Using Symbolic Geometry to Teach Secondary School Mathematics - Geometry Expressions Activities for Algebra 2 and Precalculus

Transcription

1 Using Symbolic Geometry to Teach Secondary School Mathematics - Geometry Expressions Activities for Algebra and Precalculus Irina Lyublinskaya, CUNY College of Staten Island, Staten Island, NY, USA and Valeriy Ryzhik, Lycee-Physical Technical High School, St. Petersburg, Russia Saltire Software Inc., Beaverton, OR USA

2 Copyright Saltire Software 008 ISBN

3 Table of Contents Introduction...5 Discovering Parabolas...7 Part 1 Parabola by 3 Points...7 Part The Existence of a Parabola Passing Through Three Arbitrary Points...1 Extensions:...14 Solving Systems of Equations (Inequalities) with Parameters...17 Part 1 Setting Up the Problem in Geometry Expressions...17 Part When the System Has No Solutions...18 Part 3 Investigation of the Number of Solutions...19 Part 4 Solving Systems of Equations...0 Part 5 Solving Systems of Inequalities with Parameters... Extensions:... Stained Glass Design...3 Part 1 Setting Up Problem in Geometry Expressions...4 Part Creating Stained Glass Design...5 Part 3 Finding Equations of the Curves in the Stained Glass Design...8 Part 4 Verification of the Equations with Geometry Expressions...9 Extensions...31 Translation Along Coordinate Axes...35 Part 1 Translation Along the Y-Axis...35 Part Translation Along the X-Axis...39 Part 3 Commutative Property of Translation...43 Part 4 Applications and Assessment Problems...46 Extension:...50 A Little Trig...53 Part 1 Investigating Area of the Triangle...54 Part Optimizing the Perimeter of the Rectangle...58 One Hyperbola...61 Part 1 Investigating Area of the Rectangle...6 Part Optimizing Perimeter of the Rectangle...64 Part 3 Optimization of the Diagonal of the Rectangle....66

4 Three Extrema (Circle)...69 Part 1 Length of a Tangent Segment to a Circle...70 Part Area of a Triangle Formed by a Tangent Line and the Coordinate Axes...75 Part 3 Perimeter of a Rectangle whose Diagonal is a Tangent Segment...78 Two Parabolas...83 Part 1 Optimizing Perimeter of Rectangle...84 Part Optimizing the Diagonal of the Rectangle...88 Part 3 Comparison of Points of Extrema for Perimeter, Diagonal, and Area...91

5 Introduction The National Council of Teachers of Mathematics (NCTM) Standards advocate a unified approach to mathematics education incorporating multiple strands in coherent focused elements. Emphasis is placed on the use of technology, visual thinking, and the connection between geometry and algebra. The notion of integrating algebra and geometry is at the forefront of mathematics education in the US today, potentially transformative of mathematics education in the 1 st century when technology integration is not only reasonable because of access but also because of the nature of the impact of mathematics on knowledge and directions in this century. While geometry and algebra systems have existed separately for the last couple of decades, an application that unifies the two is available through a brand new technology. This unification facilitates a variety of well-understood goals: The user interface must be simple so that students can focus on the reasoning and problem solving rather than struggle with the technology. The application must enable exploration. As much as possible, the technology must blur the artificial separation between the algebra and the geometry. It must provide easy access but also be challenging. That is, it must be easily used by students at a wide range of performance levels while also being capable of providing challenge to the most able students. It must address traditional concepts in addition to facilitating the use of more modern and realistic approaches to geometry and to real applications. The mathematical concepts addressed must be expanded in order to make a major impact on the high school curriculum. The technology that meets these goals is now available. Geometry Expressions developed by Saltire Software, is a computer application that, unlike other interactive geometry systems, can automatically generate algebraic expressions from geometric figures. A simple algebra system is embedded directly into Geometry Expressions. Generation and simplification of expressions along with simple algebraic manipulations are handled inside Geometry Expressions. From a pedagogical standpoint an interactive symbolic geometry system affords a remarkable opportunity to make concrete the concept of a variable in a readily identifiable real world setting. An educational tool takes advantage of this opportunity if it provides outputs in a symbolic form. These outputs are algebraic expressions using the input parameters. Our goal was to use Geometry Expressions to develop a set of problems for second year algebra and/or precalculus courses addressing the topics of geometric transformations of functions and optimization. In this volume you will find eight interactive problems with different levels of difficulty. In each problem the main focus is on the development of the students ability to recognize and make connections using multiple representations of the same object, such as geometric shape and function. By connecting geometric and algebraic representations the student develops a more thorough understanding of the problem. 5

