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1 Hannah Oppenheim High School Mathematics from the College Perspective There are many connections between the main areas of focus in high school mathematics. By labeling high school course as “Algebra” or “Geometry” or “Precalculus”, students do not always recognize the connections and overlapping ideas between the areas. On top of the connections between high school level mathematical topics, there are also countless connections to more advanced mathematics. Advanced mathematics can be used to investigate high school math further, provide different understandings of topics, equip students with different techniques for solving problems, explain the reasoning behind certain topics, and show more applications of what students learn. College-­level mathematics can also prove certain mathematical properties that high schoolers are taught and assume to be true without an understanding of why. Recognizing these connections and learning how to approach mathematical problems from multiple perspectives can be very beneficial for students. The first step in conveying these connections to students is for teachers to fully understand them themselves. Educators can use their more advanced mathematics education to help teach even the most basic mathematics to their students. The EngageNY website includes resources for teachers to use with their students that meet the Common Core Learning Standards. For each area of mathematics it includes modules that are comprised of lessons. There are materials for students with exercises, videos, demonstrations, and explanations of material. The lessons use repeated reasoning to help 2 students recognize the important aspects of each topic. All student material is accompanied by matching teacher documents that provide detailed guidelines for how to present the material to students and how to assess their work. I will use these lessons and resources to illustrate the connections between high school and college level mathematics. Lesson 1 of the Algebra I Module 1 begins to discuss graphing linear functions. The student outcomes listed for this lesson include: “Students define appropriate quantities from a situation (a “graphing story”), choose and interpret the scale and the origin for the graph, and graph the piecewise linear function described in the video. They understand the relationship between physical measurements and their representation on a graph.” The EngageNY teacher materials list three focus standards for mathematical practice that focus on the process of doing math and are addressed in Lesson 1. MP.1 ​Make sense of problems and persevere in solving them.​ Students are presented with problems that require them to try special cases and simpler forms of the original problem to gain better understanding of the problem. MP.3 ​Construct viable arguments and critique the reasoning of others.​ Students reason about solving equations using “if-­then” moves based on equivalent expressions and properties of equality and inequality. They analyze when an “if-­then” move is not reversible. MP.6 ​Attention to precision.​ Students formalize descriptions of what they learned before (variables, solution sets, numerical expressions, algebraic expressions, etc.) as they build equivalent expressions and solve equations. Students 3 analyze solution sets of equations to determine processes (e.g., squaring both sides of an equation) that might lead to a solution set that differs from that of the original equation. The lesson starts off by presenting students with a real life situation, such as somebody walking up and down stairs, and having them create an elevation vs. time graph (See Figure 1). Figure 1 The lesson focuses on creating a graph and understanding what it represents. It also involves interpreting linear graphs when an explanation is not given (See Figure 2). 4 Figure 2 This requires students to use everything they have learned about the behavior of graphs and what different aspects represent and apply it to a possible situation. Not only does this approach incorporate an understanding of functions and graphing, but also units and the application of functions. The answers to the problem shown in Figure 2 will vary from student to student. One such story that the teacher materials that accompany the lesson include is “A swimmer climbs a ladder to a water slide, sits for two seconds at the top of the slide, and then slides down the slide into the water. She stays steady at the same position underwater for two seconds before rising to the surface.” Teachers are encouraged to accept answers in multiple different