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CARE Curriculum Assessment Remediation Enrichment Algebra 2 Mathematics CARE Package #2 – Quadratics SPS Online Domain Algebra – Seeing Structure in Expressions Standards MAFS.912.A-SSE.2.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. a. Factor a quadratic expression to reveal the zeros of the function it defines. b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. * Students will use equivalent forms of a quadratic expression to interpret the expression’s terms, factors, zeros, maximum, minimum, coefficients, or parts in terms of the real-world situation the expression represents. Domain Standards MAFS.912.F-IF.3.8 Functions – Interpreting Functions Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry. Also assesses MAFS.912.F-IF.3.7a Graph functions expressed symbolically and show key features of the graph by hand in simple cases and using technology for more complicated cases. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. Domain Number and Quantity – The Complex Number System Standards MAFS.912.N-CN.3.7 Solve quadratic equations with real coefficients that have complex solutions. Also assesses MAFS.912.A-REI.2.4 Solve quadratic equations in one variable. a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x –p)² = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x² = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. MAFS.912.N-CN.1.2 2 Use the relation i 1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. MAFS.912. N-CN.1.1 2 Know there is a complex number i such that i 1, and every complex number has the form a bi with a and b real. Also assesses CURRICULUM Performance Task – Quadratic Forms Analysis (SPS Online) PART A - Graph the following functions either in GeoGebra, on your graphing calculator, or on the coordinate plane below. Compare and contrast the graphs, and describe the similarities and differences in terms of their key features (Axis of Symmetry, Vertex, Zeroes, etc). Comparison: _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ PART B - Below are the first three functions from the task above. f (x) x 2 4 x 3 g(x) (x 2) 2 1 h(x) (x 3)(x 1) Since the three equations all describe the same function, knowing when to use one form over the other in identifying specific features of the graph will lead to you working more efficiently. Determine the coordinates of the following points on the graph, and explain which equation is the best for finding the specified point(s) and why? Key Feature Vertex x-intercepts y-intercept Ordered Pair(s) Which Equation & Why? PART C - Make up an equation for a quadratic function whose graph satisfies the given condition below. Answers will vary, so use whatever form is most convenient. Condition Has a vertex at (-2, -5). Equation Has an y-intercept at (0, -6). Has x-intercepts at the origin and (-4,0). PART D - Describe the relationship between the solutions of the general quadratic equation ax 2 bx c 0 and the graph of f (x) ax 2 bx c . Use the functions f, i, and j from Part A above to illustrate different cases of the relationship. f (x) x 2 4 x 3 i(x) x 2 4 x 4 j(x) x 2 4 x 5 ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ASSESSMENT The Mini-MAF includes standards 912.A-SSE.2.3, 912.F-IF.3.8, 912.N-CN.3.7, and 912.N-CN.1.2 in addition to the standards that are “assessed with” the ones listed here. Use the following table to assist in remediation efforts. Questions 1–3 4–6 7–9 10 – 12 Standards 912.A-SSE.2.3 912.F-IF.3.8 912.N-CN.3.7 912.N-CN.1.2 Algebra 2 - Burger Lessons 2-1, 2-2, 2-3, 2-4 Lessons 2-1, 2-2, 2-3, 2-4 Lessons 2-4, 2-5, 2-6 Lessons 2-5, 2-9 Algebra 2 Hon - Larson Lessons 1.1, 1.2, 1.3, 1.4, 1.7 Lessons 1.1, 1.2, 1.3, 1.4, 1.7 Lessons 1.5, 1.6, 1.7, 1.8 Lessons 1.6 Answer Key with DOK Levels REMEDIATION / RETEACH Key Vocabulary for Quadratic Functions & Equations Absolute Value of a Complex Number Axis of Symmetry Completing the Square Complex Conjugate Complex Number Imaginary Number Maximum Value Minimum Value Parabola Quadratic Function o Standard Form o Vertex Form o Factored Form Roots of an Equation Vertex Zeros of a Function PERFORMANCE TASK Considerations & Solution Guide In Algebra 2, this task can be done in class near the beginning or end of a unit on parabolas. It could be given to individuals or small groups using a computer, paper, and pencil (ESE StrategiesProvide Peer Assistance/Computer Assisted Instruction; ESOL E4-PEER Pair). Students should be familiar with intercepts, and need to know what the vertex is so activating prior knowledge by reviewing the key vocabulary will be necessary. (ESE Strategy- Pre-Teach Vocabulary; ESOL