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7 Discovering Parabolas Time required 90 minutes Teaching Goals: 1. Students verify that a unique parabola with the equation y = ax + bx+ c, a 0, exists through any three given points if and only if these points are not collinear.. Students verify that an infinite number of parabolas with the equation y = ax + bx+ c, a 0, exist through any two given points. 3. Students should be able to construct the parabola defined by the coordinates of three points and find its equation with the help of the software. 4. Students should be able to construct this parabola by geometric transformations from the graph of the parabola y = x and confirm the equation given by the software. Prior Knowledge Students should know the graph of quadratic functions in two forms: y = ax + bx+ c and y = a( x x0) + y0, a 0, and they should know this graph has the shape of a parabola. Students should know geometric representations and coordinate forms of the following transformations: translation along the coordinate axes, reflection about coordinate axes, and dilation along coordinate axes. Problem: This is a problem of the existence of a quadratic function (a parabola) as a graph. Since we consider graphs of functions, situations when two points are located on the same vertical line are not considered. Part 1 Parabola by 3 Points In this part of the problem students are given three specific points, the maximum, with coordinates (, 3), and other points with coordinates (0, -1) and (3, ). They are also given the graph of an initial parabola y = x and the graph of a general parabola y = a( x x0) + y0. 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Function from the Draw toolbox. Type x^ in the Function Type dialog. 3. Select Function from the Draw toolbox. Type y[0]+a*(x-x[0])^ for the function. Click the graph of the function to highlight it, right click the mouse to invoke the context menu, and select Properties. From the Display Properties dialog choose Line Color Blue. 7

8 4. Select Point from the Draw toolbox. Construct three points anywhere on the screen. Select point A, choose Coordinate from the Constrain toolbox and type coordinates (, 3) for the point A, press Enter. Repeat this operation to constrain points B and C. If desired, change the font, size and color for the labels of the points. 4 3 A(,3) Y=y 0 +a X-x 0 C (3,) 1 Y=X B (0,-1) - Q1. Can you find a parabola that has an equation with integer numerical coefficients and will go through these three points? If it exists, what is the equation of this parabola? In order to do that: Q. Investigate which parameters in the equation of the parabola y y a x x responsible for which type of transformation of the graph. = 0 + ( 0) are 1. Click-and-drag the parabola. When you move the parabola, a small circle appears on the curve. We refer to this as the drag handle. A variable name labels the drag handle. Note which way you are able to move the curve with the parameterized handle (click-anddrag the handle).. Click another part of the parabola and another drag handle appears on the curve with another variable label. Create three drag handles for the three variables of the equation. Observe how each handle / variable is responsible for a type of transformation of the curve. 3. Record your observations in the table below: 8

9 Parameter in the equation y = y + a( x x ) 0 0 Transformation (vertical translation, horizontal translation, dilation) a Y 0 X 0 A. a dilation, y 0 vertical translation, x 0 horizontal translation Q3. Drag and move parabola until it passes through the points A, B, and C. Construct your conjecture about the equation of the parabola (round numerical values to integers). 4 3 A(,3) x 0 C (3,) a 1 Y=X B (0,-1) Y=y 0 +a X-x 0 y 0 - Students should vary parameters a, x 0, and y 0 to try to complete the task. Observe the values of the parameters in the Variables toolbox. When they think they have the solution they can round the numeric values of the coefficients with the help of the software and verify their equation by substitution. In the figure above, a 0.998, x0.003, y A. The equation of the parabola is: y = 3 ( x ). Here are steps for verification: 9

10 1. In the Variables toolbox, enter the predicted values for the parameters, a= 1, x0 =, y0 = 3. Change a variable value from the toolbox by clicking the variable row and entering a new value in the data entry window below the list.. Click the graph of the parabola, then from the Calculate toolbox click the Real tab and the Implicit equation icon. The numeric equation for the parabola will appear on the screen 4 3 A(,3) C (3,) Y=X 1 Y=3-(-+X) B (0,-1) Y=y 0 +a X-x 0-3. Substitute values of all three points and verify that all three points lie on this curve. Q4. Find geometric transformations that transform the graph of y = x into the graph of the parabola in the last problem. Is it important to preserve the order of transformations? Verify your answers with the help of the software. A. The transformations are: translation by the vector (, 3) and dilation by the factor -1. These are commutating. Students should complete the construction, verify coincidence of the image with the parabola and the fact that the order of transformation does not affect the result. Here are the steps of the construction: 1. Select Vector from the Draw toolbox and construct a vector from the origin to point A.. Click the graph of the parabola y = x, select Translation from the Construct toolbox and then click the vector. The image of the translated parabola will appear on the graph. 10

11 3. Click the translated curve, select Dilation from the Construct toolbox. Select point A and leave b in the data entry box for the dilation factor (press Enter to accept the value). 4. In the Variables toolbox setup a range of values for the dilation factor b from - to 0. Click variable b (the row will be highlighted), enter - in the Minimum value box in the lower left corner and 0 in the Maximum value box in the lower right corner. Drag the slider bar to adjust the curve so that it coincides with the graph of the parabola y = 3 ( x ). The value of b should be b A(,3) C (3,) Y=X 1 Y=3-(-+X) - -1 E B (0,-1) Y=y 0 +a X-x 0 - Q5. Do you think there exists another parabola passing through points A, B and C? Explain your answer. A. Three non-collinear points define a parabola uniquely. This can be proved analytically. 11

