06 CHAPTER Algebra. GRAPHING EQUATIONS AND INEQUALITIES Tetbook Reference Section 6. &6. CLAST OBJECTIVE Identif regions of the coordinate plane that correspond to specific conditions and vice-versa Graphing Equations Case : The graph of = a, where a is a constant, is a vertical line. Eamples a) Graph: = Solutions - - - - 0 - - - - Locate the point (,0) on the ais and draw a vertical line through the point. b) Graph: = - - - - - 0 - - - - Locate the point (-,0) on the ais and draw a vertical line through the point. c) Graph: = 0 - - - - 0 - - - - Locate the point (0, 0) on the ais and draw a vertical line through the point. Notice that this line is the same as the ais. Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities 07 Case : The graph of = b, where b is a constant, is a horizontal line. Eamples d) Graph =. - - - - 0 - - - - Solutions Locate the point (0,) on the ais and draw a horizontal line through the point. e) Graph = -. - - - - 0 - - - - Locate the point (0,-) on the ais and draw a horizontal line through the point. Case : The graph of = m + b, where m 0, is a line of slope m and -intercept b. rise Slope (m) = run Y-intercept (b) is the point (0, b) where the line crosses or intersects the ais. To use the slope and the intercept to graph a line, be sure to write the equation of the line in slope-intercept form, = m + b, if possible. Eample Eplanation f) Graph = +. Note that the equation of the line is alread in slope-intercept form with m = = and b =. 7 6 - - - - 0. Graph the -intercept (0, ).. Now use the slope to find other points on the line. From the -intercept move up three places and one place to the right. Place a point there.. From the nd point, use the slope again to get another point on the line. Move up three places and one place to the right. Place a point there.. Now draw the line using the three points.. To obtain more points on the line, proceed as in step. Houghton Mifflin Compan. All rights reserved.
08 CHAPTER Algebra Eamples Eplanations g) Graph + =. First, rewrite the equation in slope-intercept form b solving for. - - - - 0 - - - - Note that m = + = - - = - + = + and b =.. Graph the -intercept (0, ).. Now use the slope to find other points on the line. From the -intercept move down two places and three places to the right. Place a point there.. Now draw the line using the two points.. To obtain more points on the line, proceed as in step.. To get a point to the left of the -intercept, start at the - intercept and move up two spaces and to the left three spaces. Place a point there. h) What is an equation of the line below? - - - - 0 - - - - From the graph, we can determine that the -intercept is (0, -). Thus b = -. Net, we must determine how the net point was obtained. From the -intercept, if we move up (rise) and to the right (run), we get the point shown on the graph. Therefore, slope (m) is. We now have the necessar information to write an equation of the line in slope-intercept form. b = - and m = = All these equations are equivalent. ( ) = () = 6 +6 + 6 + 6 = - - 6 = same as = 6 same as - + = -6 Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities 09 Check Your Progress. Graph each of the following lines.. =. = -. =. = -6. = 6. = + Houghton Mifflin Compan. All rights reserved.
0 CHAPTER Algebra 7. = 0 8. = For Questions 9 and 0, write an equation of the line shown. 9. - - - 0 - - - - - -6 0. - - - - 0 - - - - Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities Graphing Inequalities. Graph the corresponding equation. (The corresponding equation is formed b replacing the inequalit smbol with an equal sign.) If the smbols < or > appear in the inequalit, use a dashed line when graphing the corresponding equation. If the smbols or appear in the inequalit, use a solid line when graphing the corresponding equation. The graph of the corresponding equation separates the grid into two regions, one of which will be shaded to satisf the original inequalit.. Test the inequalit and shade onl the region which makes the inequalit true. Begin b choosing a point that doesn t lie on the graphed line. (If the line does not go through the origin, the point (0, 0) is the easiest one to test the inequalit.) If the inequalit is satisfied, shade the region that includes the point. If the inequalit is not satisfied, shade the other region. Eamples i) Graph the inequalit: Solution - - - - 0 - - - - Eplanations. First, draw the graph of the equation =. Use a solid line because the inequalit smbol is.. Notice that the line = divides the grid into two regions. One region lies to the right of the line. The other region lies to the left of the line. Now choose a point to test the inequalit. We can use (0, 0). In testing this inequalit, we use onl the -value since the inequalit does not contain. Substitute 0 for. Is 0 a true statement? The answer is es. Thus, the region to the left of the line satisfies the inequalit and will be shaded. To show that the region to the right of the line = does not satisf the inequalit, let s test the point (, ). Substitute for. Is a true statement? The answer is no. This means that no point ling in the region to the right of the line = will satisf the inequalit. j) Graph: < Solution - - - - 0 - - - -. First, draw the graph of the equation =. Use a dashed line because the inequalit smbol is <.. Notice that the line = divides the grid into two regions. One region lies above the line. The other region lies below the line. Now choose a point to test the inequalit. We can use (0, ). In testing this inequalit, we use onl the -value since the inequalit does not contain. Substitute for. Is < a true statement? The answer is no, which indicates that the region above the line does not satisf the inequalit. Thus, the region below the line satisfies the inequalit and will be shaded. Houghton Mifflin Compan. All rights reserved.
