Algebra II - Chapter 2 Practice Test Answer Section

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1 Algebra II - Chapter Practice Test Answer Section SHORT ANSWER 1. ANS: g(x) is f(x) translated 3 units left and units up. Because h = + 3, the graph is translated 3 units left. Because k = +, the graph is translated units up. Therefore, g(x) is f(x) translated 3 units left and units up. PTS: 1 DIF: Average REF: 155fb74e df-9c7d f0dea OBJ: -1. Translating Quadratic Functions NAT: NT.CCSS.MTH F.BF.3 STA: MI.MIGLC.MTH A.. LOC: MTH.C MTH.C MTH.C TOP: -1 Using Transformations to Graph Quadratic Functions KEY: quadratic graph 1

2 . ANS: g(x) = 6(x 5) Identify how each transformation affects the coefficients in vertex form. reflection across x-axis a is negative. verticalstretch by a factor of 6 a = 6 translated right 5 units h = 5 Write the transformed function, using the vertex form g(x) = a(x h) + k. g(x) = 6(x 5) + 0 Substitute 6 for a, 5 for h, and 0 for k. g(x) = 6(x 5) Simplify. PTS: 1 DIF: Average REF: f df-9c7d f0dea OBJ: -1.4 Writing Transformed Quadratic Functions NAT: NT.CCSS.MTH F.BF.3 LOC: MTH.C TOP: -1 Using Transformations to Graph Quadratic Functions KEY: quadratic graph parent function

3 3. ANS: The parabola opens downward. The axis of symmetry is the line x = 1. The vertex is the point ( 1,14). The y-intercept is 10. Because a is, the graph opens downward. The axis of symmetry is given by x = ( 8) ( 4) = 8 8 = 1. x = 1 is the axis of symmetry. The vertex lies on the axis of symmetry, so x = 1. The y-value is the value of the function at this x-value. f( 1) = 4( 1) 8( 1) + 10 = = 14 The vertex is ( 1,14). Because the last term is 10, the y-intercept is 10. PTS: 1 DIF: Average REF: ae df-9c7d f0dea OBJ: -. Graphing Quadratic Functions in Standard Form NAT: NT.CCSS.MTH F.IF.7 LOC: MTH.C TOP: - Properties of Quadratic Functions in Standard Form 3

4 4. ANS: The minimum value is D: {all real numbers}; R: {y y 6.15} Step 1 Determine whether the function has a minimum or maximum value. Because a is positive, the graph opens upward and has a minimum value. Step Find the x-value of the vertex. x = b = 5 = 1.5 a () Step 3 Find the y-value of the vertex. f(1.5) = (1.5) 5(1.5) 3 = 6.15 The minimum value is The domain is {all real numbers}. The range is {y y 6.15}. PTS: 1 DIF: Average REF: be df-9c7d f0dea OBJ: -.3 Finding Minimum or Maximum Values NAT: NT.CCSS.MTH F.IF.7 STA: MI.MIGLC.MTH A.1.7 LOC: MTH.C MTH.C TOP: - Properties of Quadratic Functions in Standard Form KEY: maximum minimum 5. ANS: 4.73 meters at 0.3 seconds The solution is the vertex of the function, at (t, d(t)). Find the t-value by using 4.9 as a and.3 as b. t = b a = (.3) (4.9) = Substitute this t-value into d to find the corresponding value, d(t). d(t) = 4.9(0.3).3(0.3) + 5 d(t) = d(t) = 4.73 The maximum distance is about 4.73 meters at 0.3 seconds. PTS: 1 DIF: Average REF: 156b7c0a df-9c7d f0dea OBJ: -.4 Application NAT: NT.CCSS.MTH F.IF.4 STA: MI.MIGLC.MTH A.1.7 LOC: MTH.C TOP: - Properties of Quadratic Functions in Standard Form KEY: maximum minimum 4

