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Algebra II A Review 7

Multiple Choice Identify the choice that best completes the statement or answers the question. Identify the vertex and the axis of symmetry of the parabola. Identify points corresponding to P and Q. 1.

a. (3, 1), x = 3 P'(2, 2), Q'(2, ­2) b. (3, 1), x = 3 P'(4, 2), Q'(1, 5) 2.

c. (1, 3), x = 1 P'(4, 2), Q'(1, 5) d. (1, 3), x = 1 P'(2, 2), Q'(2, ­2)

a. (4, ­4), x = 4; P'(­3, 3), Q'(­6, 0) b. (­4, 4), x = ­4; P'(­3, 3), Q'(­6, 0) 3. Which is the graph of ?

c. (­4, 4), x = ­4; P'(­5, 3), Q'(­2, 0) d. (4, ­4), x = 4; P'(­5, 3), Q'(­2, 0)

a.

c.

b.

d.

4. Use the vertex form to write the equation of the parabola.

a. b.

c. d. .

5. Identify the vertex and the y-intercept of the graph of the function a. vertex: (­2, 5); c. vertex: (2, 5); y-intercept: ­1 y-intercept: 4

b. vertex: (­2, ­5); y-intercept: ­1

d. vertex: (2, ­5); y-intercept: ­1 . If necessary, round to the nearest

6. Use a graphing calculator to solve the equation hundredth. a. ­0.77, 0.77 c. ­0.38, 1.16 b. ­0.77, 2.32 d. ­1.16, 0.38 7. Identify the graph of the complex number a. . c.

b.

d.

8. Use a graphing calculator to determine which type of model best fits the values in the table. x y ­6 192 ­2 24 0 0 2 16 6 168 c. linear model d. none of these

a. cubic model b. quadratic model

9. Use a graphing calculator to find a polynomial function to model the data. x 1 2 3 4 5 6 7 8 9 10

f(x) a. b. c. d.

12

4

5

13

9

16

19

16

24

43

f(x) = 0.8x4 ­ 1.73x3 + 12.67x2 ­ 34.68x + 35.58 f(x) = 0.08x3 ­ 1.73x2 + 12.67x + 35.58 f(x) = 0.08x4 + 1.73x3 ­ 12.67x2 + 34.68x ­ 35.58 f(x) = 0.08x4 ­ 1.73x3 + 12.67x2 ­ 34.68x + 35.58

10. Use a graphing calculator to find the relative minimum, relative maximum, and zeros of . If necessary, round to the nearest hundredth. a. relative minimum: (­62.24, 0.36), relative maximum: (37.79, ­3.69), zeros: x = 5, ­2, 2 b. relative minimum: (0.36, ­62.24), relative maximum: (­3.69, 37.79), zeros: x = ­5, ­2, 2 c. relative minimum: (0.36, ­62.24), relative maximum: (­3.69, 37.79), zeros: x = 5, ­2 d. relative minimum: (­62.24, 0.36), relative maximum: (37.79, ­3.69), zeros: x = ­5, ­2 11. Find the zeros of a. 0, 2, 5 . Then graph the equation. c. 2, 5, ­2

b. 2, 5

d. 0, ­2, ­5

12. Find the zeros of a. 6, multiplicity 2; ­3, multiplicity 6 b. 2, multiplicity 6; ­3, multiplicity 6 c. 6, multiplicity 2; 6, multiplicity ­3 d. 2, multiplicity 6; 6, multiplicity ­3 Solve the equation by graphing. 13. a. x = 49 14. a. 0, 0.49, ­2.06 b. no solution 15. a. 3 Short Answer 16. Graph . b. ­3 b. x = 11

and state the multiplicity.

c. x = 18 c. 0, ­0.49, 2.06 d. 0.49, ­2.06 c. ­3, 3

d. no solution

d. no solution

17. Graph

. Identify the vertex and the axis of symmetry.

18. Graph

. What is the minimum value of the function?

