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Purdue MA 11100 - Lecture Notes
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1Graphs• Points and Ordered Pairs• Quadrants• Solutions of Equations• Nonlinear Equations2.1(2, 4)(0, 0) origin(4, 2)-5 -4 -3 -2 -1 1 2 3 4 565423-41-2-1-3Points and Ordered PairsxyFirst axisSecond axisLabel points (x, y). Order is important.-5-6Plot the points (3, 4), (–2, 0), (4.5, –2) and (–1, –5 ). ExampleSolution(–1, –5)(3, 4)(–2, 0)(4.5, –2)65423-41-2-1-3-5 -4 -3 -2 -1 1 2 3 4 5-5QuadrantsThe axes divide the plane into four regions called quadrants, as shown below.First quadrantSecond quadrantThird quadrantFourth quadrantI IV III II xyWhich quadrant?• (-2, 4)• (5, 2)• (-6, -1)• (-2, 10)• (0, 4)• (-2, 0)Solutions of EquationsIf an equation has two variables, its solutions are pairs of numbers. When such a solution is written as an ordered pair, the first number listed in the pair generally replaces the variable that occurs first alphabetically.2ExampleSolutionDetermine whether (2, 5) and (–2, 1) are solutions to y = 2x + 1. y = 2x + 15 2(2) + 14 + 15 = 5True, so (2, 5) is a solution.y = 2x + 11 2(–2) + 1–4 + 11 = –3False, so (–2 , 1) isnot a solution.Solutions?Yes 666?606?)2(3)0(2No 636?366?)1(3)3(2Yes 666?246?323)2(2=++≠−+−+−=++)2,0()1,3()32,2(−Are the following ordered pairs solutions of the equation a)b)c)?632=+yxSolutions??223+= mn−25,21)0,34Are the following solutions of a)b)No 41125243?2522123?25Yes 0022?023423?0≠++=+−+−ExampleGraph the equation y = x + 1. (0,1)(3, 4)(–2, –1 )(x, y)14–1 03–2 yxSolution(3, 4)(0,1)xy(–2, –1)-2 -1 1 2 3 4 5 6612345-2-1ExampleGraph the equation1.2y x= −(0,0)(–4, 2)(4, –2 )(x, y)02–2 0–44yxSolution(0,0)xy(–4, 2)(4,–2 )-5 -4 -3 -2 -1 1 2 3 4 5-32-23-11123+−= xyx y02-241-24-5yx31253=−yxx y042-12/50-6/5yxNonlinear EquationsThere are many equations for which the graph is not a straight line. Graphing these nonlinear equations often requires plotting many points to see the general shape of the graph. We refer to any equation whose graph is a straight line as a linear equation.ExampleGraph the equation21.y x= −(0, –1)(1, 0)(–1, 0)(2, 3)(–2, 3)(x, y)–10033 01–12–2yxSolution(0,-1)xy(1,0)(-1,0 )(2,3)(-2,3)-5 -4 -3 -2 -1 1 2 3 4 5436251-3-1ExampleGraph the equation3.y x= +(0, 3)(1, 4)(–1, 4)(2, 5)(–2, 5)(x, y)34455 01–12–2yxSolution(0,3)xy(1,4)(-1,4 )(2,5)-5 -4 -3 -2 -1 1 2 3 4 51-34-2-165(-2,5)1212−= xyx y01-12-23-1-1/2-1/2117/2yx1−= xyx y01-12-23102132xy4Using CourseCompass• Plotting Points• Finding Solutions•


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Purdue MA 11100 - Lecture Notes

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