5 6. ANS: x = 15 or x = 3 h( x) = x + 1x 45 x + 1x 45 = 0 Set the function equal to 0. (x + 15)(x 3) = 0 Factor: Find factors of 45 that add to 1. x + 15 = 0 or x 3 = 0 Apply the Zero-Product Property. x = 15 or x = 3 Solve each equation. PTS: 1 DIF: Basic REF: c df-9c7d f0dea OBJ: -3. Finding Zeros by Factoring NAT: NT.CCSS.MTH A.REI.4 NT.CCSS.MTH F.IF.8 LOC: MTH.C TOP: -3 Solving Quadratic Equations by Graphing and Factoring KEY: solve quadratic equations 7. ANS: 6 seconds h( t) = 16t + v 0 t + h 0 Write the general projectile function. h( t) = 16t + 96t + 0 Substitute 96 for v 0 and 0 for h 0. The toy rocket will hit the ground when its height is zero. 16t + 96t = 0 Set h( t) equal to 0. 16t( t 6) = 0 Factor. The GCF is 16t. 16t = 0 or ( t 6) = 0 Apply the Zero Product Property. t = 0 or t = 6 Solve each equation. The toy rocket will hit the ground in 6 seconds. Notice that the height is also zero when t = 0, the instant that the toy rocket is launched. PTS: 1 DIF: Average REF: d df-9c7d f0dea OBJ: -3.3 Application STA: MI.MIGLC.MTH A1.. MI.MIGLC.MTH A.1.7 LOC: MTH.C MTH.C TOP: -3 Solving Quadratic Equations by Graphing and Factoring KEY: solve quadratic equations 5

6 8. ANS: f(x) = x + x 48 x = 6 or x = 8 Write the zeros as solutions for two equations. x 6 = 0 or x + 8 = 0 Rewrite each equation so that it is equal to 0. 0 = (x 6)(x + 8) 0 = x + x 48 Multiply the binomials. f(x) = x + x 48 Replace 0 with f(x) Apply the converse of the Zero-Product Property to write a product that is equal to 0. PTS: 1 DIF: Average REF: a df-9c7d f0dea OBJ: -3.5 Using Zeros to Write Function Rules NAT: NT.CCSS.MTH A.APR. LOC: MTH.C MTH.C TOP: -3 Solving Quadratic Equations by Graphing and Factoring 9. ANS: x = 6 ± 5 x 1x + 36 = 0 ( x 6) = 0 Factor the perfect square trinomial. x 6 = ± 0 Take the square root of both sides. x = 6 ± 0 Add 6 to each side. x = 6 ± 5 Simplify. PTS: 1 DIF: Average REF: 1575c8a df-9c7d f0dea OBJ: -4.1 Solving Equations by Using the Square Root Property NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C TOP: -4 Completing the Square 6

7 10. ANS: x = 9 or x = 11 x = 99 x x + x = 99 Collect variable terms on one side. x Ê + x + Á ˆ Ê = 99 + Á ˆ Ê b Add Á ˆ to each side. x + x + 1 = 100 Simplify. (x + 1) = 100 Factor. x + 1 = ±10 Take the square root of each side. x + 1 = 10 or x + 1 = 10 Solve for x. x = 9 or x = 11 PTS: 1 DIF: Basic REF: 1579ca df-9c7d f0dea OBJ: -4.3 Solving a Quadratic Equation by Completing the Square NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C TOP: -4 Completing the Square KEY: complete the square solve quadratic equations 11. ANS: f(x) = (x 6) + 5; vertex: (6, 5) f(x) = x + 4x 67 Factor to make the coefficient of the first term 1. f(x) = (x 1x +? ) 67? Set up to complete the square. f(x) = (x Ê 1x + 1 ˆ Á Ê ˆ Ê b Á Add and subtract Á ˆ. f(x) = (x 6) 67 ( )( 6) Simplify and factor. f(x) = (x 6) + 5 Simplify. The equation is now in vertex form, f(x) = a( x h) + k, and the vertex is (h, k) or (6, 5). PTS: 1 DIF: Average REF: 1579f df-9c7d f0dea OBJ: -4.4 Writing a Quadratic Function in Vertex Form NAT: NT.CCSS.MTH F.IF.4 LOC: MTH.C MTH.C TOP: -4 Completing the Square 7

