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Transcript
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:
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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
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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
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.