12 Part The Existence of a Parabola Passing Through Three Arbitrary Points In this part of the problem students investigate how the relative positions of three arbitrary points affects the existence of a parabola passing through these points. Given three points on the plane, AB,, and C, and a parabola defined by the equation y = y + a( x x ) Open a new file. If the axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Function from the Draw toolbox. Type y[0]+a*(x-x[0])^ for the function. Click the graph of the function to highlight it, right click the mouse to invoke the context menu, and select Properties. From the Display Properties dialog choose Line Color Blue, Line Style Solid. 3. Select Point from the Draw toolbox. Construct three points anywhere on the screen. 4. Select point A and the parabola. Select Incident from the Constrain toolbox and point A will appear on the curve. Repeat the same procedure with the points B and C. Q1. How can you position points A, B, and C, so that the parabola no longer exists, e.g. the shape of the parabola changes into a different shape? A. When points lie on the same line, the parabola becomes a line. Since we use the equation of parabola in the form y = y0 + a( x x0) the line will always be horizontal in the form y = constant and a = 0. Students can verify this with following the steps: 1. Move all points to lie on the line y = some constant (approximately). The parabola will look like a horizontal line.. Observe the value of parameter a in the Variables toolbox. The value of a is close to zero. Assume that a = 0 and enter 0 for the value of a. 1

13 6 4 A B Y=y 0 +a X-x 0 C Check the real coordinates of points A, B, and C and. Make sure to use the Real tab in the Calculate toolbox. Select each point, one at a time and choose Coordinates from the Calculate toolbox. 4. Click the line, and select Implicit equation (Real tab). The equation of the line will appear on the screen confirming the fact that the line is not a quadratic function. 13

14 6 Y= A ( ,) 4 ( ,) B C Y=y 0 +a X-x 0 ( ,) Extensions: 1. The existence of a parabola: given a parabola defined by the equation y = ax + bx+ c and three points, A, B, C, follow steps similar to those above to determine that if three points are collinear, then a = 0, when the points lie on an inclined line with slope b, and therefore the parabola does not exist. 14

15 (3.9874, ) Y=1+ X 8 ( , ) C 6 A Y=X a+x b+c 4 ( , ) B The existence of a parabola through any two points: given two points, the choice of a 3 rd point defines a unique parabola, so there are an infinite number of parabolas since there are an infinite number of choices for the 3 rd point. In this case the teacher can provide students with conditions based on real problems, for example: a) given the point of a basketball shot and the position of the basket, make the basket b) make a pass with a volleyball at its highest point (vertex) c) trace a soccer ball kick so that it hits the right upper corner of the goal. 15

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17 Solving Systems of Equations (Inequalities) with Parameters Time required minutes Teaching Goals: 1. Students apply a graphical method for solving systems of equations (inequalities) with parameters.. Students apply geometric transformations of graphs of functions in order to solve systems of equations (inequalities) with parameters. 3. Students obtain solutions of systems of equations (inequalities) in symbolic form. Prior Knowledge Students should know the graphical method of solving equations and inequalities without parameters. Students should know graphs of linear functions and the equation of a circle with the center at the origin. Students should know that the equation y = f(x) + a corresponds to the translation of the graph of the function y = f(x) by a along the y-axis. Problem: Investigate the following system of equations with a parameter and find its solutions: y = x + a x + y = 1 Part 1 Setting Up the Problem in Geometry Expressions 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Line Segment from the Draw toolbox and construct segment AB with point A at an arbitrary position on the y-axis. Do not put point A on the origin. Click the selection arrow to return to select mode. 3. Select the line segment and then choose Slope from the Constrain toolbox. In the data entry box type 1 for the slope followed by the Enter key. 4. Repeat steps -3 for the other segment, AC, with the slope -1. As a result the graph should look like the function y = x on the restricted domain shown below. 17

18 5. Drag point A up and down the y-axis. If the construction is made correctly the shape of the graph should remain the same and the point A should remain on the y-axis. 6. Select Circle from the Draw toolbox and construct a circle with the center at the origin. Return to select mode and double-click the label of the center of the circle and re-name it to be point O. 7. Select the circle and choose Radius from the Constrain toolbox. In the edit box enter 1 to fix the radius of the circle to be constant and equal to Select point O, click Coordinate in the Constrain toolbox and type (0, a). 9. The drawing is shown on the picture below. The problem is now set up and ready for investigation. Part When the System Has No Solutions Q1: Consider a > 1. Check that the system of equations has no solutions. A: It is expected that students will be moving the point A up and observing the values of parameter a in the Variables toolbox. Students can see that when a > 1 the function no longer intersects the circle, so there are no solutions. Q: Are there any other values of parameter a when the system has no solutions? A: It is expected that students will move point A down until the graph of the function does not intersect the circle and their answer will be: yes, there are values of a when there are no solutions. Q3. Can you find these values? A: In order to answer this question, students should realize that they need to determine values of parameter a when the line segments are tangent to the circle. In order to do that, the distance from the center of the circle to the segment should be equal to the radius of the circle. 18

