If that is not possible, eplain wh not

14 ANSWERS Two distinct lines in the plane either are parallel or the intersect. In eercises 55 to 58, graph each pair of equations on the same set of aes, and find the point of intersection, where possible , , , , Graph and 2 on the same set of aes. What do ou observe? 60. Graph 2 1 and 2 1 on the same set of aes. What do ou observe? 61. Graph 2 and 2 1 on the same set of aes. What do ou observe? 62. Graph 3 1 and 3 1 on the same set of aes. What do ou observe? Graph 2 and on the same set of aes. What do ou observe? Graph and 3 2 on the same set of aes. What do 3 3 ou observe? 203

15 ANSWERS Use our graphing utilit to graph each of the following equations Write an equation whose graph will have no intercept but will have a intercept at (0, 6). 70. Write an equation whose graph will have no intercept but will have an intercept at ( 5, 0). Answers 1. (0, 5), (1, 3), ( 1, 7) 3. (0, 8), (1, 1), ( 1, 15) (0, 4), (5, 0), 1, 5 7. (0, 0), (1, 3), ( 1, 3)

16 ; intercept (0, 2); ; intercept (0, 6); intercept (4, 0) intercept (3, 0) ; intercept (0, 2); ; intercept (5, 0) intercepts: (0, 0) ; intercept (0, 2); ; intercept ( 8, 0) intercepts: (0, 0) 39. D: R ; R: R 41. D: R ; R: R 43. D: 4 ; R: R 45. D: R ; R: X ma 40, Y ma X ma 450, Y ma X min 18, Y ma An viewing window that shows the origin 205

17 55. Intersection: (5, 1) 57. Intersection: (1, 2) 59. The line corresponding to 61. The two lines appear to be parallel. 2 is steeper than that corresponding to. 63. The lines appear to be 65. perpendicular

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