contexts, as long as there is some understanding of the zero and subzero elevation. Additionally, students are introduced to piecewise linear functions. The lesson is set up in a way that has students plot their own piecewise linear functions based on a scenario before they are even introduced to the term. This is done so students create their own understanding of how the functions work, rather than being told in a way they might not initially comprehend. It 5 is not until the students have worked with this type of function and understand what the changes in direction mean that they are given a definition and explanation for piecewise linear functions. By examining the teacher materials that are provided with the lesson, it is clear that the emphasis in the lesson is not on the exact function or the equations that accompany a graph. Instead, the main focus is how certain behaviors can be represented and understanding what the behaviors mean. At one point students are asked, given the elevation vs. time graph of someone walking down stairs that they created (See Figure 1), over which time interval does the person descend the stairs the fastest. Students would then calculate the rates by dividing the person’s change in elevation by the number of seconds it took and compare that to other rates. What students may not realize is that they are using algebra to calculate change in distance with respect to time, or the slope of the graph. While students have some knowledge about slope from discussing linear functions in middle school, they have a deeper understanding of the meaning well before being introduced in a calculus class. By concluding that the person descends faster from 8.5-­10 seconds than 0-­6 seconds, students can then look at the graph and make certain observations about graphical representations of rate of change. For instance, a horizontal line means there is no change in distance over a period of time. Additionally, the steeper a segment of a graph is, the greater the the magnitude of the rate of change. Lesson 2 of Algebra I Module 1 explores graphing quadratic functions. The student outcomes for the lesson include: “Students represent graphically a non-­linear relationship between two quantities and interpret features of the graph. They will understand the relationship 6 between physical quantities via the graph.” Two focus standards for mathematical practice that are addressed in Lesson 2. MP.1 ​Make sense of problems and persevere in solving them.​ Students are presented with problems that require them to try special cases and simpler forms of the original problem to gain better understanding of the problem. MP.4 ​Model with mathematics.​ Students have numerous opportunities in this module to solve problems arising in everyday life, society, and the workplace from modeling bacteria growth to understanding the federal progressive income tax system. The lesson starts off with a video of a man jumping from 36 feet above ground into 1 foot of water. Students are asked to plot a graphical representation in change in elevation with respect to time for the jump (See Figure 3). 7 Figure 3 The teacher materials that accompany the lesson emphasize that students are graphing a function of time ​t​, not a physical trajectory of the diver. This is a possible misconception by students. Some graphed functions do look similar to the physical trajectories they are associated with, but not all do. For instance, the diver’s motion was straight down into the pool, Figure 3 does not have a vertical component that matches this. However, if the diver jumped outward into a pool, the shape of the graph may match the shape of the graph in Figure 3, but the ​x​-­variable would be position, not time. For this reason, it would be beneficial for students if teachers explained the difference between a position graph ​f(x)​ and a time graph ​f(t)​. The ​x​-­variable in ​f(x)​ is the position of an object, where the ​x​-­variable in ​f(t)​ is time. 8 One problem presents students with an elevation versus time graph of a ball rolling down a ramp (See Figure 4). Figure 4 The problem addresses the changing speed of the ball as it rolls down the ramp. It asks at what point is the speed of the ball the fastest. This incorporates the idea that change in elevation with respect to time is linked to speed. This directly relates to ideas discussed in calculus such as slopes and derivatives. Another ramp problem presents students with a diagram of an actual ramp, rather than an elevation vs. time graph of the ball (See Figure 5). Figure 5 9 Students explore the average speed of a ball rolling down the ramp over a certain time