19 The following steps should be followed in order to do that: 1. Select the center of the circle and one of the segments (hold the CTRL key to select two objects simultaneously). Choose Distance/Length from the Calculate toolbox, Symbolic tab.. The distance value will appear on the screen. C 1 1 D - a B O A (0,a) Student should solve ( a < ). a = 1 for a and decide what values of a will lead to no solutions 4. After students solve this problem, delete the measurement from the screen. Part 3 Investigation of the Number of Solutions In this part of the problem students investigate conditions when the system of equations has one, two, three, or four solutions. The main task of this part of the problem is not to solve the system of equations, but to consider the problem qualitatively. Q4: How many solutions can this system have when parameter a is changing? A: Expected answer: the system can have one, two, three, four or no solutions. 19

20 Q5: What are the values of parameter a when the system has one, two, three or four solutions. A: Students should move point A and observe the values of parameter a (the y coordinate of point A) in the Variables toolbox or on the graph. They should have seen already that when two segments were tangent to the circle, the system had two solutions, and they already determined the value of a at this tangency to be a =. The next set of values for a where the function intersects the circle in points only, occurs when 1< a < 1. When a = -1 there are three points of intersection and when a = 1 there is one point of intersection. The system has 4 solutions when < a < 1, where each segment intersects the circle twice. The teacher could expand question 5 with more specific details, such as: what are the values of parameter a when at least one of the solutions is in the 1 st quadrant ( nd, 3 rd, 4 th ); or when is the solution larger than the absolute value of 0.5 (or any other specific value), etc. Part 4 Solving Systems of Equations Q6: Find the solution of the system of equations for values of parameter a in the cases where the solutions exist. A: One of the approaches that students might take is to set up a diagram when all four solutions exist and to find intersection points and their coordinates using Geometry Expressions. Here are the steps for this procedure: 1. Select the circle and one of the line segments (hold the CTRL key to select both objects simultaneously). Choose Intersection from the Construct toolbox. Points of intersection will appear on the screen.. Repeat the same procedure for the second segment. For convenience you may change the label of the points of intersections. 3. Select one of the points of intersection and choose Coordinates from the Calculate (Symbolic) toolbox. The coordinates of this point (in terms of parameter a) will appear on the screen. 4. Repeat that for each point of intersection. 0

21 C B 1 D a- -a a+ -a, a+ -a K -1 a- -a, L O -1 M A (0,a) 1 3 N -a - -a -a+ -a a+ -a, a- -a, - -3 Comment: in order to find points of intersection between the curves, Geometry Expressions calculates equations for the objects on the graph. It finds the equation x + y = 1 for the circle and linear equations without restrictions for the line segments. Thus, when the function is moved up and no longer has all four points of intersection with the circle, the program will still show the points of intersection as if the line segments were infinite lines. 1

22 3 C 1-1+X +Y =0 D B O X- Y+ a=0-1 - X+ Y- a=0 -A (0,a) Q7: Check that the number of solutions that you observed graphically in part 3 is consistent with the number of solutions you observed symbolically in part 4. A: It is expected that students will substitute special values of a in the solutions and show that a) when a = - 1, points L and M are the same solutions, and K and N are different; b) when a = 1 all points K and N have the same solutions, but points L and M will evaluate to (1,0) and (-1, 0) and will look like two additional solutions; however, they are extraneous solutions due to the fact that Geometry Expressions found intersections of the circle with the extended lines and not the actual line segments. The same case applies if students substitute values in the range -1< a <. Students should be able to exclude extraneous solutions supplied by the software. Part 5 Solving Systems of Inequalities with Parameters This problem can be expanded to solving the following inequalities or system of inequalities with a parameter: x + a ± 1 x ; x + a ± 1 x ; 1 x x + a 1 x Extensions: For all these extensions the approach will be similar to one considered in the problem above. 1. Given y = x on a restricted domain, x + ( y a) = 1. Given y = x a on a restricted domain, x + y = 1 3. Given y = x on a restricted domain, ( x a) + y = 1 More generally, both shapes translate on an arbitrary vector (a,b).

23 Stained Glass Design Time required minutes Teaching Goals: 1. Students apply graphic methods to design various shapes on the plane.. Students apply geometric transformations of graphs of functions in order to design various shapes on the plane. 3. Students find equations of all curves obtained by geometric transformations in general form in either Cartesian or parametric form as defined by the teacher. 4. Students will determine domains of the independent variable for the equations of all curves for the design of the shape they construct in either Cartesian or parametric form as defined by the teacher. 5. Students will verify these results with the help of the software. Prior Knowledge Students should know graphs of linear functions and the equation of a circle in Cartesian and parametric forms. Students should know the coordinate form of a translation by a vector along each coordinate axis. Students should know that the equation y = f(x a) corresponds to the translation of the curve with the equation y = f(x) by vector (a, 0) along the x-axis. Students should know that equation y = f(x) + b corresponds to the translation of the curve with the equation y = f(x) by vector (0, b) along the y-axis. Students should know that a counter clockwise rotation has a positive angle and a clockwise rotation has negative angle, with angles in the range [0, π]. Students should know that the equation y = - f(x) corresponds to the reflection of the curve with the equation y = f(x) about the x-axis. Students should know that equation y = f(-x) corresponds to the reflection of the curve with the equation y = f(x) about the y-axis. Problem: Construct a stained glass design as shown in the picture below. This design is created by using the curves given by the following equations: y = x on [-3, 3] and x + y = a on [0, a ]. Try to recreate a similar design and to create other designs that could be created by these curves. 3

