3x - 79 + _ b. 6x - 5x
Lesson
38
Warm Up 38 1. inequality 2. x2 + 10x + 21 3. 65 Lesson Practice 38 40 a. 4x2 - 3x - 79 + _ 3x
b. 6x3 - 5x2 - 49x + 60 c. 2x2 + 17x − 84 d. Yes e. No 15x2 + 75x _ f. 62x + 160 ; it has a nonzero remainder.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–1
Saxon Algebra 2
Lesson Practice 38 1. y = (x + 10)2 2. y = (x - 1)2 - 3 3. x2 - 15x + 6 ⎡-.25 4. X = ⎢ ⎣ .5
-.125⎤ -.25 ⎦
5. Yes 6. No 2 -x - 6x - 7 7. __ 3 2
x + 6x + 11x + 6
8. x = 3, y = 2 9. 110
38
13. 1 and only 1; Possible explanation: The graph of a quadratic function is a parabola that opens either up or down and extends without end to the left and to the right. So for any quadratic function, the graph intersects the y-axis in at least one point. And, the graph intersects the y-axis in at most one point because no vertical line can intersect the graph of a function in more than one point.
10. 5.5 11. Slope: -0.3; y-intercept: -0.1 12. Direct variation; The equation is C = πd.
14. The remainder is zero because the polynomial is equivalent to (x + a)(x + b). 15. $16: 35; $24: 5; $32: 4 16. Student B incorrectly subtracts 36x instead of adding. 17. (2, -6) 18. B
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–2
Saxon Algebra 2
Lesson 19.
4
y
22. a.
2 y = -_ x+2 3
-4
-2
O
38
7x + y = 5 14x + 2y = 10
x 2
b. Possible answer: infinitely many solutions, all the like terms would become opposites if the first equation was multiplied by -2
3 y =_ x-1 2
-4
3 2 a. _ and - _ 2 3
b. -1 c. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. 3 r _ = = 1.5 20. _ 2 2
21. The student incorrectly reasoned that the events draw an ace and draw a spade are mutually exclusive. The events are not mutually exclusive because the ace of spades is both an ace and a spade. The sum 4 + 13 includes the ace of spades twice. To get the correct answer, subtract 1 from that sum. The correct answer is 16.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
c. infinitely many solutions, consistent and dependent 23. 205 ninth graders, 190 tenth graders, 180 eleventh graders
LSN 38–3
Saxon Algebra 2
Lesson 24.
A change in the value of h shifts the parent function to the right when h is positive and to the left when h is negative. A change in the value of k shifts the parent function up when k is positive and down when k is negative. When the absolute value of a is greater than 1, the graph is stretched away from the x-axis, appearing narrower than the parent function. When the absolute value of a is between 0 and 1, the graph is compressed towards the x-axis, appearing wider than the parent function.
38
25. When a polynomial is factored and is equal to zero, since the product of the terms equals zero, at least one of its terms equals zero. Setting each term containing a variable equal to zero enables you to solve for a possible value of the variable. 1.5d 26. _ 2 t + 1.5t
27. The point (0, 3) is the highest point on the parabola; the y-values do not extend above 3. 8
y
4 x -8
-4
4
8
-4 -8
28. C 29. y = 3x + 29.5 1 30. _ 36
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–4
Saxon Algebra 2
38
Warm Up 38 1. inequality 2. x2 + 10x + 21 3. 65 Lesson Practice 38 40 a. 4x2 - 3x - 79 + _ 3x
b. 6x3 - 5x2 - 49x + 60 c. 2x2 + 17x − 84 d. Yes e. No 15x2 + 75x _ f. 62x + 160 ; it has a nonzero remainder.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–1
Saxon Algebra 2
Lesson Practice 38 1. y = (x + 10)2 2. y = (x - 1)2 - 3 3. x2 - 15x + 6 ⎡-.25 4. X = ⎢ ⎣ .5
-.125⎤ -.25 ⎦
5. Yes 6. No 2 -x - 6x - 7 7. __ 3 2
x + 6x + 11x + 6
8. x = 3, y = 2 9. 110
38
13. 1 and only 1; Possible explanation: The graph of a quadratic function is a parabola that opens either up or down and extends without end to the left and to the right. So for any quadratic function, the graph intersects the y-axis in at least one point. And, the graph intersects the y-axis in at most one point because no vertical line can intersect the graph of a function in more than one point.
10. 5.5 11. Slope: -0.3; y-intercept: -0.1 12. Direct variation; The equation is C = πd.
14. The remainder is zero because the polynomial is equivalent to (x + a)(x + b). 15. $16: 35; $24: 5; $32: 4 16. Student B incorrectly subtracts 36x instead of adding. 17. (2, -6) 18. B
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–2
Saxon Algebra 2
Lesson 19.
4
y
22. a.
2 y = -_ x+2 3
-4
-2
O
38
7x + y = 5 14x + 2y = 10
x 2
b. Possible answer: infinitely many solutions, all the like terms would become opposites if the first equation was multiplied by -2
3 y =_ x-1 2
-4
3 2 a. _ and - _ 2 3
b. -1 c. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. 3 r _ = = 1.5 20. _ 2 2
21. The student incorrectly reasoned that the events draw an ace and draw a spade are mutually exclusive. The events are not mutually exclusive because the ace of spades is both an ace and a spade. The sum 4 + 13 includes the ace of spades twice. To get the correct answer, subtract 1 from that sum. The correct answer is 16.
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
c. infinitely many solutions, consistent and dependent 23. 205 ninth graders, 190 tenth graders, 180 eleventh graders
LSN 38–3
Saxon Algebra 2
Lesson 24.
A change in the value of h shifts the parent function to the right when h is positive and to the left when h is negative. A change in the value of k shifts the parent function up when k is positive and down when k is negative. When the absolute value of a is greater than 1, the graph is stretched away from the x-axis, appearing narrower than the parent function. When the absolute value of a is between 0 and 1, the graph is compressed towards the x-axis, appearing wider than the parent function.
38
25. When a polynomial is factored and is equal to zero, since the product of the terms equals zero, at least one of its terms equals zero. Setting each term containing a variable equal to zero enables you to solve for a possible value of the variable. 1.5d 26. _ 2 t + 1.5t
27. The point (0, 3) is the highest point on the parabola; the y-values do not extend above 3. 8
y
4 x -8
-4
4
8
-4 -8
28. C 29. y = 3x + 29.5 1 30. _ 36
© 2009 Saxon®, an imprint of HMH Supplemental Publishers Inc. All rights reserved.
LSN 38–4
Saxon Algebra 2