interval, as well as the average rate of horizontal and vertical change. While this problem also involves the speed of a ball rolling down a ramp, it involves a more geometric approach towards the situation. Students use their knowledge of right triangles and the Pythagorean Theorem to examine the behavior of the ball rolling down the ramp. Students make the connection that the Pythagorean Theorem, even though traditionally used when measuring side lengths of a right triangle, can also be applied to the average speed of the ball in each direction. This means that the ​a ​= the average rate of vertical change of a ball with respect to time ​t​, ​b ​= the average rate of horizontal change of a ball with respect to time ​t​, and ​c ​= the average speed of a ball with respect to time ​t​ (along the hypotenuse of the ramp). An important aspect of all of the exercises is students’ ability to articulate what they did and justify the choices they made. This allows them to express their understanding of the instructions, story, and graphing procedures. An understanding of the reasoning behind all graphing choices contributes to a student’s ability to apply what he/she learned to other situations. This is the same approach that is taken in most college math courses. Especially in proof based classes, the focus is on the explanations behind mathematical properties and proving them to be true.. Lesson 3 of Algebra I Module 1 introduces students to graphing exponential functions. The student outcomes for this lesson include: “Students choose and interpret the scale on a graph to appropriately represent an exponential function. Students plot points representing the number of bacteria over time, given that bacteria grow by a constant factor over evenly 10 spaced time intervals.” Two focus standards for mathematical practice that are addressed in Lesson 3. MP.4 ​Model with mathematics.​ Students have numerous opportunities in this module to solve problems arising in everyday life, society, and the workplace from modeling bacteria growth to understanding the federal progressive income tax system. MP.6 ​Attention to precision.​ Students formalize descriptions of what they learned before (variables, solution sets, numerical expressions, algebraic expressions, etc.) as they build equivalent expressions and solve equations. Students analyze solution sets of equations to determine processes (e.g., squaring both sides of an equation) that might lead to a solution set that differs from that of the original equation. After watching a short video clip on bacteria growth, students plot points on a graph of the number of bacterial versus time in minutes (See Figure 6). Figure 6 11 Students then plot points on a graph of the number of bacterial versus time in hours, rather than minutes (See Figure 7). Figure 7 Through this process students are introduced to a graph of an exponential function by creating their own. They can observe and interpret the behavior of an exponential function in terms of a biological application. By plotting both graphs, students are able to recognize that the graphs have the same form even though the length of time the graph represents significantly changes. It is also important to note that the bacteria are increasing at an increasing rate, and the percent rate of increase is constant. A constant increase implies a linear function, but a constant rate of increase does not. The consistent shapes of the graphs in different units of time can be explained by this constant rate of increase. This lesson also provides an overview in understanding different graphical representations of functions the students have learned about. In one problem, students are presented with three stories and four different graphs (See Figure 8). 12 Figure 8 This encompasses a student’s abilities to interpret and create graphs. Story 1 can be matched with graph (d) by recognizing that a horizontal line represents no change in population over a period of time. Story 2 can be matched with graph (b) by recognizing that the function needs to always be increasing, but does not have a constant rate of change as graph (a) represents. Instead of using a process of elimination in this case, calculus students would recognize this as an opportunity to draw tangent lines at different points on the line and compare the slopes of each tangent line. The larger the slope of the tangent line, the greater the rate of change and the faster the population size grows. Calculus students can also recognize that the concavity (concave down, increasing) of the graph means that the population is increasing at a decreasing rate. Story 3 is not represented by any of the four graphs given, but there are many different possibilities. It 13 can be any graph that is always decreasing and ends up at ​y​=0. It can be linear, concave up, concave down, etc. (See Figures 9, 10, 11). Figure 9