24 Determine equations of all curves on your design with their domain and verify them with the help of the software. Comment: the idea of this problem is to be able to send this information to another person who should be able to recreate this design based on the information provided (not by mindless copying). Part 1 Setting Up Problem in Geometry Expressions 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Function from the Draw toolbox. In the Type: box click the down arrow on the right and select Parametric from the list. In the next row, X=, enter: T ; in the next row, Y=, enter: abs(t); for Start: enter: -3; for End: enter: Click the equation of the function, right-click and select Hide from the context menu. Select the curve, right-click and select Properties from the context menu. Click the Line Style row and it s down arrow to change it to Dot. 4. Select Function from the Draw toolbox. In the Type: box click the down arrow on the right and select Parametric from the list. In the next row, X=, enter: a*cos(t) ; in the next row, Y=, enter: a*sin(t); for Start: enter: 0; for End: enter: Click the equation of the function, right-click and select Hide from the context menu. Select 4

25 the curve, right-click and select Properties from the context menu. Click the Line Style row and it s down arrow to change it to Dot Part Creating Stained Glass Design Q1: How can you construct a square from the given curves? A: It is expected that students will use translation along y-axis and reflection about x-axis according to the following steps, for example: 1. Click the straight line function and select Translation from the Construct toolbox. Construct a vector of translation from the origin down along the y-axis. The translated function will appear on the screen.. Select the end point of the translation vector and click Coordinate in the Constrain toolbox. Enter (0,-3) for the coordinates of the endpoint of the translation vector. 3. Select the translation vector, right-click and select Hide from the context menu. 4. Select the translated function. Right-click and select Properties from the context menu. Change the Line Color to Blue and the Line Style to Solid 3. 5

26 5. Click the translated function and select Reflection from the Construct toolbox. Click the x- axis and the function will be reflected over the x-axis. Adjust the line color and style as in step 4 above A B (0,-3) Q: How can you construct a quarter circle with its center at the bottom vertex of the square? A: Students may suggest rotating the given quarter circle by 4 π about the origin and then translating the center of the circle down to the point (-3, 0). Here are the steps: 1. Select the arc and click Rotation from the Construct toolbox. Click the origin and the box will appear for entering the angle. (If the Symbols toolbox is not displayed, click View / Tool Panels / Symbols) π can be found in the lower right corned of the dialog. Enter π/4.. Select the angle of rotation label and Hide it. 3. Select the rotated arc and click Translation from the Construct toolbox. Construct the translation vector from the origin to the point (0,-3). Hide the intermediate copy of the arc and the translation vector. 4. Change the new arc s line color to Red and the line style to Solid 3. Q3. How can you construct a quarter circle with the center in the top vertex of the square? 6

27 A: Students may suggest similar steps as above, but the simplest way to do this is to reflect the newly constructed arc about the x-axis. The following steps will accomplish this: 1. Click the arc with its center at the bottom vertex of the square. Select Rotation from the Construct toolbox and select the x-axis. The image will appear on the screen.. Select the arc and right-click Properties to change the line color to Red and the line style to Solid A B (0,-3) Q4: How can you construct a quarter circle with the center in the left vertex of the square? A: Students may suggest rotating the given quarter circle by -π/4 about the origin and then translating the center of the circle down to the point (-3, 0). Here are the steps: 1. Select the original arc and click Rotation from the Construct toolbox. Click the origin and the angle edit box will appear. From the Symbols toolbox, find π and type: -π/4.. Select the angle of rotation label and Hide it. 3. Select the rotated arc and click Translation from the Construct toolbox. Construct the translation vector from the origin to the point (-3, 0). Hide the intermediate copy of the arc and the translation vector. 4. Select the arc and right-click Properties to change the line color to Red and the line style 7

28 to Solid 3. Q5. How can you construct a quarter circle with the center in the right vertex of the square? A: Students may suggest similar steps as above, but the simplest way to do this is to reflect the arc with the center at the left vertex about the y-axis. The following steps will accomplish this: 1. Click the arc and select Reflection from the Construct toolbox. Click the y-axis and the image will appear on the screen.. Select the image and right-click Properties to change the line color to Red and the line style to Solid Click-and-drag the original quarter circle arc to adjust the transformed arcs to the desired size. 3 1 a -5 D (-3,0) A B (0,-3) Part 3 Finding Equations of the Curves in the Stained Glass Design In this part of the problem students have to determine the equation of each curve or function produced by transformation of the original curves. Based on the students prior knowledge, the teacher can determine in which form students will write the equations of the curves: Cartesian or parametric. Students should also determine the domain of the independent variable for the 8

