Download The parabola defined by the equation y = ax2 + bx + 1

Problem Set Facilitation and Submission
Professor Lubna Al-Shammaa
Margaret Cooper
Daljit Dhaliwal
Ross Dolan
PED 4187
22/01/2010
Problem:
The parabola defined by the equation y = ax2 + bx + 1 passes through the point (1, 2). For
what values of “a” does the parabola intersect the x-axis at two distinct points?
Important Mathematical Ideas
This problem focuses on the analysis of problems involving quadratic functions. The
complete solution to the problem involves the following demonstration of the ideas from
the Grade 11 Math Curriculum for both MCR3U and MCF3M:
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Determining the number of zeros (i.e. x-intercepts) of a quadratic function, using a
variety of strategies (e.g., inspecting graphs; factoring; calculating the discriminant)
Demonstrate an understanding of functions, and make connections between the
numeric, graphical, and algebraic representations of quadratic functions;
Solve quadratic equations by selecting and applying a factoring strategy.
Determine, through investigation, and describe the connection between the factors
used in solving a quadratic equation and the x-intercepts of the graph of the
corresponding quadratic relation
Explore the algebraic development of the quadratic formula and apply the formula
to solve quadratic equations, using technology
Relate the real roots of a quadratic equation to the x-intercepts of the
corresponding graph, and connect the number of real roots to the value of the
discriminant (e.g., there are no real roots and no x-intercepts if b2 – 4ac < 0)
Determine the real roots of a variety of quadratic equations, and describe the
advantages and disadvantages of each strategy (i.e., graphing; factoring; using the
quadratic formula.
Problem Solution:
At the point where the parabola intersects the x –axis, y = 0. We can re-write the equation
as:
ax 2  bx  1  0
… (1)
At point (1, 2), the equation can be written as:
a  b 1  2
therefore, a  b  1
Or, b = 1 – a
… (2)
… (3)
Substituting the value of b into the equation (1), where y = 0, yields:
ax 2  (1  a) x  1  0
… (4)
The problem now requires us to solve for the possible values of “a”. Once we have arrived
at this point, there are several methodologies which may be used to solve the problem.
Among these are:
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1.
2.
3.
4.
Using the Quadratic Equation
Factoring using Algebra tiles
Using a Graphing calculator
Using a software package such as Geometer’s Sketchpad
We will explore each of these methods in order to determine the values of “a”.
1. Quadratic equation formula:
The quadratic formula is of the form:
 b  b 2  4ac
x
2a
Substituting the values of a and b into equation the quadratic equation yields:
x=
 (1  a)  (1  a) 2  4a
2a
 (1  a)  1  a 2  2a  4a
2a
 (1  a)  1  a 2  6a
x 
2a
x 
For a parabola to have two real roots, the term under the square root (which is also
known as the discriminant: D = b2 – 4ac) must be greater than zero
Therefore,
b 2  4ac  0
(1-a) – 4(a)(1) > 0
2
a 2  6a 1  0
Solving this quadratic equation in terms of “a” using the formula from equation (4), we
get two distinct values of a:
 (6)  6 2  4
2
a  3 2 2
a 
The two distinct values (in decimal form) that we get by solving the quadratic equation
are: 5.82; 0.18. Therefore, the equation a2 – 6a +1 = 0 is true when a = 5.82 or 0.18.
This equation gives us two values of a but doesn’t indicate any other constraints.
Therefore, it is important to mention here that a must be non-zero otherwise the
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equation y = ax2 + bx + 1 would be reduced to a linear form: y = bx + 1, which would
yield only a single x-intercept.
Since we are looking for all the values of a for which the inequality a2 – 6a +1> 0 holds,
we use examination to see that a2 – 6a +1> 0 when a > 5.82 or a < 0.18.
Further analysis of the quadratic equation was done by plotting the parabolic function
in geometer’s sketchpad with different values of “a”.
Area Model or Algebra Tiles:
The quadratic equation was solved above using the traditional formula. However, we can
also determine the roots of a quadratic equation by using algebra tiles or the Area Model.
To solve for the values of a in the equation a2 – 6a + 1 = 0, we use the area model as
follows:
+2√2
a
a
-3
-2√2
-3
a2
-3a
-3a
9
-2√2a
6√2
2√2a
-6√2
-8
The resulting equation found from this method is:
(a – 3 + 2 2 )(a – 3 – 2 2 ) = 0
Solving for a in the above equation yields two possible values for a: 3  2 2
According to the area model, the area in the middle would be positive if both factors
(sides) are positive or both factors are negative. Thus (a – 3 + 2 2 ) and (a – 3 – 2 2 )
must both be positive or negative. Thus, either “a” is greater than both 3  2 2 or it is less
than both 3  2 2 . This then means that “a” is greater than 3 + 2 2 (5.82) or it is less
than 3 - 2 2 (0.18). Again, the constraint that “a” cannot equal zero must be taken into
account.
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Graphical methods that can be employed to find the range of values of “a” for which there
are two distinct points of intersection, are Geometer’s Sketchpad or graphing calculator.
