Discover Parabola Time required 45 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 the 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 parabola y = x and confirm the equation given by the software. Prior Knowledge Students should know graph of quadratic function in two forms: 0 0 y = ax + bx+ c and y = a( x x ) + y, a 0, and they should know this graph has a shape of 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 existence of as a graph of quadratic function (parabola). 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, maximum with coordinates (, 3), and points with coordinates (0, ) and (3, ). They are also given graph of initial parabola y = x and graph of arbitrary parabola y = a( x x0) + y0 1. Open new file. If system of coordinate does not appear in the blank document, create it by clicking on the Toggle Grid and Axes button in the top toolbar.. Select Function tool from the Draw menu. Type x^ for the function. 3. Select Function tool from the Draw menu. Type y[0]+a*(x-x[0])^ for the function. Click on the graph of the function to highlight it and then choose Edit/Properties and choose Line Color lue. 4. Select Point tool from the Draw menu. Construct three points anywhere on the screen. Select point A, choose Coordinate from the Constrain menu and type coordinates (, 3) for the point A, press Enter. Repeat that for the points and C. If desired, change the font, size and color for the labels of the points.
4 3 A(,3) C (3,) 1 Y=X - 1 3 4 5 6 7 (0,) - Q1. Can you find a parabola that has 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 = y0 + a( x x0) are responsible for which type of transformation of the graph of parabola. In order to do that 1. Click on the curve of parabola and start dragging it. Simultaneously note what parameter will appear by the circle and what type of transformation is happening with the curve.. Click on the other part of the curve and do the same. 3. Record your observations in a table below: 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,, and C. Construct your conjecture about the equation of the parabola (round numerical values to whole). 4 3 A(,3) x 0 C (3,) a 1 Y=X - (0,) 1 3 4 5 6 7 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 window of the Tools menu. When they think they have the solution, they come up with the conjecture, they can round the numeric values of the coefficients received with the help of software and verify their equation by substitution. For the picture shown above, a 0.998, x0.003, y0 3.009. A. The equation of parabola is: y = 3 ( x ). Here are steps of verification: 1. In the variables menu, enter predicted values for the parameters, a= 1, x0 =, y0 = 3.. Click on the graph of the parabola and choose Real tab, then Implicit equation tool under Calculate menu. The numeric equation for the parabola will appear on the screen
4 3 A(,3) C (3,) Y=X 1 Y=3-(-+X) - (0,) 1 3 4 5 6 7-3. 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 graph of this parabola. Is it important to preserve the order of transformations? Verify your answers with the help of software. A. The transformations are: translation by vector (, 3) and dilation by factor. These are commutating. Students should complete construction, verify coincidence of the image with the parabola and the fact that order of transformation does not affect the result. Here are the steps of construction: 1. Select Vector Tool from the Draw menu and construct vector from the origin to the point A.. Click on the graph of the parabola y = x, select Translation tool from the Construct menu and then click on the vector. The image of the translated parabola will appear on the graph. 3. Choose the image, select Dilation tool from the Construct menu. Select point A and leave b in the open box for the dilation factor. 4. In the Variables menu setup range of values for the dilation factor b from - to 0. Drag the bar to adjust the value to see that the curve will coincide with the graph of parabola y = 3 ( x ) when b =.
4 3 b A(,3) C (3,) Y=X 1 Y=3-(-+X) - E 1 3 4 5 6 7 (0,) - Q5. Do you think there exists another parabola passing through points A, and C? Explain your answer. A. Three non-collinear points define parabola uniquely, can be proved analytically. Part Conditions of Existence of Parabola Passing Through Three Arbitrary Points In this part of the problem students investigate how relative position of three arbitrary points affects existence of parabola passing through these points. Given three points on the plane: A,, and C, and arbitrary parabola defined by the equation y = y + a( x x ). 0 0 1. Open new file. If system of coordinate does not appear in the blank document, create it by clicking on the Toggle Grid and Axes button in the top toolbar.. Select Function tool from the Draw menu. Type y[0]+a*(x-x[0])^ for the function. Click on the graph of the function to highlight it and then choose Edit/Properties and choose Line Color lue, Line Style Solid. 3. Select Point tool from the Draw menu. Construct three points anywhere on the screen. 4. Select point A and graph of the parabola. Select Incident tool from the Constrain menu
and point A will merge to the curve. Repeat the same procedure with the points and C. Q1. What are conditions on the relative position of the points A,, and C, so that parabola no longer exists, e.g. the shape of parabola changes into a different shape? A. When points lie on the same line, parabola does not exist, the curve 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 that following the steps: 1. Move all points to lie on the line y = const (approximately). The parabola will look like a horizontal line.. Observe the value of parameter a in the Variables window. The value of a is close to zero. Make conjecture that a = 0 and enter 0 for the value of a. 6 4 A C -6-4 6-3. Measure real coordinates of points A,, and C and measure real equation of the line. Make sure to use the tab Real in the Calculate menu. Select each point, one at a time and choose Coordinates. 4. Choose graph of the line, select Implicit equation and equation of the line will appear on the screen confirming the fact that this is not an equation of the quadratic function.
6 Y= A (-.0747664,) 4 (1.0093458,) C (3.0093458,) -6-4 6 - Extensions: 1. Existence of parabola. Given parabola defined by the equation y = ax + bx+ c and three points, A,, C. Follow steps similar to the part to determine that if three points are collinear, then a = 0, and points lie on inclined line with slope b. so parabola does not exist.
14 1 10 (3.9874,8.9748001) Y=1+ X 8 (.4454303,5.8908605) C 6 A Y=X a+x b+c 4 (1.0093458,3.0186916) 0-8 -6-4 6 8 10. Existence of parabola through any two points. If given two points choice of 3 rd point defines unique parabola, so there are infinite number of parabolas since there is infinite number of choices for the 3 rd point. In this case teacher will provide students with conditions based on real problems, for example: a) given the point of basketball take off and position of the basket, make the basket b) make a pass in a volleyball in its highest point (vertex) c) in soccer ball should hit to the right upper corner of the goal, etc.