G1Activating Prior Knowledge). This task is effective after students have graphed parabolas in vertex form, but have not yet explored graphing other forms. However, most Algebra 2 students should have experienced and mastered all of the forms in Algebra 1. PART A (2 Points) Graph the following functions either in GeoGebra, on your graphing calculator, or on the coordinate plane below. Compare and contrast the graphs, and describe the similarities and differences in terms of the key features of the graphs (Axis of Symmetry, Vertex, Zeroes, etc.). Comparison Statements: All of the graphs are all parabolas that open upwards, each with the same axis of symmetry ( x 2 ). The functions f, g, and h all name the same function but they are in different forms (Standard / Vertex / Factored or Intercept) The functions i and j have exactly the same shape as f, g, and h. However, i and j have been translated up from f by 1 unit and 2 units respectively. The functions f, g, and h have two x-intercepts, while the function i has one x-intercept, and the function j has none. Comments Part A is to be accomplished with the aid of a computer as a GeoGebra applet is embedded on the task page (ESE Strategies - Computer Assisted Instruction; ESOL G10-Visualization). Students can also use a graphing calculator or graph the functions by hand if the technology is not available to them. By using the technology, the students will quickly generate graphs and be able to compare their features. Without technology, the task length of time will increase and graphing skills will be reinforced. The fact that the first three functions have the same graph might not be obvious to the students; mathematical residue and excitement should be left behind when they realize that equivalent expressions produce the same graph. In writing the comparison between the functions, encourage students to use graphic organizers and provide if necessary (ESE Strategies – Provide Graphic Organizer; ESOL G10-Visualization). PART B (2Points) Below are the first three functions from the task above. f (x) x 2 4 x 3 g(x) (x 2) 2 1 h(x) (x 3)(x 1) Since the three equations all describe the same function, knowing when to use one over the other in identifying specific features of the graph will lead to you working more efficiently. Determine the coordinates of the following points on the graph, and explain which equation is the best for finding the specified point(s) and why? Key Feature Vertex Ordered Pair(s) (2, 1) x-intercepts (1, 0) & (3, 0) y-intercept (0, 3) Which Equation & Why? I would use g(x) since it is in Vertex form and the vertex(h, k) can be identified simply by identifying h and k from the equation. I would use h(x) since it is expressed in a factored form and solving the equation h(x)=0 to find the x-intercepts would be quick using the zero-product property. I would use f(x) since substituting zero in for x would wipe out the quadratic and linear terms leaving me with only the constant. Knowing that, I just need to look at the constant value to find my y-intercept. PART C (1 point) Make up an equation for a quadratic function whose graph satisfies the given condition below. Answers will vary, so use whatever form is most convenient. Condition Has a vertex at (-5, -7). Equation q(x) 4(x 5) 2 7 Has an y-intercept at (0, -2). q(x) x 2 10x 2 Has x-intercepts at the origin and (-7,0). q(x) 5x(x 7) Comments Parts B and C lead to important whole group discussions about the value of different forms of equations, and should culminate in a discussion of how we can convert between forms and when we might want to do so. f (x) ax 2 bx c g(x) a(x h) 2 k h(x) a(x r1 )(x r2 ) PART D (2 Points) Describe the relationship between the solutions of the general quadratic equation ax 2 bx c 0 and the graph of f (x) ax 2 bx c . Use the functions f, i, and j from Part A above to illustrate different cases of the relationship. f (x) x 2 4 x 3 2 i(x) x 4 x 4 j(x) x 2 4 x 5 Description The real solutions of the general quadratic equation ax 2 bx c 0 are the values of the x2 intercepts for the function f (x) ax bx c . For example, the function f has x-intercepts at the 2 points (1, 0) and (3, 0) and the solutions to the equation x 4x 3 0 are {1, 3}. But not all quadratic functions have two x-intercepts, nor do all quadratic equations have two real solutions. Sometimes they have one, like in the case of the function i, and sometimes they have no x intercepts, like in the case of the function j. When there is only 1 x-intercept, like in the case of the function i, the graph comes in and bounces offof the x-axis at (2, 0) and the equation x 2 4x 4 0 has 1 real solution, namely {2}. Finally, when there are no x-intercepts, like for 2 the function j, the equation x 4x 5 0 will have no real solutions but will have two complex solutions that are non-real. In this case the solutions are {2 + i, 2 – i}. Comments the connection between the x-intercepts (zeroes) of a quadratic function and the Part D drives home solutions (roots) to corresponding quadratic equation. (ESE Strategies - Computer Assisted Instruction; ESOL G10-Visualization). A natural and important extension is a discussion of the discriminant and its usefulness in determining the nature of the zeroes and/or roots. In the Extra for Experts enrichment section that follows, a geometric interpretation of complex roots will be explored. Remediation/Reteaching Resources–Quadratic Functions Resource Key Features of a Quadratic Function Parabola - Standard Form Investigation Parabola - Factored Form Investigation Transformations of a Quadratic Function Computer Aided Instruction / Visualization Students / Teachers should use this GeoGebra applet to transform a quadratic function and observe changes in domain, range, and key features of the graph. They should describe the roles of a, h, and k in determining the vertex, domain, and range of a quadratic function expressed in vertex form. This investigation focuses on the parameters a, b, and c in the standard form of a quadratic function and how changes in these value change the graph. Students will get practice with sliders and dynamic geometry tools as they investigate, build conceptual understanding, and solve problems given conditions of specific parabolas. Similar to the investigation above, but with an emphasis on the parameters a, p, and q in the factored form of a quadratic function. Students / Teachers should use this GeoGebra applet to explore shifts, compressions and stretches of functions. The default function is a quadratic, but it can be changed by the user to any function. ENRICHMENT Extra for Experts– Graphing Complex Solutions (SPS Online) The graph of f (x) x 2 4 x 5 opens upward and has a vertex above the x-axis. Therefore, it has no real x-intercepts and the solutions to the equation x 2 4x 5 0 turn out to be complex numbers. If you were to stretch out the x-axis into the complex number plane, you get the three dimensional graph shown below. Below the vertex, there is another parabola. It is in a plane that is perpendicular to the real x-axis. The two complex roots are the intercepts where this second parabola intersects the complex xplane. Problem Set 1) Set f (x) equal to zero, and solve to demonstrate that the roots really are 2 i and 2 i . 2) Show that f (2 i) really does equal zero. 3) Show that f (2 3i) is a real number. 4) Show that f (1 i) is not a real number f a bi 5) Make a conjecture about the values of a and b for which is a real number. Explain how you arrived at your conjecture. 6) Prove that your conjecture from above is true. Given: a,b R and f (x) x 2 4 x 5 Prove: If a 2 , then f a bi is a real number. Extra for Experts – Solution Guide 1. Set f (x) equal to zero, and solve to demonstrate that the roots really are 2 i and 2 i . f (x) x 2 4 x 5 2. Show that f (2 i) really does equal zero. f (x) x 2 4 x 5 x 4 x 5 0 f (2 i) (2 i) 2 4(2 i) 5 2 f (2 i) 4 4i i 2 8 4i 5 4 16 4(1)(5) 2 4 4 x 2 4 2i x 2 x 2i f (2 i) 4 1 8 5 x 3. Show that f (2 3i) is a real number. f (2 i) 0 4. Show that f (1 i) is not a real number. f (x) x 2 4 x 5 f (x) x 2 4 x 5 2 f (2 3i) (2 3i) 4(2 3i) 5 f (2 i) 4 12i 9i 2 8 12i 5 f (1 i) (1 i) 2 4(1 i) 5 f (2 i) 1 2i i 2 4 4i 5 f (2 i) 4 9 8 5 f (2 i) 1 1 4 5 2i f (2 i) 8 f (2 i) 1 2i 5. Make a conjecture about the values of a and b for which f a bi is a real number. Explain how you arrived at your conjecture. Conjecture: If a 2 , then f a bi is a real number. First substitute a bi into f , and show that f a bi will be real as long as 2abi 4bi 0 . f (x) x 2 4 x 5 f (a bi) (a bi) 2 4(a bi) 5 f (a bi) a 2 2abi b 2i 2 4a 4bi 5 f (a bi) a b i 4a 5 2abi 4bi 2 2 2 2abi 4bi 0 ab 2b 0 ab 2b f (a bi) real# 2abi 4bi of b is in order to leave the expression So if a 2 , it should not matter what the value 2abi 4bi 0 , thus making f a bi a real number. 6. Prove that your conjecture from above is true. Given: a,b R and f (x) x 2 4 x 5 Prove: If a 2 , then f a bi is a real number. 2 in for a and find f 2 bi Substitute f (x) x 2 4 x 5 f (2 bi) (2 bi) 2 4(2 bi) 5 f (a bi) 4 4bi b 2i 2 8 4bi 5 f (a bi) 1 b 2 4bi 4bi f (a bi) real# (4b 4b)i f (a bi) real# 0i f (a bi) real# Since it can be shown that for the function f (x) x 2 4 x 5 , f 2 bi will always lead to a number that when written in a bi form, will have b 0, thus making it a real number.