C A R E

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 Functions – Interpreting Functions

Standards MAFS.912.F-IF.3.8 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

Use the relation

i2 1 and the commutative, associative, and distributive properties to add,

subtract, and multiply complex numbers.

Also assesses MAFS.912. N-CN.1.1

Know there is a complex number

i such that

i2 1, and every complex number has the form

abi with a and b real.

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:

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PART B - Below are the first three functions from the task above.

f (x) x 2 4x 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 Ordered Pair(s) Which Equation & Why?

Vertex

x-intercepts

y-intercept

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 Equation

Has a vertex at (-2, -5).

Has an y­intercept at (0, -6).

Has x­intercepts at the origin and (­4,0).

PART D - Describe the relationship between the solutions of the general quadratic equation

ax2 bx c 0 and the graph of

f (x) ax2 bx c . Use the functions f, i, and j from Part A

above to illustrate different cases of the relationship.

f (x) x 2 4x 3

i(x) x 2 4x 4

j(x) x 2 4x 5

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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 Standards Algebra 2 - Burger Algebra 2 Hon - Larson

1 – 3 912.A-SSE.2.3 Lessons 2-1, 2-2, 2-3, 2-4 Lessons 1.1, 1.2, 1.3, 1.4, 1.7

4 – 6 912.F-IF.3.8 Lessons 2-1, 2-2, 2-3, 2-4 Lessons 1.1, 1.2, 1.3, 1.4, 1.7

7 – 9 912.N-CN.3.7 Lessons 2-4, 2-5, 2-6 Lessons 1.5, 1.6, 1.7, 1.8

10 – 12 912.N-CN.1.2 Lessons 2-5, 2-9 Lessons 1.6

Answer Key with DOK Levels

REMEDIATION / RETEACH

Key Vocabulary for Quadratic Functions & Equations

Absolute Value of a Complex Number Parabola

Axis of Symmetry Quadratic Function

Completing the Square o Standard Form

Complex Conjugate o Vertex Form

Complex Number o Factored Form

Imaginary Number Roots of an Equation

Maximum Value Vertex

Minimum Value 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 Strategies-

Provide 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 G1-

Activating 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 4x 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 Ordered Pair(s) Which Equation & Why?

Vertex (2, 1) 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.

x-intercepts (1, 0) & (3, 0) 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.

y-intercept (0, 3) 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 Equation

Has a vertex at (-5, -7).

q(x) 4(x 5)2 7

Has an y­intercept at (0, -2).

q(x) x 2 10x 2

Has x­intercepts 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) ax2 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

ax2 bx c 0 and the graph of

f (x) ax2 bx c . Use the functions f, i, and j from Part A above to illustrate

different cases of the relationship.

f (x) x 2 4x 3

i(x) x 2 4x 4

j(x) x 2 4x 5

Description

The real solutions of the general quadratic equation

ax2 bx c 0 are the values of the x-

intercepts for the function

f (x) ax2 bx c . For example, the function f has x-intercepts at the

points (1, 0) and (3, 0) and the solutions to the equation

x2 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 off of the x-axis at (2, 0) and the equation

x2 4x 4 0 has 1 real solution, namely {2}. Finally, when there are no x-intercepts, like for

the function j, the equation

x2 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

Part D drives home the connection between the x-intercepts (zeroes) of a quadratic function and the

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 Computer Aided Instruction / Visualization

Key Features of a Quadratic Function

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.

Parabola - Standard Form Investigation 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.

Parabola - Factored Form Investigation Similar to the investigation above, but with an

emphasis on the parameters a, p, and q in the

factored form of a quadratic function.

Transformations 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 4x 5 opens upward and has a vertex above the x-axis. Therefore, it has

no real x-intercepts and the solutions to the equation

x2 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 x-

plane.

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

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.

6) Prove that your conjecture from above is true.

Given:

a,bR and

f (x) x 2 4x 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 4x 5

x 2 4x 5 0

x 4 16 4(1)(5)

2

x 4 4

2

x 4 2i

2

x 2 i

2. Show that

f (2 i) really does equal zero.

f (x) x 2 4x 5

f (2 i) (2 i)2 4(2 i) 5

f (2 i) 4 4i i2 8 4i 5

f (2 i) 4 1 8 5

f (2 i) 0

3. Show that

f (2 3i) is a real number.

f (x) x 2 4x 5

f (2 3i) (2 3i)2 4(2 3i) 5

f (2 i) 4 12i 9i2 8 12i 5

f (2 i) 4 9 8 5

f (2 i) 8

4. Show that

f (1 i) is not a real number.

f (x) x 2 4x 5

f (1 i) (1 i)2 4(1 i) 5

f (2 i) 1 2i i2 4 4i 5

f (2 i) 11 4 5 2i

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

abi into

f , and show that

f a bi will be real as long as

2abi 4bi 0.

f (x) x 2 4x 5

f (a bi) (a bi)2 4(a bi) 5

f (a bi) a2 2abi b2i2 4a 4bi 5

f (a bi) a2 b2i2 4a 5 2abi 4bi

f (a bi) real# 2abi 4bi

2abi 4bi 0

ab 2b 0

ab 2b

So if

a 2 , it should not matter what the value of b is in order to leave the expression

2abi 4bi 0, thus making

f a bi a real number.

6. Prove that your conjecture from above is true.

Given:

a,bR and

f (x) x 2 4x 5

Prove: If

a 2 , then

f a bi is a real number.

Substitute 2 in for a and find

f 2 bi

f (x) x 2 4x 5

f (2 bi) (2 bi)2 4(2 bi) 5

f (a bi) 4 4bi b2i2 8 4bi 5

f (a bi) 1 b2 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 4x 5,

f 2 bi will always lead

to a number that when written in

abi form, will have

b 0, thus making it a real

number.