19. Graph

. Does the function have a maximum or minimum value? What is this value?

20. Graph

.

21. Use the graph of . a. If you translate the parabola to the right 2 units and down 7 units, what is the equation of the new parabola in vertex form? b. If you translate the original parabola to the left 2 units and up 7 units, what is the equation of the new parabola in vertex form? c. How could you translate the new parabola in part (a) to get the new parabola in part (b)?

Other 22. A baseball player hits a fly ball that is caught about 4 seconds later by an outfielder. The path of the ball is a parabola. The ball is at its highest point as it passes the second baseman, who is 127 feet from home plate. About how far from home plate is the outfielder at the moment he catches the ball? Explain your reasoning.

Algebra II A Review 7 Answer Section

MULTIPLE CHOICE 1. ANS: STA: 2. ANS: STA: 3. ANS: 4. ANS: 5. ANS: 6. ANS: 7. ANS: 8. ANS: STA: 9. ANS: STA: 10. ANS: STA: 11. ANS: STA: 12. ANS: STA: 13. ANS: STA: 14. ANS: STA: 15. ANS: STA: B OBJ: 5-1.1 Quadratic Functions and Their Graphs MI A.3.1.1 | MI A3.1.2 | MI A3.1.3 B OBJ: 5-1.1 Quadratic Functions and Their Graphs MI A.3.1.1 | MI A3.1.2 | MI A3.1.3 C OBJ: 5-3.1 Using Vertex Form STA: MI A2.3.3 A OBJ: 5-3.1 Using Vertex Form STA: MI A2.3.3 B OBJ: 5-3.1 Using Vertex Form STA: MI A2.3.3 C OBJ: 5-5.2 Solving by Graphing D OBJ: 5-6.1 Identifying Complex Numbers B OBJ: 6-1.2 Modeling Data with a Polynomial Function MI A.3.1.1 | MI A3.1.2 | MI A3.1.3 D OBJ: 6-1.2 Modeling Data with a Polynomial Function MI A.3.1.1 | MI A3.1.2 | MI A3.1.3 B OBJ: 6-2.1 The Factored Form of a Polynomial MI A3.1.2 | MI A3.1.3 A OBJ: 6-2.2 Factors and Zeros of a Polynomial Function MI A3.1.2 | MI A3.1.3 B OBJ: 6-2.2 Factors and Zeros of a Polynomial Function MI A3.1.2 | MI A3.1.3 D OBJ: 6-4.1 Solving Equations by Graphing MI A1.2.5 A OBJ: 6-4.1 Solving Equations by Graphing MI A1.2.5 A OBJ: 6-4.1 Solving Equations by Graphing MI A1.2.5

SHORT ANSWER 16. ANS:

OBJ: 5-2.1 Graphing Parabolas 17. ANS:

STA: MI G1.7.2

vertex:

, axis of symmetry:

OBJ: 5-2.1 Graphing Parabolas 18. ANS:

STA: MI G1.7.2

minimum: 4 OBJ: 5-2.2 Finding Maximum and Minimum Values 19. ANS: STA: MI G1.7.2

maximum value; 8 OBJ: 5-2.2 Finding Maximum and Minimum Values 20. ANS: STA: MI G1.7.2

OBJ: 5-3.1 Using Vertex Form 21. ANS: a. b. c. left 4 units, up 14 units

STA: MI A2.3.3

OBJ: 5-3.1 Using Vertex Form OTHER

STA: MI A2.3.3

22. ANS: The outfielder is about 254 feet from home plate. The parabola is symmetric about its axis, which is a vertical line. The second baseman is at a point on this axis of symmetry. So the outfielder is the same distance from the second baseman as the second baseman is from home plate. 127 + 127 = 254, so the distance from home plate to the outfielder is about 254 feet. OBJ: 5-1.1 Quadratic Functions and Their Graphs STA: MI A.3.1.1 | MI A3.1.2 | MI A3.1.3

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