29 equations of each curve. This should be completed independently and then verified with the help of the software. A. A Square is formed by two curves: y = x 3, [ 3,3] produced by the equation y = x, [ 3,3] translated by vector (0, - 3) and y = x + 3, [ 3,3] produced by equation y = x, [ 3,3] reflected about x-axis and translated by vector (0, 3). The inner design is made of four quarter circles. When the circles are tangent, the radius a of the circle is equal to half of 3 the side of the square, a =, and the x-coordinates of the points where the arcs intersect the squares are -1.5 and 1.5. The student can create the equations of the arcs as bounded circles centered at the origin, translated to the vertices of the square. Thus, the equations of the arcs of the circles follow. At the bottom vertex (0,-3): the circle 9 x + y =, translated by vector (0,-3) becomes 9 x + ( y+ 3) =, on [-1.5, 1.5] ; at the top vertex (0, 3): the bottom arc reflected about x-axis 9 9 becomes x + ( y 3) = on [-1.5, 1.5]; at the left vertex (-3, 0): the circle x + y =, 9 translated by vector (-3,0) becomes ( x+ 3) + y =, on [ 1.5, ] ; and at the right 9 vertex (3, 0): the left arc reflected about y-axis becomes ( x 3) + y =, on [3 1.5,1.5]. Students may have difficulty finding the domains of the functions. Part 4 Verification of the Equations with Geometry Expressions Here are the steps to find the equations using Geometry Expressions: 1. Select each curve, one at a time, and choose Implicit Equation from the Calculate toolbox. The equation of the curve will appear on the screen. 9

30 9+X -6 Y+Y -a =0 3 Y=- -3+ X 9+6 X+X +Y -a = X+X +Y -a =0-5 D (-3,0) A Y=-3+ X - -3 B (0,-3) 9+X +6 Y+Y -a =0 Students should substitute the value of a and verify that the equations given by the software are the same as ones they derived. In order to verify points of intersections, students can construct a square and a circle in a new document and determine points of intersection using the following steps: 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Point from the Draw toolbox and construct four points for the vertices of the square. 3. Select one of the points, choose Coordinate from the Constrain toolbox and type (0, 3). Constrain the other points with the coordinates (0, -3), (3, 0) and (-3, 0). 4. Select Line Segment from the Draw toolbox and connect the points to construct a square. 5. Select Circle from the Draw toolbox and construct a circle with the center at the point (3, 0). With the circle selected choose Radius from the Constrain toolbox and enter a for the value. 6. Holding the CTRL key, select the circle and a segment in the 1 st quadrant. Choose Intersection from the Construct toolbox. The point of intersection will appear. 7. Select the intersection point, choose Coordinate from the Calculate (Symbolic) toolbox and the coordinates of the intersection point will be displayed. 30

31 B 4 3- a, a (0,3) G -6 C (-3,0) - F a A (3,0) H (0,-3) D Comment: students can substitute the value of a and find the other intersection points based on symmetry. Extensions 1. Students can consider the following cases for the stained glass designs 3 a. when a < (arcs do not intersect) b. when 3 < a < 3(each pair of arcs intersect) c. when a > 3. Investigate case b where each pair of arcs intersects. Try to make the sum of the areas of the quadrilateral formed outside the arcs and the regions formed by the overlapping sectors in a given proportion, for example, 1:4 to the total area of the square. This problem should be solved by approximation, using a square to approximate the shape formed in the center outside the arcs. 31

32 Students can use circles and segments and construct part of the design in a new file and then find coordinates of the intersection point for the vertex of the central square in terms of parameter a. 6 a a, a -6 a AJ' - N' AF O' a AI' P 4 6 V' AF' S' - AH' 3. 3

33 3. Investigate the case c where a > 3. In this case concave quadrilateral shapes are formed at the vertices of the square with a convex quadrilateral at the origin. Students can again use circles and segments, construct the design and approximate the curved shapes with squares to look for specific ratio of areas for the design. a 4 a a a, a a - 33

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35 Translation Along Coordinate Axes Time required 45 minutes Teaching Goals: 1. Students verify that the graph of the function y = f(x) + b is derived from the graph of the function y = f(x) by translation along the y-axis by the vector (0, b).. Students verify that the graph of the function y = f(x a) is derived from the graph of the function y = f(x) by translation along the x-axis by the vector (a, 0). 3. Students verify the commutative property of horizontal and vertical translations of the graph of a function. 4. Students determine that two translations by vectors (a, 0) and (0, b) is equivalent to the translation by the vector (a, b). 5. Students determine that the graph of the function y = f(x a) + b is derived from the graph of function y = f(x) by translation by the vector (a, b) 6. Students determine the position of the translated graph of a function depending on the sign of parameters a and b. 7. Students verify these results with the help of the software. Prior Knowledge Students should know the definition of a vector. Students should know the coordinate representation of vectors. Students should know how to add vectors in vector form and in coordinate form. Students should know the coordinate form of translation by a vector, including special cases along each coordinate axis. Problem: Given the function y = x, investigate translation of the graph of the function. Part 1 Translation Along the Y-Axis In this part of the problem students investigate vertical translations and discover that the graph of the function y = f(x) + b is achieved by translating the graph of the function y = f(x) by vector (0, b) for any function f(x). They start working with the graph of the function y = x and then generalize for an arbitrary f(x). Here are the steps of construction and questions. 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar. 35