Figure 10
Figure 11 Graph (a) represents the population increasing at a constant rate and graph (c) represents the population size increasing, then decreasing, and then increasing again at a slower rate than before. As discussed in previous problems, the larger, steeper slope of the first increasing portion of graph (c) means the population is increasing at a faster rate than the second increasing portion of graph (c). While the EngageNY materials do not directly reference connections to advanced mathematics for teachers to utilize, there are plenty of opportunities for educators to expand upon the content. There are other resources that clearly link high school mathematics to college level mathematics. Zalman Usiskin’s ​Mathematics for High School Teachers: An Advanced Perspective​ is a text intended for current or prospective mathematics teachers. The book looks at mathematics that stem from high school mathematics courses and addresses them from a more advanced mathematics point of view. 14 One problem in Usiskin’s book is, given an equation for a line and a point off that line, find the shortest distance between the point and the line. This is a more complex problem than it may seem and can be solved using multiple different techniques. Students can approach this from many different perspectives. The multiple methods allow students to utilize different areas of mathematics. They also provide an opportunity for students to check their work and verify that the results they got from one method match the results using a different method. One method involves an algebraic and geometric approach. Let’s assume the students are given a point (-­1,3) to be off the line y = 4x − 1 . The point on the line y = 4x − 1 that is closest to (-­1,3) can be found by drawing a line through (-­1,3) that is perpendicular to y = 4x − 1 on a graph. The fact that drawing the perpendicular line provides the shortest line between the point and the line can be proven true by creating a triangle on the graph of y = 4x − 1 and (-­1,3). Let’s call (-­1,3) point ​A​. Let ​B​ be the point on the line where the perpendicular intersects y = 4x − 1 and let ​C​ be any other point on y = 4x − 1 . Connect points ​A​, ​B​, and ​C​ with lines to create Δ ABC​. Δ ABC is a right triangle with legs AB ​and BC ​ and hypotenuse AC ​(See Figure​ ​12). Figure​ ​12 15 It is known that the hypotenuse of a right triangle is longer than either of the legs of the right triangle. Therefore, AC > AB ​and the distance of a perpendicular line drawn between (-­1,3) and y = 4x − 1 is shorter than a non-­perpendicular line. If the slope of y = 4x − 1 is 4, the slope of its perpendicular is the negative inverse, -­¼. Therefore, the perpendicular line, written in slope-­intercept form, is y =− 14 x + b . Since it is known that the point (-­1,3) is on the line y =− 14 x + b , these values can be plugged in for ​x ​and ​y and solved for b=11/4. This means the perpendicular line is y =− 14 x + 114 . It is also known that the point on the line y = 4x − 1 that is closest to (-­1,3) also lies on y =− 14 x + 114 . So students can set 4x − 1 =− 14 x + 114 and solve for ​x​, and then ​y​ to get the point (15/17, 43/17). This procedure to solve the problem uses a basic knowledge of geometry and algebraic tools. Another approach is one calculus students would find familiar. It involves minimizing the distance from the point (-­1,3) off the line to the point (​x,y)​ on the line. The point (​x,y​) represents the point on the line closest to (-­1,3). A student would start off with an equation, d=
√(x − x ) + (y − y ) , that represents the distance between the two points. This derives 2
1
2
2
1
2
from the Pythagorean Theorem, a2 + b2 = c2 . This geometric perspective involves the line connecting (-­1,3) and (​x,y​) being the hypotenuse of a triangle. The ​a​ component of the theorem is one of the legs of the triangle and represents the change in ​x​-­values between the two points, the ​b​ component is the other leg of the triangle and represents the change in ​y​-­values between the two points. 16 Again, let’s use a point (-­1,3) to be off the line y = 4x − 1 . These values would be plugged into d =
√(x − x ) + (y − y ) and simplified to get f (x) = √17x − 30x + 17 . The 2
1
2
2
1
2
2
steps to get this solution can be seen below: d=
√
(x 2 − x 1)2 + (y 2 − y 1)2 =
√(− 1 − x) + (3 − y) =
√(x + 1) + (3 − 4x + 1) =
√(x + 1) + (4 − 4x) 2
