940 CHAPTER 9 Systems and Matrices CAUTION A linear inequality must be in slope-intercept form (solved for y) to determine, from the presence of a * symbol or a + symbol, whether to shade the lower or the upper half-plane. In Figure 17, the upper half-plane is shaded, even though the inequality is 3x - 2y … 6 (with a 6 symbol) in standard form. Only when we write the inequality as y Ú 3 2 x - 3 Slope-intercept form does the 7 symbol indicate to shade the upper half-plane. Nonlinear Inequalities in Two Variables If a nonlinear inequality has a boundary graph that is a function, we use the same methods to graph the solution set as when graphing the solution set of a linear inequality. If a nonlinear inequality has a boundary graph that is not a function, or cannot be solved for y, then using a test point is recommended. EXAMPLE 2 Graphing a Nonlinear Inequality in Two Variables Graph y 6 -1x + 122 + 3. SOLUTION The boundary of the graph is the parabola y = -1x + 122 + 3. This parabola has vertex 1-1, 32 and opens downward. Because points on this curve do not satisfy the inequality, it is customary to make the boundary dashed, as in Figure 18. Because the inequality y 6 -1x + 122 + 3 is already solved for y, we determine from the presence of the 6symbol that the solution set is the region below the boundary. See Figure 18. As an alternative method for determining which region to shade, or as a check, we can choose a test point not on the boundary and substitute into the inequality. The point 10, 02 is a good choice if it does not lie on the boundary, because the substitution is easily done. y 6 -1x + 122 + 3 Original inequality 0 6 -10 + 122 + 3 Use 10, 02 as a test point. 0 6 -1 + 3 Simplify: -10 + 122 = -1122 = -1. 0 62 True The test point 10, 02 lies below the boundary, so all points that satisfy the inequality must also lie below the boundary. This agrees with the result above. If we were to substitute the coordinates of a point above the boundary, such as 12, 32, a false statement would result. y 6 -1x + 122 + 3 Original inequality 3 6 -12 + 122 + 3 Use 12, 32 as a test point. 3 6 -9 + 3 Simplify: -12 + 122 = -1322 = -9. 3 6 -6 False S Now Try Exercise 21. Recall that an inequality containing 6 or 7 is a strict inequality and does not include the boundary in its solution set. This is indicated with a dashed boundary, as shown in Example 2. A nonstrict inequality contains … or Ú and does include its boundary in the solution set. This is indicated with a solid boundary, as shown in Example 1. −7 6 −6 6 To graph the inequality in Example 2 using a graphing calculator, solve for y, and then direct the calculator to shade below the boundary, y = -1x + 122 + 3. y < –(x + 1)2 + 3 x 2 –4 –1 –3 3 y 0 (0, 0) (2, 3) Figure 18
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