Comments
Description
Transcript
Graphing Equations by Hand
Chapter 3 Introduction to Graphing 3.1 Graphing Equations by Hand 1. To plot the point A(2, −4), move 2 units to the right, then move 4 units down. The remaining points are plotted in a similar manner. y y 5 5 C(−3, 2) 2 −5 5 x −5 5 −4 −5 B(−4, −3) A(2, −4) −5 A(2, −4) 3. To plot the point A(3, −2), move 3 units to the right, then move 2 units down. The remaining points are plotted in a similar manner. 127 x CHAPTER 3. INTRODUCTION TO GRAPHING 128 y y 5 5 C(−2, 3) 3 −5 −2 5 x −5 x 5 A(3, −2) A(3, −2) B(−3, −4) −5 −5 5. To plot the point A(−2, 2), move 2 units to the left, then move 2 units up. The remaining points are plotted in a similar manner. y y 5 5 A(−2, 2) A(−2, 2) 2 −5 −2 5 x −5 5 C(2, −2) B(−2, −3) −5 x −5 7. Start at the origin, move 4 units to the left then move 3 units up to reach the point P . y 5 P (−4, 3) 3 −5 −4 5 −5 Second Edition: 2012-2013 x 3.1. GRAPHING EQUATIONS BY HAND 129 9. Start at the origin, move 3 units to the right then move 4 units down to reach the point P . y 5 3 −5 5 x −4 P (3, −4) −5 11. Substitute (x, y) = (5, 20) into the equation y = 5x − 5. y = 5x − 5 Original equation. 20 = 5(5) − 5 Substitute: 5 for x, 20 for y. 20 = 25 − 5 20 = 20 Multiply: 5(5) = 25. Subtract: 25 − 5 = 20. Because this last statement is a true statement, (5, 20) satisfies (is a solution of) the equation y = 5x − 5. For contrast, consider the point (8, 36). y = 5x − 5 36 = 5(8) − 5 Original equation. Substitute: 8 for x, 36 for y. 36 = 40 − 5 36 = 35 Multiply: 5(8) = 40. Subtract: 40 − 5 = 35. Note that this last statement is false. Hence, the pair (8, 36) is not a solution of y = 5x − 5. In similar fashion, readers should also check that the remaining two points are not solutions. 13. Substitute (x, y) = (1, 1) into the equation y = −5x + 6. y = −5x + 6 Original equation. 1 = −5(1) + 6 1 = −5 + 6 Substitute: 1 for x, 1 for y. Multiply: −5(1) = −5. 1=1 Add: −5 + 6 = 1. Second Edition: 2012-2013 CHAPTER 3. INTRODUCTION TO GRAPHING 130 Because this last statement is a true statement, (1, 1) satisfies (is a solution of) the equation y = −5x + 6. For contrast, consider the point (2, −3). y = −5x + 6 Original equation. −3 = −5(2) + 6 −3 = −10 + 6 Substitute: 2 for x, −3 for y. Multiply: −5(2) = −10. −3 = −4 Add: −10 + 6 = −4. Note that this last statement is false. Hence, the pair (2, −3) is not a solution of y = −5x+6. In similar fashion, readers should also check that the remaining two points are not solutions. 15. Substitute (x, y) = (7, 395) into the equation y = 8x2 + 3. y = 8x2 + 3 2 Original equation. 395 = 8(7) + 3 Substitute: 7 for x, 395 for y. 395 = 8(49) + 3 Square: (7)2 = 49. 395 = 392 + 3 395 = 395 Multiply: 8(49) = 392. Add: 392 + 3 = 395. Because this last statement is a true statement, (7, 395) satisfies (is a solution of) the equation y = 8x2 + 3. For contrast, consider the point (1, 12). y = 8x2 + 3 2 Original equation. 12 = 8(1) + 3 Substitute: 1 for x, 12 for y. 12 = 8(1) + 3 Square: (1)2 = 1. 12 = 8 + 3 12 = 11 Multiply: 8(1) = 8. Add: 8 + 3 = 11. Note that this last statement is false. Hence, the pair (1, 12) is not a solution of y = 8x2 + 3. In similar fashion, readers should also check that the remaining two points are not solutions. 17. Substitute (x, y) = (8, 400) into the equation y = 6x2 + 2x. y = 6x2 + 2x 2 Original equation. 400 = 6(8) + 2(8) Substitute: 8 for x, 400 for y. 400 = 6(64) + 2(8) 400 = 384 + 16 Square: (8)2 = 64. Multiply: 6(64) = 384; 2(8) = 16 400 = 400 Add: 384 + 16 = 400. Second Edition: 2012-2013 3.1. GRAPHING EQUATIONS BY HAND 131 Because this last statement is a true statement, (8, 400) satisfies (is a solution of) the equation y = 6x2 + 2x. For contrast, consider the point (2, 29). y = 6x2 + 2x Original equation. 