Section 3.1 Equations and the Rectangular Coordinate System 199 3c. x y = ux + 2u (x, y) -4 y = u-4 + 2u = u-2u = 2 (-4, 2) -3 y = u-3 + 2u = u-1u = 1 (-3, 1) -2 y = u-2 + 2u = u0u = 0 (-2, 0) -1 y = u-1 + 2u = u1u = 1 (-1, 1) 0 y = u0 + 2u = u2u = 2 (0, 2) Now plot the solutions and connect them to form the graph. –4, 2 –3, 1 2 0, 2 –1, 1 2 4 –2 –6 –4 –2 –4 x y –4, 2 –3, 1 2 0, 2 –1, 1 2 4 –2 –6 –4 –2 –4 x y Notice the graph of y = ux + 2u has a “V-shape” pattern. Student Check 3 Graph each equation by completing a table of solutions. a. y = x - 1 b. y = x2 c. y = ux - 1u Verifying Solutions Graphically In Objective 1, we determined if an ordered pair is a solution of an equation by substituting the values of x and y in the equation. We can also examine the graph of an equation to determine if an ordered pair is a solution of an equation. If a point lies on the graph, then it is a solution of the equation. If a point does not lie on the graph, then it is not a solution of the equation. Objective 4 Examples The graph of the equation y��= x2 - 4 is provided. Use the graph to determine if each ordered pair is a solution of the equation. 4a. (-4, 0) 4b. (0, -4) 4c. (2, 0) 4d. (0, 2) 4e. (-3, 5) Solutions After plotting the points, we find that (0, -4), (2, 0), and (-3, 5) lie on the graph; thus, they are solutions of the equation. The points (-4, 0) and (0, 2) are not solutions of the equation because they do not lie on the graph. Objective 4 ▶ Determine graphically if an ordered pair is a solution of an equation. x y 4 2 4 –2 –4 4 2 –3, 5 –4 4 x –2 y 0, 2 –4, 0 2, 0 0, –4
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