8 1. ANS: x = ±3i 7x + 63 = 0 7x = 63 Add 63 to both sides. x = 9 Divide both sides by 7. x = ± 9 Take square roots. x = ±3i Express in terms of i. PTS: 1 DIF: Average REF: 157e8eea df-9c7d f0dea OBJ: -5. Solving a Quadratic Equation with Imaginary Solutions NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C MTH.C TOP: -5 Complex Numbers and Roots KEY: complex numbers 13. ANS: x = 7, y = 1 3 8x = 56 Equate the real parts. x = 7 Solve for x. 3 = 9y Equate the imaginary parts. 1 3 = y Solve for y. PTS: 1 DIF: Average REF: 157eb5fa df-9c7d f0dea OBJ: -5.3 Equating Two Complex Numbers LOC: MTH.C TOP: -5 Complex Numbers and Roots KEY: complex numbers 14. ANS: x = i or 4 10i x 8x = 0 Set f(x) = 0. x 8x = 116 Rewrite. x 8x + 16 = Ê b Add Á ˆ to both sides of the equation. (x 4) = 100 Factor. x 4 = ± 100 Take square roots. x = 4 ± 10i Simplify. PTS: 1 DIF: Average REF: 1580f df-9c7d f0dea OBJ: -5.4 Finding Complex Zeros of Quadratic Functions NAT: NT.CCSS.MTH N.CN.7 NT.CCSS.MTH A.REI.4 LOC: MTH.C MTH.C TOP: -5 Complex Numbers and Roots KEY: complex numbers 8

9 15. ANS: 8 16i 16i 8 = 8 + (16)i Rewrite as a + bi. = 8 (16)i Find a bi. = 8 16i Simplify. PTS: 1 DIF: Basic REF: a df-9c7d f0dea OBJ: -5.5 Finding Complex Conjugates NAT: NT.CCSS.MTH N.CN.3 STA: MI.MIGLC.MTH L.1.5 LOC: MTH.C TOP: -5 Complex Numbers and Roots KEY: complex numbers 16. ANS: 7 ± 13 x = x + 7x + 9 = 0 Set f(x) = 0. x = b ± b 4ac a x = 7 ± (7) 4(1)(9) (1) x = x = 7 ± ± 13 Write the Quadratic Formula. Substitute 1 for a, 7 for b, and 9 for c. Simplify. Write in simplest form. PTS: 1 DIF: Average REF: 15837ab df-9c7d f0dea OBJ: -6.1 Quadratic Functions with Real Zeros NAT: NT.CCSS.MTH A.REI.4 STA: MI.MIGLC.MTH A1..9 LOC: MTH.C MTH.C TOP: -6 The Quadratic Formula KEY: quadratic formula 9

10 17. ANS: The equation has two real solutions. x + 7x = Make sure the equation is in standard form, ax + bx + c = 0. x + 7x + = 0 b 4ac Evaluate the discriminant. = ( 7) 4( 1) ( ) =49 8 = 41 The discriminant is positive. The equation has two real solutions. PTS: 1 DIF: Average REF: a df-9c7d f0dea OBJ: -6.3 Analyzing Quadratic Equations by Using the Discriminant NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C MTH.C TOP: -6 The Quadratic Formula KEY: quadratic formula 18. ANS: Step 1 Graph the boundary of the related parabola y = 4x 3x 5 with a dashed line for < or > and a solid line for or. If the coefficient of x is positive, the vertex is the minimum value. If the coefficient of x is negative, the vertex is the maximum value. Step Shade below the parabola for < or and shade above the parabola for > or. PTS: 1 DIF: Average REF: 158a7ab df-9c7d f0dea OBJ: -7.1 Graphing Quadratic Inequalities in Two Variables LOC: MTH.C TOP: -7 Solving Quadratic Inequalities KEY: quadratic formula 10

11 19. ANS: x or x 1 Use a graphing calculator to graph each side of the inequality. Use y 1 = x + x 6 and y = 4. Identify the values of x for which y 1 y. x y 1 y The parabola is at or above the line when x is less than or equal to or greater than or equal to 1. So, the solution set is x or x 1. The table supports the answer. The number line shows the solution set. PTS: 1 DIF: Average REF: 158cdd df-9c7d f0dea OBJ: -7. Solving Quadratic Inequalities by Using Tables and Graphs LOC: MTH.C TOP: -7 Solving Quadratic Inequalities 11

