HORIZONTAL AND VERTICAL LINES
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1 the graph of the equation AlgebraDate 4.2 Notes: Graphing Linear Equations In Lesson (pp ) ou saw examples of linear equations in one variable. The solution of A an solution equation of such an equation as 2x º 1 = 3 is a real number. Its graph is a point on the real makes number the equation line. The true. equation The graph + 2 = of 3xan in Example equation2 is a linear equation in two variables. Its graph is a straight line. in two variables x and is an ordered pair (x, ) that in x and is the set of all points (x, ) that are solutions of the equation. In this lesson ou will see that the graph of a linear equation is a line. G R APHING A LINEAR EQUATION GOAL 2 HORIZONTAL AND VERTICAL LINES STEP 1 Rewrite the equation in function form, if necessar. All STEP linear 2equations Choose in a xfew and values can be of written x and make in the a form table Ax of + values. B = C. When A = 0 the equation reduces to B = C and the graph is a horizontal line. When STEP 3 Plot the points from the table of values. A line through these B = 0 the equation reduces to Ax = C and the graph is a vertical line. points is the graph of the equation. E Q UATIONS OF HORIZONTAL AND VERTICAL LINES b x x a x In the coordinate plane, the graph of = b is a horizontal line. In the coordinate plane, the graph of x = a is a vertical line. E Q UATIONS OF HORIZONTAL AND VERTICAL LINES EXAMPLE 5 Graphing = b Graph the equation = 2. The equation does not have x as a variable. The -value is alwas 2, regardless of the value of x. For instance, here are some points ( 3, 2) 3 (0, 2) (3, 2)
2 EXAMPLE 1 Verifing Solutions of an Equation Use the graph to decide whether the point lies on the graph of x + 3 = 6. Justif our answer algebraicall. a. (1, 2) b. (º3, 3) x a. The point (1, 2) is not on the graph of x + 3 = 6. This means that (1, 2) is not a solution. You can check this algebraicall. x + 3 = 6 Write original equation (2) 6 Substitute 1 for x and 2 for. 7 6 Simplif. Not a true statement (1, 2) is not a solution of the equation x + 3 = 6, so it is not on the graph. b. The point (º3, 3) is on the graph of x + 3 = 6. This means that (º3, 3) is a solution. You can check this algebraicall. x + 3 = 6 Write original equation. º3 + 3(3) 6 Substitute º3 for x and 3 for. 6 = 6 Simplif. True statement (º3, 3) is a solution of the equation x + 3 = 6, so it is on the graph
3 Use the graph to decide whether the point lies on the graph of 4x + = 8. Justif our answer algebraicall. 1. (1, 4) 2. (3, 4)
4 EXAMPLE 2 Graphing an Equation Use a table of values to graph the equation + 2 = 3x. Rewrite the equation in function form b solving for. + 2 = 3x = 3x º 2 Write original equation. Subtract 2 from each side. Choose a few values for x and make a table of values. Choose x. With this table of values ou have found the five solutions (º2, º8), (º1, º5), (0, º2), (1, 1), and (2, 4). Plot the points. Note that the appear to lie on a straight line. The line through the points is the graph of the equation Substitute to find the corresponding -value. º2 = 3(º2) º 2 = º8 º1 = 3(º1) º 2 = º5 0 = 3(0) º 2 = º2 1 = 3(1) º 2 = 1 2 = 3(2) º 2 = x
5 3. Use a table of values to graph the equation 5x = 2. Use the graph to decide whether the point lies on the graph of x 2 = 5. Justif our answer algebraicall. 4. (3, -4) 5. (5, 0)
6 EXAMPLE 3 Graphing a Linear Equation Use a table of values to graph the equation 3x + 2 = 1. 1 Rewrite the equation in function form b solving for. This will make it easier to make a table of values. 3x + 2 = 1 2 = º3x + 1 Write original equation. Subtract 3x from each side. = º 3 2 x + Divide each side b Choose a few values of x and make a table of values. Choose x. º2 0 2 Evaluate º 5 2 With this table of values ou have found three solutions. º2, 7 2, 0, 1 2, 2, º Plot the points and draw a line through them. The graph of 3x + 2 = 1 is shown at the right x
7 6. Use a table of values to graph the equation 3x 2 = 6. EXAMPLE 4 Using the Graph of a Linear Model An Internet Service Provider estimates that the number of households h (in millions) with Internet access can be modeled b h = 6.76t , where t represents the number of ears since Graph this model. Describe the graph in the context of the real-life situation. Make a table of values. Use 0 t 6 for 1996º2002. t h Households (millions) h Internet Access Years since 1996 t From the table and the graph, ou can see that the number of households with Internet access is projected to increase b about 7 million households per ear.
8 7. The number n of U.S. citizens (in millions) who exercise walk can be modeled b n = 1.38t where t represents the number of ears since Sketch a graph of this model. Describe the graph in the context of the real-life situation. 8. The number of U.S. citizens (in millions) who in-line skate is n = 3.47t where t represents the number of ears since Sketch a graph of this model. Describe the graph of this line in the context of the real-life situation.
9 EXAMPLE 5 Graphing = b STUDENT HELP Graph the equation = 2. Stud Tip The equations = 2 and 0x + 1 = 2 are equivalent. For an value of x, the ordered pair (x, 2) is a solution of = 2. The equation does not have x as a variable. The -value is alwas 2, regardless of the value of x. For instance, here are some points that are solutions of the equation: (º3, 2), (0, 2), (3, 2). 3 ( 3, 2) 1 (0, 2) (3, 2) x The graph of the equation is a horizontal line 2 units above the x-axis. 9. Sketch the graph of = -1. EXAMPLE 6 Graphing x = a Graph the equation x = º3. The x-value is alwas º3, regardless of the value of. For instance, here are some points that are solutions of the equation: (º3, º2), (º3, 0), (º3, 3). x 3 ( 3, 3) 1 ( 3, 0) 1 x The graph of the equation is a vertical line 3 units to the left of the -axis. ( 3, 2)
10 10. Sketch the graph of x = Describe the graph of = Describe the graph of x = Which point does not lie on the graph of = 3? A. (-3, 3) B. (-1, 3) C. (0, 3) D. (3, -3) D. (3, 3) 14. Complete the following sentence: An ordered pair that makes an equation in two variables true is called a(n).
11 15. True or false: The graph of the equation x = 3 is a horizontal line. Explain our reasoning. Use a table of values to graph the equation x 3 = 12
12 17. x = = -2
13 Tell whether the point is a solution of the equation 4x - = (-1, 3) 20. (1, 3) 21. (1, 0) 22. (1, 4)
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