Graphing Calculator:
Steps for using graphing calculator to plot: f(a) = a2 – 6a + 1
 Entering the equation in Y1, where Y1= x2 – 6x + 1 (x has been used here instead
of “a” because of alphabetic limitations on the TI-83 calculator)
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To set scale of the calculator to see the parabola, press ZOOM and 6
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Press 2nd → TRACE → 1, and screen will look like:
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Select 2:ZERO to get the following screen:
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To enter left-bound and right-bound values, move the cursor to the left and right of
the corresponding intercepts and press ENTER.
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The next screen will say GUESS. You can either GUESS or press ENTER. You will
get one x-intercept.
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Repeating through the above steps again, we can get the other x-intercept.
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Geometer’s Sketchpad:
If we are to graph the function f(a) = a2 - 6a + 1, as defined by the determinant in the
previous examples, we see the relationship in Graph 1:
Graph 1: Plot of function f(a) = a2 - 6a + 1
From referencing Graph 1, we can clearly determine that when a > 5.8 and when a< 0.18,
this will yield a positive value for the function f(a) = a2 - 6a + 1 .
If we again apply the discriminant: D = b2 – 4ac, we found that: a 2  6a 1  0 is a
condition to have two real roots. Hence, referencing Graph 1, the conditions that satisfy
this constraint are: a > 5.8 and a < 0.18. Where we have a = 5.8 and a = 0.18, this
condition will produce only a single root for the function ax2 + bx + 1.
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The interval a < 5.8 and a> 0.18 is the range for which values of “a” where we will not have
any real roots as the discriminant: D = b2 – 4ac < 0.
Testing the Findings and Conclusion
In order to test out our findings, we will graph a number of functions for various values of
and b. These functions are tabulated in Table 1.
a
.1
.18
0
1
3
5.82
8
20
b
.9
.82
1
0
-2
-4.82
-7
-19
Function
.1x2 + .9x +1
.18x2 + .82x +1
x+1
x2 + 1
3x2 -2x + 1
5.82x2 -4.82x +1
8x2 – 7x + 1
20x2 – 19x + 1
x-intercepts
two
one
one
none
none
one
two
two
Table 1: Functions derived from various values of a
Testing the various values of a will yield a family of functions as depicted in Graph 2.
Geometer’s Sketchpad was used to create Graph 2.
In order to fully demonstrate the family of functions that resulted, we animated the solution
with Geometer’s Sketchpad. The following is a screenshot of that animation.
The GSP file will be included in the submission of this problem set.
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Graph 2: Family of functions for f(x) = ax2 + bx + 1
From Graph 2, we can interpret the following:
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All functions have the co-ordinates (1,2) and (0,1) in common.
The values of a = 5.82; 0.18 yield parabolas which intersects the x-axis at only one
point.
The value of a=0 yields a straight line with form y = x + 1 and has a single intercept
with the x-axis.
Testing any value of a > 0.18 and a < 5.82, yields no intercept with the x-axis, as
expected.
Only values of a > 5.82 or a < 0.18 produce a function in the form of a parabola
with two intercepts on the x-axis; and a≠0
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Facilitation Discussion
During the classroom facilitation, we had the privilege of having Rob McLeish present. Rob
McLeish is currently an instructional coach with the OCDSB with focus on Secondary
Mathematics.
The process of facilitation for the problem went as follows:
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The problem was introduced by our team and written clearly on the board.
The team handed out TI-83 graphing calculators and instruction sheets as to how
they could be used.
The students were arranged in groups of three and given broad sheets of paper on
which to write their solutions.
The facilitation team went around the room to assist the groups in preparing their
answers on the sheets.
When the teams were finished, each was asked to post their solutions on the wall.
All of the teams then introduced their own solutions and explained their rationale
for their solution.
Outcome and Feedback from the facilitation:
The following results were observed:
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Our facilitation team provided the TI-83 graphing calculator and instructed them
how to use it. However, no team chose to use this technology in their solution
approach.
Rob McLeish indicated that he was dissatisfied with our collective approach to the
solving of the problem.
Rob McLeish was critical of our use of the quadratic equation. He indicated this
was not intuitive as a solution and we had defaulted to using a procedure without
understanding the process required to achieve a Level 4 solution. He encouraged
us to use algebra tiles for the solution.
Rob was critical of our approach with the facilitation of the problem solving. He
indicated that we should have coached the students to encourage thought
surrounding the solution. Instead, we allowed the groups to be procedural and go
directly to the quadratic equation.
Rob encouraged us to differentiate the lesson for visual, auditory, and kinesthetic
learners. This would be using visual aids such as Geometer’s Sketchpad for the
visual learners and algebra tiles or other manipulatives for the kinesthetic learners.
He indicated that we should animate the solution in Geometer’s Sketchpad. We did
this for our final submission.
Only one group of students thought through the case where “a” was equal to zero,
which would yield only one x-intercept.
One group factored the solution, but only came up with the case where “a” was
greater than 5.82. They did not get the other intercept where “a” was 0.18.