12 0. ANS: 5 x 8 Step 1 x 3x 8 = 1 Write the related equation. x 3x 40 = 0 Write the equation in standard form. (x + 5)(x 8) = 0 Factor. Step x + 5 = 0 or x 8 = 0 Find the critical values. x = 5 or x = 8 The critical values are 5 and 8. Step 3 The critical values divide the number line into three intervals: x < 5, 5 <x < 8, or x > 8. Test an x-value in each interval in the original inequality to determine which intervals make the inequality true. The solution is 5 x 8. PTS: 1 DIF: Average REF: 158d df-9c7d f0dea OBJ: -7.3 Solving Quadratic Inequalities Using Algebra NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C TOP: -7 Solving Quadratic Inequalities KEY: quadratic inequality 1

13 1. ANS: 6 x 16 The profit must be at least $ x + 330x Find the critical values by solving the related equation. 15x + 330x 815 = x + 330x 1415 = 0 5(3x 66x + 83) = 0 x = ( 66) ± ( 66) 4(3)(83) (3) 66 ± 960 x = 6 x 5.84 or x Use a number line and test an x-value in each of the three regions created by the critical points. Try x = 10: 15(10) + 330(10) 815 = Round. 6 x 16 Try x = 18: 15(0) + 330(0) 815 = is not 600 Try x = 4: 15(4) + 330(4) 815 = is not 600 PTS: 1 DIF: Average REF: 158f3f6e df-9c7d f0dea OBJ: -7.4 Problem-Solving Application NAT: NT.CCSS.MTH A.REI.4 LOC: MTH.C TOP: -7 Solving Quadratic Inequalities KEY: quadratic inequality MSC: DOK 4. ANS: + 11i To add complex numbers, add the real parts and the imaginary parts. To subtract complex numbers, subtract the real parts and the imaginary parts. (8 + i) (6 9i) = (8 (6)) + ( ( ))i = + 11i PTS: 1 DIF: Average REF: 159db4a df-9c7d f0dea OBJ: -9.3 Adding and Subtracting Complex Numbers NAT: NT.CCSS.MTH N.CN. STA: MI.MIGLC.MTH L.1.5 LOC: MTH.C TOP: -9 Operations with Complex Numbers 13

14 3. ANS: i 6i( 3 5i) 18i 30i Distribute. 18i 30( 1) Use i = i Write in a + bi form. PTS: 1 DIF: Basic REF: 15a54e df-9c7d f0dea OBJ: -9.5 Multiplying Complex Numbers NAT: NT.CCSS.MTH N.CN. STA: MI.MIGLC.MTH L.1.5 LOC: MTH.C TOP: -9 Operations with Complex Numbers 4. ANS: 5 5i 8 = 5 i ˆ Á 14 Rewrite i 8 as a power of i. = 5( 1) i = 1. = 5 Simplify. PTS: 1 DIF: Average REF: 15a795e df-9c7d f0dea OBJ: -9.6 Evaluating Powers of i NAT: NT.CCSS.MTH N.CN. STA: MI.MIGLC.MTH L.1.5 LOC: MTH.C TOP: -9 Operations with Complex Numbers 5. ANS: i 3 i 3 + 5i = ( 3 i) ( 3 + 5i) (3 5i) (3 5i) Multiply by the conjugate i 3i + 5i = 9 15i + 15i 5i Distribute i 5 = Use i = i Simplify. PTS: 1 DIF: Average REF: 15a4b4aa df-9c7d f0dea OBJ: -9.7 Dividing Complex Numbers NAT: NT.CCSS.MTH N.CN.3 STA: MI.MIGLC.MTH L.1.5 LOC: MTH.C TOP: -9 Operations with Complex Numbers 14

15 6. ANS: The real axis is the x-axis, and the imaginary axis is the y-axis. Think of a + bi as x + yi. Thus the complex number 5 i is at (5, ). PTS: 1 DIF: Basic REF: 1598c8de df-9c7d f0dea OBJ: -9.1 Graphing Complex Numbers NAT: NT.CCSS.MTH N.CN.4 LOC: MTH.C TOP: -9 Operations with Complex Numbers KEY: graph complex number MSC: DOK 15