2
2
2
2
since y = 4x − 1 2
= √x 2 + 2x + 1 + 16 − 32x + 16x 2 f (x) = √17x2 − 30x + 17 The next step is to minimize f (x) . Calculus students would recognize minimize to mean take the derivative of the function. They would get f ′(x) = 12 √17x34x−30
2−6x+17 . Students would then set ​f’(x)​=0 and solve for ​x​=15/17, plug that into y = 4x − 1 and solve for ​y​=43/17. Therefore, the point (15/17, 43/17) the the closest point to (-­1,3) on the line y = 4x − 1 . As shown, this problem, just like many others, can be solved from different perspectives using both high school and college level mathematics. While a student is not required to have a knowledge of calculus in order to complete the problem, even the first approach that uses geometry and algebra hints at certain elements of calculus and can be explained by calculus. Ira J. Papick, however, suggests that this advanced knowledge is required by educators in order for them to successfully communicate idea to their students. 17 Papick discusses the notion of “mathematical knowledge for teaching” in ​Strengthening the Mathematical Content Knowledge of Middle and Secondary Mathematics Teachers​. He presents a list of questions students often ask that algebra teacher should be able to answer using their deeper mathematical knowledge. 1. My teacher from last year told me that whatever I do to one side of the equation, I must do the same thing to the other side to keep the equality true. I can’t figure out what I’m doing wrong by adding 1 to the numerator of both fractions in the equality 12 = 24 and getting 22 = 34 . 2. Why does the book say that a polynomial anxn + an−1xn−1 + ... + a1x + a0 = 0 if and only if each ai = 0 , and then later says that 2x2 + 5x + 3 = 0 ? 3. You always ask us to explain our thinking. I know that two fractions can be equal, but their numerators and denominators don’t have to be equal. What about if ab = dc , and they are both reduced to simplest form. Does a = c and b = d , and how should we explain this? 4. I don’t understand why (− 3) × (− 5) = 15 . Can you please explain it to me? 5. The homework assignment asked us to find the next term in the list of numbers 3, 5, 7,...? John said the answer is 9 (he was thinking of odd numbers), I said the answer is 11 (I was thinking odd prime numbers), and Mary said the answer is 3 (she was thinking of a periodic pattern). Who is right? √
6. We know how to find 22 , but how do we find 22.5 or 2 2 ? 18 +x−6
7. My algebra teacher said x x−2
= [(x+3)(x−2)]
= x + 3 , but my sister’s boyfriend (who (x−2)
2
is in college) says that they are not equal, because the original expression is not defined at 2, but the other expression equals 5 when evaluated at 2. While many of these questions can be answered using an authoritarian approach and declaring certain mathematical properties to be true, it is important for students to learn why these rules of mathematics are true. The first question Papick presents requires the student to have an understanding of common denominators. By adding 1 to the numerator of both fractions, the student is really 1 1
2
2+1
2 1
3
adding 12 + 12 = 24 + 14 to get 22 = 34 . This is because 1+1
2 = 2 + 2 = 2 and 4 = 4 + 4 = 4 . The student is not actually adding the same thing to both sides, he or she is adding 12 to one side and 1
4
to the other. In order to add 1 to each side, the student would have to find a common denominator between 12 and 11 , and 24 and 11 . All of the values this student is working with are rational numbers. The teacher would be able to make deeper connections between the student’s question and the additive properties of Q . For instance, how rational numbers under addition are closed (the sum of any rational numbers is always a rational number) and rational numbers under addition are commutative. There are many different approaches for answering question #4. Diana Brown presents six different methods in her paper “Why is a negative number times a negative number a positive number?” Method 1 involves utilizing money representation. Given a scenario such as “an employer removes the $100 mortgage deduction from your paycheck for 12 months,” students can figure out the answer by multiplying − 100 ×− $12 = $1, 200 . The gaining of money can be observed through the positive result after multiplying two negative numbers. Method 2 is a 19 mathematical illustration. Most students will agree that (− 1)x =− x and (− 1)(0) = 0 . Start with with rewriting (− 1)(0) = 0 as (− 1)(0) = (− 1 + 1) and then follow the steps below: (− 1)(0) = (− 1)(− 1 + 1) = (− 1)(− 1) + (− 1)(1) = (− 1)(− 1) + (− 1) . Therefore, (− 1)(0) = (− 1)(− 1) + (− 1) and 0 = x + (− 1) . We know x must equal 1, so (− 1)(− 1) = 1 and a negative times a negative equals a positive value. A third method involves a proof. First, let ​a​ and ​b​ to be any two real numbers. Consider x​ to be defined as x = ab + (− a)(b) + (− a)(− b) . Next, factor out − a to get x = ab + (− a)[(b) + (− b)] = ab + (− a)(0) = ab + 0 = ab You could also factor out b to get x = [a + (− a)]b + (− a)(− b) = (0)(b) + (− a)(− b) = 0 + (− a)(− b) = (− a)(− b) So x = ab and x = (− a)(− b) Therefore, ab = (− a)(− b) . Since we know the product of two positive numbers, (ab) , is positive, the product of two negative numbers, [(− a)(− b)] , must be positive as well. Other methods involve word representations, pattern recognition, and using technology. With each method, teachers utilize their advanced knowledge to explain a basic property of numbers. 20 The axioms of a ring come into play when conducting a proof. The axioms include: ● Addition is associative ● Addition is commutative ● The ring contains an additive identity, 0 ● Every element of the ring has an additive inverse These axioms are the rules of arithmetic needed to conduct a proof of why a negative number times a negative number is a positive number. For instance, in the proof above, it is essential to know that (− a) is the additive inverse of a and (− b) is the additive inverse of b . James Tanton outlines arguments for this complicated question that utilize the axioms of a ring. There are many different ways to approach question #5. On one hand, the three students are correct. If the pattern was odd numbers, the next number in the sequence would be 9. If the pattern was odd prime numbers, the next number would be 11. If the pattern was periodic, the next number could be 3. However, the next number could also be found using polynomials. Since the first three numbers in the sequence are 3, 5, 7, in order to find the next term, we would want to find a polynomial p(x) such that p(1) = 3, p(2) = 5, and p(3) = 7 . Depending on what degree polynomial you are working with, you will get different results for the following terms. For instance, say you are dealing with a first-­degree polynomial in the form p(x) = ax + b , then the only possible polynomial that would yield the correct results for p(1), p(2), and p(3) would be p(x) = 2x + 1 . Therefore, the next term in the list would be 9 since p(4) = 9 . If you were dealing with third degree polynomials in the form p(x) = ax3 + bx2 + cx + d , then you would have to solve for the system of equations 21 a + b + c + d = 3 8a + 4b + 2c + d = 5 27a + 9b + 3c + d = 7 Since there are four unknowns and three equations, you must choose the value for one variable in order to determine the others. In linear algebra, this is known as the free variable. When you solve this system, you get b =− 6a , c = 2 + 11a , and d = 1 − 6a . This means p(x) = ax3 + (− 6a)x2 + (2 + 11a)x + (1 − 6a) = ax3 − 6ax2 + 11ax + 2x − 6a + 1 Let the free variable, a , be 1. Then b =− 6 , c = 13 , and d =− 5 . Therefore, the polynomial when a = 1 is p(x) = x3 − 6x2 + 13x − 5 . Using this polynomial you can find p(4), p(5) , and so on to continue the list of numbers. Using this method, there are infinite possibilities for the next term in the series of numbers. Many different approaches can be taken when integrating advanced mathematics into high school mathematics lessons. Educators can use their knowledge to improve upon their own explanations and techniques, and, in turn, benefit their students’ understanding of the ideas. By making connections, students have the opportunity to expand their knowledge, discover new applications of mathematics, and explore areas of the subject beyond the classroom material. Teachers and students should all have the skills necessary to connect different mathematical concepts. It is a learning tool that is valuable for students at all levels of education. 22 References Algebra I Module 1. Retrieved from https://www.engageny.org/resource/algebra-­i-­module-­1 Brown, D. (2005, January 1). Why is a negative number times a negative number a positive number? Retrieved April 1, 2015, from http://jwilson.coe.uga.edu/EMAT6680/Brown/6690/negneg.htm Papick, I. J. (2011). Strengthening the mathematical content knowledge of middle and secondary mathematics teachers. ​Notices of the AMS,​ ​58​(3), 389-­392. Tanton, J. (2012, October 1). Why negative times negative positive. Retrieved April 1, 2015, from http://www.jamestanton.com/ Usiskin, Z., Peressini, A., Marchisotto, E., & Stanley, D. (2003). Functions. In ​Mathematics for high school teachers: An advanced perspective​. Upper Saddle River, N.J.: Pearson Education.