11_6 - Chapter 11 Systems of Equations and Inequalities...

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Chapter 11 1438 Systems of Equations and Inequalities 11.6 Systems of Nonlinear Equations 1. y = 3 x + 2 The graph is a line. x -intercept: 0 = 3 x + 2 3 x = - 2 x = - 2 3 y -intercept: y = 3 0 ( 29 + 2 = 2 2. y = x 2 - 4 The graph is a parabola. x -intercepts: 0 = x 2 - 4 x 2 = 4 x = - 2, x = 2 y -intercept: y = 0 2 - 4 = - 4 The vertex has x -coordinate: x = - b 2 a = - 0 2 1 ( 29 = 0. The y -coordinate of the vertex is y = 0 2 - 4 = - 4. 3. y 2 = x 2 - 1 x 2 - y 2 = 1 The graph is a hyperbola. x -intercepts: 0 2 = x 2 - 1 1 = x 2 x = - 1, x = 1 y -intercepts: 0 - y 2 = 1 - y 2 = 1 y 2 = - 1, no real solution There are no y -intercepts.
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Section 11.6 Systems of Nonlinear Equations 1439 4. x 2 + 4 y 2 = 4 x 2 4 + y 2 = 1 The graph is an ellipse. x -intercepts: x 2 + 4 0 ( 29 2 = 4 x 2 = 4 x = - 2, x = 2 y -intercepts: 0 2 + 4 y 2 = 4 4 y 2 = 4 y 2 = 1 y = - 1, y = 1 5. y = x 2 + 1 y = x + 1 (0, 1) and (1, 2) are the intersection points. Solve by substitution: x 2 + 1 = x + 1 x 2 - x = 0 x ( x - 1) = 0 x = 0 or x = 1 y = 1 y = 2 Solutions: (0, 1) and (1, 2) 6. y = x 2 + 1 y = 4 x + 1 (0, 1) and (4, 17) are the intersection points. Solve by substitution: x 2 + 1 = 4 x + 1 x 2 - 4 x = 0 x ( x - 4) = 0 x = 0 or x = 4 y = 1 y = 17 Solutions: (0, 1) and (4, 17)
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Chapter 11 Systems of Equations and Inequalities 1440 7. y = 36 - x 2 y = 8 - x (2.59, 5.41) and (5.41, 2.59) are the intersection points. Solve by substitution: 36 - x 2 = 8 - x 36 - x 2 = 64 - 16 x + x 2 2 x 2 - 16 x + 28 = 0 x 2 - 8 x + 14 = 0 x = 8 ± 64 - 56 2 = 8 ± 2 2 2 = 4 ± 2 If x = 4 + 2 , y = 8 - 4 + 2 ( 29 = 4 - 2 If x = 4 - 2 y = 8 - 4 - 2 ( 29 = 4 + 2 Solutions: 4 + 2 , 4 - 2 ( 29 and 4 - 2 , 4 + 2 ( 29 8. y = 4 - x 2 y = 2 x + 4 (–2, 0) and (–1.2, 1.6) are the intersection points. Solve by substitution: 4 - x 2 = 2 x + 4 4 - x 2 = 4 x 2 + 16 x + 16 5 x 2 + 16 x + 12 = 0 ( x + 2)(5 x + 6) = 0 x =- 2 or x 6 5 y = 0 or y = 8 5 Solutions: - 2,0 ( 29 and - 6 5 , 8 5
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Section 11.6 Systems of Nonlinear Equations 1441 9. y = x y = 2 - x (1, 1) is the intersection point. Solve by substitution: x = 2 - x x = 4 - 4 x + x 2 x 2 - 5 x + 4 = 0 ( x - 4)( x - 1) = 0 x = 4 or x = 1 y = - 2 or y =1 Eliminate (4, –2); it does not check. Solution: (1, 1) 10. y = x y = 6 - x (4, 2) is the intersection point. Solve by substitution: x = 6 - x x = 36 - 12 x + x 2 x 2 - 13 x + 36 = 0 ( x - 4)( x - 9) = 0 x = 4 or x = 9 y = 2 or y = - 3 Eliminate (9, –3); it does not check. Solution: (4, 2) 11. x = 2 y x = y 2 - 2 y (0, 0) and (8, 4) are the intersection points. Solve by substitution: 2 y = y 2 - 2 y y 2 - 4 y = 0 y ( y - 4) = 0 y = 0 y = 4 x = x = 8 Solutions: (0, 0) and (8, 4)
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Chapter 11 Systems of Equations and Inequalities 1442 12. y = x - 1 y = x 2 - 6 x + 9 (2, 1) and (5, 4) are the intersection points. Solve by substitution: x 2 - 6 x + 9 = x - 1 x 2 - 7 x + 10 = 0 ( x - 2)( x - 5) = 0 x = 2 or x = 5 y = 1 or y = 4 Solutions: (2, 1) and (5, 4) 13. x 2 + y 2 = 4 x 2 + 2 x + y 2 = 0 (–2, 0) is the intersection point. Substitute 4 for x 2 + y 2 in the second equation: 2 x + 4 = 0 2 x = - 4 x 2 y = 4 - ( - 2) 2 = 0 Solution: (–2, 0) 14. x 2 + y 2 = 8 x 2 + y 2 + 4 y = 0 (–2, –2) and (2, –2) are the intersection points. Substitute 8 for x 2 + y 2 in the second equation: 8 + 4 y = 0 4 y 8 y 2 x = ± 8 - ( - 2) 2 2 Solution: (–2, –2) and (2, –2)
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Section 11.6 Systems of Nonlinear Equations 1443 15. y = 3 x - 5 x 2 + y 2 = 5 (1, –2) and (2, 1) are the intersection points.
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This homework help was uploaded on 04/07/2008 for the course MAC 1105 taught by Professor Any during the Spring '08 term at FIU.

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11_6 - Chapter 11 Systems of Equations and Inequalities...

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