36 . Select Function from the Draw toolbox. Type: sqrt(x) for the function. 3. Draw a second function and enter: sqrt(x)+b for the function. Select the curve, right-click and select Properties from the context menu to change the line color to Blue. 4 Y= X+b -6 - Y= X Q1. As you drag the graph of function y = x + bup and down along the y-axis, what do you observe about the sign of parameter b with respect to the position of this graph relative to the graph y = x? A: It is expected that students will make the following conjecture: b > 0 when the graph of y = x + b is above the graph of y = x ; and b < 0 when the graph of y = x + b is below the graph of y = x. In order to see that, students can observe the value of b in the Variables toolbox while dragging the graph of the function y = x + b. Q. What is the geometric transformation from the graph of y = x to y = x + b. A. It is expected that students will bring up translation by a vertical vector. In order to verify this with the software, follow these steps. 36

37 1. Select Vector from the Draw toolbox. Construct a vector AB anywhere on the plane. Click the selection arrow.. Click the graph of the function y = x. Select Translation from the Construct toolbox, select vector AB and the translation of the graph will appear on the screen. 3. Select the translated graph, right-click, select Properties from the context menu, and choose Line Style Solid, Line Color Red. Q3. What are the coordinates of the vector of translation that transforms graph of y = x to the graph of y = x + b? Make a conjecture and verify it with the help of the software. A. The teacher may help students come up with a conjecture by pointing to the characteristic points of the graph, in this case point (0, 0) which is the vertex of the semi-parabola. This hint should help them determine the coordinates of the vector of translation. Students will also use the software to help find the answer and verify their conjecture. 1. Select vector AB, choose Coefficients from the Constrain toolbox. The coefficients of the vector will appear in the edit box. Accept the default values by pressing Enter. u0 v0 6 B 4 u 0 v 0 Y= X+b A - Y= X By moving the endpoint of the vector B students can overlap the translated image with the 37

38 graph of the function y = x + b and observe the values of u 0, v 0, and b in the Variables toolbox to answer the question and to verify their conjecture. The expected answer: u 0 = 0, v 0 = b. 3. In order to verify this, students can double click the coefficients and substitute these values. The translated image will overlap with the graph of the function y = x + b. Drag point B to confirm that. Q4. What vector will translate the graph of function y = x + b back to the graph of function y = x? Make a conjecture and verify with the software. A. The vector of translation is (0, -b). Q5. Generalize these results for an arbitrary function f(x) and confirm your conclusion with the help of the software. A. Students should conclude that the graph of function y = f(x) + b is derived from graph of function y = f(x) by translation along the y-axis by the vector (0, b). They should also conclude that the graph of function y = f(x) is derived from graph of function y = f(x) + b by translation along the y-axis by the vector (0, -b). They can verify this by editing the equation of the function: 1. Double click the equation of the function and change it to: f(x). The graph of the arbitrary function will appear on the screen.. Click the image of the translated function and choose Implicit Equation from the Calculate toolbox. The equation y = b + f(x) will appear on the screen. 38

39 10 Y=b+f(X) Y=f(X) B A 0 b Y= X+b Similarly, students can verify inverse translation with the help of the software. Part Translation Along the X-Axis In this part of the problem students investigate horizontal translations and discover that the graph of function y = f(x a) is always derived by translation from the graph of the function y = f(x) by the vector (a, 0) for any function f(x). They start working with the graph of the function y = x and then generalize for an arbitrary f(x). Here are the steps of construction and questions. 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Function from the Draw toolbox. Type: sqrt(x) for the function. 3. Select Function from the Draw toolbox. Type: sqrt(x - a) for the function. Select the function, right-click and select Properties from the context menu to change the line color to Blue. 39

40 3 1 Y= X-a -1 Y= X Q1. As you drag the graph of the function y = x a left and right along the x-axis, what do you observe about the sign of parameter a with respect to the position of this graph relative to the graph y = x? A: It is expected that students will make the following conjecture: if a > 0, the graph of y = x a is shifted to the right of the graph of y = x ; if a < 0, the graph of y = x a is shifted to the left of the graph of y = x. In order to see this, students can observe the value of a in the Variables toolbox while dragging the graph of the function y = x a. Q. What is the geometric transformation from the graph of y = x to y = x a. A. It is expected that students will bring up translation by a horizontal vector. In order to verify this with the software, follow these steps: 1. Select Vector from the Draw toolbox. Construct a vector AB anywhere on the plane. Click the selection arrow.. Click the graph of the function y = x. Select Translation from the Construct toolbox, 40

41 select vector AB and the image of the graph will appear on the screen. 3. Select the translated image, right-click and select Properties from the context menu to make Line Style Solid, and Line Color Red. Q3. What are the coordinates of the vector of translation that transform the graph of y = x to the graph of y = x a? Make a conjecture and verify it with the help of the software. A. The teacher may help students come up with a conjecture by pointing to the characteristic points of the graph, in this case point (0, 0) which is the vertex of the semi-parabola. This hint should help them determine the coordinates of the vector of translation. Students will also use the software to help find the answer and verify their conjecture. 1. Select vector AB, choose Coefficients from the Constrain toolbox. The coefficients of the vector will appear in the edit box. Accept the default values by pressing Enter. u0 v0 3 A B u 0 v 0 1 Y= X-a -1 Y= X By moving the endpoint of the vector B students can overlap the translated image with the graph of the function y = x a and observe the values of u 0, v 0, and a in the Variables toolbox to answer the question and to verify their conjecture. 41