CHAPTER Algebra Eamples k) Graph: - and Solution - - - - 0 - - - - Eplanations. First, draw the graphs of the equations = - and =. Use a solid line because the inequalit smbols are and.. Notice that the lines divide the grid into four regions. We will test all four regions. Note that both inequalities must be satisfied. We can use the points: (0, 0), (, ), (-, -), and (-, ). Test Point - (0, 0) 0 - False 0 False (, ) - False True (-, -) - - True - False (-, ) - - True True Shade the region for which both inequalities are true. Note that we can also use the intersection of sets to determine the solution. In which case, we would shade onl the region that includes all points where 0 and. l) Graph: + > Solution - - - - 0 - - - -. First, draw the graph of the equation + =. Use a dashed line because the inequalit smbol is >.. Now choose a point to test the inequalit. We can use (0, 0). Substitute 0 for and 0 for. Is 0 + 0 > a true statement? The answer is no, which indicates that this region containing the point (0, 0) does not satisf the inequalit. Thus, the other region satisfies the inequalit and will be shaded. m) Graph: 0; 0; and + < 6 Solution - - - - 0 - - - -. First, draw the graphs of the corresponding equations. Use a dashed line when graphing + =6 because the inequalit smbol is <.. Note that we can use the intersection of sets to determine the solution. The inequalit 0 is true for all points to the right of the -ais. The inequalit 0 is true for all points above the - ais. The inequalit + < 6 is true for all points below the line + = 6. The intersection is shaded. When using test points, note that there are seven regions to test. Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities Eample n) Give the conditions for the shaded region. Solution - - - - 0 - - - - 0; 0; > - 6 Eplanation. Find all the boundaries of the shaded region. The vertical bound is = 0 and its shading is 0. The horizontal bound is = 0 and its shading is 0. The dashed line has -intercept at (0, ) and slope ¾. Thus, the equation of the line is = ¾ +. Rewriting the equation in standard form we have = -6 and its shading is > -6. = + ( ) = + ( ) = + 6 6 6 = 6 6 =. List the conditions: 0; 0; > -6 Check Your Progress.. Graph:. Graph: < -. Graph: > -. Graph: Houghton Mifflin Compan. All rights reserved.
CHAPTER Algebra. Graph: + 6. Graph: > - 7. Graph: > -6 8. Graph: + - Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities For Questions 9 and 0, give the conditions for the shaded regions. 9. 0. - - - - 0 - - - - - - - - 0 - - - - For Questions and, give the conditions for the shaded regions... - - - - 0 - - - - - - - - 0 - - - - Houghton Mifflin Compan. All rights reserved.
6 CHAPTER Algebra See If You Remember SECTIONS... Simplif: [ ( ) ]. Write an equivalent epression: ( + z ). Write an equivalent epression: + ( + ). Name the propert: + 0 =. ( 0 ) ( 7 0 ) 6. 0.000, 00 7. Simplf: 6 8. Solve: ( ) + Houghton Mifflin Compan. All rights reserved.
SECTION. Graphing Equations and Inequalities 7 9. Find f ( - ) if f () = + 0. If A = LW and L = and W = 6., find A.. Is - a solution to + < 7?. Write in scientific notation: 0.000000000. In a vacuum, light travels at a speed of 99,79,8 meters per second. Write this number in scientific notation.. Find f ( ) if f ( ) = +. Simplif: π + π 6. Write in standard form: 8.9 0 Houghton Mifflin Compan. All rights reserved.