Additionally, even though the solution was incorrect, the class did not challenge the
posting.
All other student groups defaulted to the use of the quadratic equation to solve for
the roots of “a”.
None of the groups graphed the solution for a using f(a) = a2 - 6a + 1
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None of the groups graphed the equation f(x) = ax2 + bx + 1 to test various values
of “a”
Rob McLeish indicated that the class demonstrated a level 2 understanding of the
problem solution.
Suggestions for improving the Facilitation:
The following are areas that we would improve from the facilitation exercise:
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We should introduce the use of manipulatives for solving this problem.
We must also encourage use of calculators, spreadsheets, algebra tiles,
Geometer’s Sketchpad. However, students must be comfortable first with
manipulatives and technology to fully explore the solution to Level 4.
We would pose the question differently in order to make the question more leading
to encourage the student to go on past a Level 2 answer. The conditions or
constraints of the question were too subtle to indicate there was a more complex
solution to the problem.
The question should have been more explicitly stated to accommodate different
learners. This is definitely a good accommodation for ESL students to explicitly
state without ambiguity as to what the expectations of the question are.
The students should have had more time to accommodate a richer solution to be
discovered within the time period allowed.
This was also the first class for working on problem sets together, so there was not
the opportunity for the students to become fully comfortable with each other. This
hampered productivity.
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Curriculum Connections:
The given problem is highly applicable to the Grade 11 Academic course, MCR3U. It is
possible that this problem could also be given to MCF3M as the curriculum for this course
covers the application of the quadratic equation and the use of determinants for evaluating
the number of real roots.
This would not be a suitable problem for the MBF3C course as the scope of the problem
exceeds the curriculum expectations for that course.
The following are the curriculum expectations for the various Grade 11 courses
which utilize the quadratic equation.
Grade 11 (MCR3U):
Overall Expectations:
2. Determine the zeros and the maximum or minimum of a quadratic function, and solve
problems involving quadratic functions, including problems arising from real-world
applications;
Specific Expectations:
2.1 determine the number of zeros (i.e. x-intercepts) of a quadratic function, using a
variety of strategies (e.g., inspecting graphs; factoring; calculating the discriminant)
2.2 determine the maximum or minimum value of a quadratic function whose equation is
given in the form f(x) = ax + bx + c, using an algebraic method (e.g., completing the
square; factoring to determine the zeros and averaging the zeros)
2.3 solve problems involving quadratic functions arising from real-world applications and
represented using function notation
2.4 determine, through investigation, the transformational relationship among the family of
quadratic functions that have the same zeros, and determine the algebraic representation
of a quadratic function, given the real roots of the corresponding quadratic equation and a
point on the function
2.5 solve problems involving the intersection of a linear function and a quadratic function
graphically and algebraically
Grade 11 (MCF3M)
Overall Expectations:
1. Expand and simplify quadratic expressions, solve quadratic equations, and relate the
roots of a quadratic equation to the corresponding graph;
2. Demonstrate an understanding of functions, and make connections between the
numeric, graphical, and algebraic representations of quadratic functions;
Specific Expectations:
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1.1 poses problems involving quadratic relations arising from real-world applications and
represented by tables of values and graphs, and solve these and other such problems
1.2 represent situations (e.g., the area of a picture frame of variable width) using quadratic
expressions in one variable, and expand and simplify quadratic expressions in one
variable
1.4 solve quadratic equations by selecting and applying a factoring strategy
1.5 determine, through investigation, and describe the connection between the factors
used in solving a quadratic equation and the x-intercepts of the graph of the corresponding
quadratic relation
1.6 explore the algebraic development of the quadratic formula and apply the formula to
solve quadratic equations, using technology
1.7 relate the real roots of a quadratic equation to the x-intercepts of the corresponding
graph, and connect the number of real roots to the value of the discriminant (e.g., there
are no real roots and no x-intercepts if b – 4ac < 0)
1.8 determine the real roots of a variety of quadratic equations (e.g., 100x = 115x + 35),
and describe the advantages and disadvantages of each strategy (i.e., graphing; factoring;
using the quadratic formula)
2.11 sketch the graph of a quadratic function whose equation is given in the standard form
f(x) = ax + bx + c by using a suitable strategy (e.g., completing the square and finding the
vertex; factoring, if possible, to locate the x-intercepts), and identify the key features of the
graph (e.g., the vertex, the x- and y-intercepts, the equation of the axis of symmetry, the
intervals where the function is positive or negative, the intervals where the function is
increasing or decreasing)
Grade 11(MBF3C)
Overall Expectations:
1.
make connections between the numeric, graphical, and algebraic representations
of quadratic relations, and use the connections to solve problems;
Specific Expectations:
1.8 determine, through investigation, and describe the connection between the factors of a
quadratic expression and the x-intercepts of the graph of the corresponding quadratic
relation
1.9 solve problems, using an appropriate strategy (i.e., factoring, graphing), given
equations of quadratic relations, including those that arise from real-world applications
(e.g., break-even point)
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