42 The expected answer: u 0 = a, v 0 = In order to verify this, students can double click the coefficients and substitute these values. The translated image will overlap with the graph of the function y = x a. Drag point B to confirm this. Q4. What vector will translate the graph of the function y = x a back to the graph of the function y = x? Make a conjecture and verify with the software. A. The vector of translation is (-a, 0). Q5. Generalize these results for an arbitrary function f(x) and confirm your conclusion with the help of the software. A. Students should conclude that the graph of function y = f(x a) is derived from the graph of function y = f(x) by translation along the y-axis by a vector (a, 0). They should also conclude that the graph of function y = f(x) is derived from the graph of function y = f(x a) by translation along the x-axis by a vector (-a, 0). They can verify this by editing the equation of the function: 1. Double click the equation of the function and change it to: f(x). The graph of the arbitrary function will appear on the screen. Click the image of the translated function and choose Implicit Equation from the Calculate toolbox. The equation y = f(x - a) will appear on the screen. 4

43 8 6 Y=f(X) 4 Y=f(X-a) A a 0 B Y= X-a Similarly, students can verify the inverse translation with the help of the software Part 3 Commutative Property of Translation In this part of the problem students will investigate the commutative property of translations given by vectors (0, b) and (a, 0) along the coordinate axes of the graph of an arbitrary function y = f(x). Here are the steps of construction and questions. 1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.. Select Function from the Draw toolbox. Type f(x) for the function. 3. Select Vector from the Draw toolbox. Construct two vectors, AB and CD. Click the selection arrow. Select vector AB, choose Coefficients from the Constrain toolbox, type: (0, b) for the coefficients and press Enter. Select vector CD, choose Coefficients from the Constrain toolbox, type: (a, 0) for the coefficients and press Enter. Q1. Make a conjecture about relative position of the curves transformed by two methods 1) translate the graph of the function y = f(x) by vector AB and then translate the resulting graph by the vector CD, or ) translate the graph of the function y = f(x) by vector CD and then translate the resulting graph by the vector AB. Verify your conjecture with the help of the 43

44 software. A. Students will formulate their conjecture prior to working on the computer, then complete constructions on the computer and determine coincidence of the graphs, confirming or rejecting their conjectures. Here are the steps of the constructions. At this point, it s useful to adjust the default Curve setting to the thinnest black line so that you can see when two curves are the same. From the drop-down Edit menu at the top of the screen select Settings and the Geometry tab. In the Curve box check that the Line Color is Black and the Line Style is Solid 1. Method 1: 1. Click the graph of the function y = f(x), select Translation from the Construct toolbox, then click vector AB. Click the resulting image, right-click and select Properties from the context menu to change the line style to Dot.. Click the translated image, select Translation from the Construct toolbox. Then click the vector CD. Click the translated image, right-click and select Properties from the context menu and choose Line Style Solid 4, Line Color Light Blue (or any pastel shade). Method : 3. Click the graph of the function y = f(x) and select Translation from the Construct toolbox, then click vector CD. Click the resulting image, right-click and select Properties from the context menu to change the line style to Dot. 4. With the graph still selected, choose Translation from the Construct toolbox. Then click vector AB. Observe the image constructed by method coincides with the image constructed by method 1. Students should drag point B up and down and point D left and right to verify that this conjecture works for vectors of various lengths and directions. 44

45 Q. What is the equation of this curve? A. It is expected that students will apply prior knowledge from parts 1 and, and will be able to write in either order: (0, ) (,0) ( ) b a y f x y f( x) b y f( x a) b = = + = +, or (,0) (0, ) ( ) a b y = f x y = f( x a) y = f( x a) + b. Students then can check their findings using the software click the curve and select Implicit Equation from the Calculate toolbox. Q3. Is it possible to derive a graph of the function y = f(x a) + b from the graph of the function y = f(x) in one step? A. It is expected that students will conjecture that a translation by a vector (a, b) will transform one graph into another in one step. Students can check that using the software. In order to do that students will need to add two vectors of translation and then translate the original graph of function by the resultant vector to verify that it will coincide with the function y = f(x a) + b 45

46 Part 4 Applications and Assessment Problems Q1. Given two functions, f ( x) = x a and gx ( ) = x+ b. a. Can graph of one of the functions be derived from the graph of the other function by translations? b. If yes, by how many ways? c. If yes, write the vector of translation for the graph of f to be transformed to the graph of g and visa versa. d. Confirm your results with the help of the software A. a. The graph of one of the functions can be derived from the other by translation along the x-axis b. There are two translations, one from graph of f to g, and another one from g to f, if a + b 0. If a + b = 0 and thus f(x) = g(x) then there is only one translation. c. The vectors of translation are (± a + b, 0) d. Here are the steps of construction: 1. Click Function from the Draw toolbox and type sqrt(x-a) for the function f(x). Click OK. Select the graph of the function, right-click and select Properties from the context menu, select Line Style Solid, Line Color Blue.. Click Function from the Draw toolbox and type sqrt(x+b) for the function g(x). Click OK. Select graph of the function, right-click and select Properties from the context 46