2 29 = 6(2) + 2(2) Substitute: 2 for x, 29 for y. 29 = 6(4) + 2(2) 29 = 24 + 4 Square: (2)2 = 4. Multiply: 6(4) = 24; 2(2) = 4 29 = 28 Add: 24 + 4 = 28. Note that this last statement is false. Hence, the pair (2, 29) is not a solution of y = 6x2 + 2x. In similar fashion, readers should also check that the remaining two points are not solutions. 19. First, complete the table of points that satisfy the equation y = 2x + 1. y y y y y y y x −3 −2 −1 0 1 2 3 = 2(−3) + 1 = −5 = 2(−2) + 1 = −3 = 2(−1) + 1 = −1 = 2(0) + 1 = 1 = 2(1) + 1 = 3 = 2(2) + 1 = 5 = 2(3) + 1 = 7 y = 2x + 1 −5 −3 −1 1 3 5 7 (x, y) (−3, −5) (−2, −3) (−1, −1) (0, 1) (1, 3) (2, 5) (3, 7) Next, plot the points in the table, as shown in the image on the right. This first image gives us enough evidence to believe that if we plotted all points satisfying the equation y = 2x + 1, the result would be the graph on the right. y y 10 −10 10 10 −10 x −10 y = 2x + 1 10 −10 Second Edition: 2012-2013 x CHAPTER 3. INTRODUCTION TO GRAPHING 132 21. First, complete the table of points that satisfy the equation y = |x − 5|. y y y y y y y x 2 3 4 5 6 7 8 = |2 − 5| = 3 = |3 − 5| = 2 = |4 − 5| = 1 = |5 − 5| = 0 = |6 − 5| = 1 = |7 − 5| = 2 = |8 − 5| = 3 y = |x − 5| 3 2 1 0 1 2 3 (x, y) (2, 3) (3, 2) (4, 1) (5, 0) (6, 1) (7, 2) (8, 3) Next, plot the points in the table, as shown in the image on the right. This first image gives us enough evidence to believe that if we plotted all points satisfying the equation y = |x − 5|, the result would be the graph on the right. y y 10 10 y = |x − 5| −10 10 x −10 −10 10 −10 23. First, complete the table of points that satisfy the equation y = (x + 1)2 . y y y y y y y 2 = (−4 + 1) = 9 = (−3 + 1)2 = 4 = (−2 + 1)2 = 1 = (−1 + 1)2 = 0 = (0 + 1)2 = 1 = (1 + 1)2 = 4 = (2 + 1)2 = 9 x −4 −3 −2 −1 0 1 2 y = (x + 1)2 9 4 1 0 1 4 9 (x, y) (−4, 9) (−3, 4) (−2, 1) (−1, 0) (0, 1) (1, 4) (2, 9) Next, plot the points in the table, as shown in the image on the left. This first image gives us enough evidence to believe that if we plotted all points Second Edition: 2012-2013 x 3.1. GRAPHING EQUATIONS BY HAND 133 satisfying the equation y = (x + 1)2 , the result would be the graph on the right. y y = (x + 1)2 10 −10 10 x y 10 −10 −10 10 −10 25. First, complete the table of points that satisfy the equation y = −x − 5. y y y y y y y = −(−3) − 5 = −2 = −(−2) − 5 = −3 = −(−1) − 5 = −4 = −(0) − 5 = −5 = −(1) − 5 = −6 = −(2) − 5 = −7 = −(3) − 5 = −8 x −3 −2 −1 0 1 2 3 y = −x − 5 −2 −3 −4 −5 −6 −7 −8 (x, y) (−3, −2) (−2, −3) (−1, −4) (0, −5) (1, −6) (2, −7) (3, −8) Next, plot the points in the table, as shown in the image on the right. This first image gives us enough evidence to believe that if we plotted all points satisfying the equation y = −x − 5, the result would be the graph on the right. Second Edition: 2012-2013 x CHAPTER 3. INTRODUCTION TO GRAPHING 134 y y 10 10 −10 10 x −10 10 −10 −10 x y = −x − 5 27. Enter the equation y = x2 − 6x + 5 in Y1 in the Y= menu, using the following keystrokes. ∧ X,T,θ,n 2 − 6 × X,T,θ,n + 5 ENTER Press 2ND WINDOW, then set TblStart = 0 and ΔTbl = 1. Press 2ND GRAPH to produce the table. Enter the data from your calculator into a table on your graph paper. Use a ruler to construct the axes of your coordinate system, label and scale each axis. Plot the points in the table, as shown in the image on the right. y 10 x y (x, y) 0 1 2 3 4 5 6 5 0 −3 −4 −3 0 5 (0, 5) (1, 0) (2, −3) (3, −4) (4, −3) (5, 0) (6, 5) −10 10 −10 Second Edition: 2012-2013 x 3.1. GRAPHING EQUATIONS BY HAND 135 The plotted points provide enough evidence to convince us that if we plotted all points that satisfied the equation y = x2 −6x+5, we would get the following result. y 10 y = x2 − 6x + 5 −10 10 x −10 29. Enter the equation y = −x2 + 2x + 3 in Y1 in the Y= menu using the following keystrokes. (-) X,T,θ,n ∧ 2 + 2 × X,T,θ,n + 3 ENTER Press 2ND WINDOW, then set TblStart = −2 and ΔTbl = 1. Press 2ND GRAPH to produce the table. Enter the data from your calculator into a table on your graph paper. Use a ruler to construct the axes of your coordinate system, label and scale each axis. Plot the points in the table, as shown in the image on the right. Second Edition: 2012-2013 CHAPTER 3. INTRODUCTION TO GRAPHING 136 y 10 x y (x, y) −2 −1 0 1 2 3 4 −5 0 3 4 3 0 −5 (−2, −5) (−1, 0) (0, 3) (1, 4) (2, 3) (3, 0) (4, −5) −10 10 x −10 The plotted points provide enough evidence to convince us that if we plotted all points that satisfied the equation y = −x2 +2x+3, we would get the following result. y 10 −10 10 x y = −x2 + 2x + 3 −10 31. Enter the equation y = x3 − 4x2 − 7x + 10 in Y1 in the Y= menu using the following keystrokes. X,T,θ,n ∧ 3 − × 4 + 1 X,T,θ,n 0 ∧ 2 − 7 × X,T,θ,n ENTER Press 2ND WINDOW, then set TblStart = −3 and ΔTbl = 1. Press 2ND GRAPH to produce the table. Second Edition: 2012-2013 3.1. GRAPHING EQUATIONS BY HAND 137 Enter the data from your calculator into a table on your graph paper. Use a ruler to construct the axes of your coordinate system, label and scale each axis. Plot the points in the table, as shown in the image on the right. y x y (x, y) −3 −2 −1 0 1 2 3 4 5 6 −32 0 12 10 0 −12 −20 −18 0 40 (−3, −32) (−2, 0) (−1, 12) (0, 10) (1, 0) (2, −12) (3, −20) (4, −18) (5, 0) (6, 40) 50 −10 10 −50 The plotted points provide enough evidence to convince us that if we plotted all points that satisfied the equation y = x3 − 4x2 − 7x + 10, we would get the following result. y 50 y = x3 − 4x2 − 7x + 10 −10 10 x −50 Second Edition: 2012-2013 x CHAPTER 3. INTRODUCTION TO GRAPHING 138 33. Enter the equation y = −x3 + 4x2 + 7x − 10 in Y1 in the Y= menu using the following keystrokes. (-) X,T,θ,n ∧ 3 + − × 4 1 0 X,T,θ,n ∧ 2 + 7 × ENTER Press 2ND WINDOW, then set TblStart = −3 and ΔTbl = 1. Press 2ND GRAPH to produce the table. Enter the data from your calculator into a table on your graph paper. Use a ruler to construct the axes of your coordinate system, label and scale each axis. Plot the points in the table, as shown in the image on the right. x y (x, y) −3 −2 −1 0 1 2 3 4 5 6 32 0 −12 −10 0 12 20 18 0 −40 (−3, 32) (−2, 0) (−1, −12) (0, −10) (1, 0) (2, 12) (3, 20) (4, 18) (5, 0) (6, −40) y 50 −10 10 −50 The plotted points provide enough evidence to convince us that if we plotted all points that satisfied the equation y = −x3 + 4x2 + 7x − 10, we would get the following result. Second Edition: 2012-2013 x X,T,θ,n 3.1. GRAPHING EQUATIONS BY HAND 139 y 50 −10 10 x y = −x3 + 4x2 + 7x − 10 −50 √ 35. First, enter the equation y = x + 5 into Y1 in the Y= menu using the following keystrokes. The results are shown in the first image below. 2ND √ X,T,θ,n + 5) ENTER Next, select 2ND WINDOW and adjust the settings to match those in the second image. below. Finally, select 2ND GRAPH to open the TABLE. Copy the results from the third image below into the tables on your graph paper. x −5 −4 −3 −2 −1 0 1 2 y= √ x+5 0 1 1.4 1.7 2 2.2 2.4 2.6 (x, y) (−5, 0) (−4, 1) (−3, 1.4) (−2, 1.7) (−1, 2) (0, 2.2) (1, 2.4) (2, 2.6) x 3 4 5 6 7 8 9 10 y= √ x+5 2.8 3 3.2 3.3 3.5 3.6 3.7 3.9 (x, y) (3, 2.8) (4, 3) (5, 3.2) (6, 3.3) (7, 3.5) (8, 3.6) (9, 3.7) (10, 3.9) Next, plot the points in the tables, as shown in the image on the left. This first image gives us enough√evidence to believe that if we plotted all points satisfying the equation y = x + 5, the result would be the graph on the right. Second Edition: 2012-2013