Graphing Quadratic Fraction By: Bianca Siregar Math SL 11 a) The quadratic equation you found which models the path of the basketball. First Shot: y = −0.79x 2 +1.58x + 0.71 Second shot: y = −0.7x 2 +1.26x + 0.73 b) Answer the question: which of these shots actually goes in the basket (based on the parabola you found describing the path of the basketball). The First Shots is the shot that actually goes in the basket c) An explanation of how you found your equations and the math you used to do so. When trying to make the parabola I found there are 2 ways to get the equation of the parabola. The two ways are: I. II.
Using Slider Manually
Using Slider Using slider is much easier than the manual calculation. We only set the point and line in the GeoGebra 1. The same procedure as the manual technique, insert both picture in the GeoGebra. Before setting the position of the image, I stretch out the y and x -axis so I could get smaller division number (from 1 to 0.2). I also change the view from normal into grid view to support my view in seeing the intersection; we could see the intersection by reducing the opacity of the image 2. Insert new point, click in every ball in the picture and in the hoop as the parabola’ s point 3. Insert 3 sliders in the graph 4. Input the equation in vertex form y = a(x − h)2 + k 5. Change the name of the slider from a, b and c into a, h, and k 6. Set the sensitivity of each slider to make sure we could get the parabola line hitting the point The result of graphing using the slider is shown as below
Shot 1 Result: As can be seen, the line of the parabola pass through the hoop, which means the ball, is successfully in. I also get the same result with the manual technique, the ball when in, even though the equation is slightly different.
Shot 2 Result: From the graph above, we could see the parabola didn’t go into the basket. The line went through the hoops ball, we could suspect if this happen, the ball will bounce back and not successfully go in the hoop. There is still a possibility that the ball could go into the basket, if the person didn’t throw the ball to hard which makes the ball bounce back. Manually:
1. First of all, as what the instruction said I insert both picture in the GeoGebra. Before setting the position of the image, I stretch out the y and x -axis so I could get smaller division number (from 1 to 0.2). This could help me to determine the points of the parabola. I also change the view from normal into grid view to support my view in seeing the intersection; we could see the intersection by reducing the opacity of the image. 2. Then by looking at the picture I take the clearest and estimate intersections as my starting point. For example, I wrote the y-intercept, since the first ball is touching in y-axis. I took the other 2-points, and from this points I use it to find the equation of the graph, which will pass the intersection by using elimination of quadratic formula standard form (y = ax 2 + bx + c) These are my calculations: Shot 1: Points:
(0, 0.9) (0.4,1.39) (0.8,1.6) C = 0.9 1.39 = a(0.4)2 + b(0.4) + 0.9 → 0.16a + 0.4b = 0.49 1.6 = a(0.8)2 + b(0.8) + 0.9 → 0.64a + 0.8b = 0.7 Elimination 0.16a + 0.4b = 0.49 × 4 → 0.64 +16b = 1.96 0.64a + 0.8b = 0.7 ×1 → 0.64 + 0.8b = 0.7 0.8b = 1.2b b = 1.5
Find a 0.16a + 0.6 = 0.49 0.16a = −0.11 a = −0.6785
Equation of the Parabola: y = −0.6785x 2 +1.5x + 0.9 Shot 2:
(0, 0.62) (0.15, 0.8) (0.7, 0.1) C = 0.62 0.8 = a(0.15)2 + b(0.15) + 0.62 → 0.0225a + 0.15b = 0.18 0.1 = a(0.7)2 + b(0.7) + 0.62 → 0.49a + 0.7b = −0.52 Elimination 0.0225a + 0.15b = 0.18 × 7 → 0.1575a +1.05b = 1.26 0.49a + 0.7b = −0.52 ×15 → 7.35a +1.05b = −7.8 −7.1925a = 9.06 a = −1.259 ≈ −1.26
Find B 0.49(−1.26)+ 0.7b= −0.52 −0.6174 + 0.7b = −0.52 b = −0.1391
Equation of the Parabola: y = −1.26x 2 + 0.1391x + 0.62 3. Then, I determine the point of the hoop to see whether the line of the parabola will pass through it or not. The y-point of the hoop should equal to the equation of the parabola when the value of is being inserted by x-point of the hoop. Shot 1 Hoop Point: (1.9,1.2) Therefore, when 1.2 is being inserted to the formula, the equation should be equal to the y value: 1.9
y = ax 2 + bx + c 1.2 = (−0.6875)(1.9)2 + (1.5)(1.9) + 0.9 1.2 = 1.268 After it being proven through the calculation it shows that the line of the parabola passes through the point of the hoop. Therefore, the first shot is actually successful.
Shot 2 Hoop Point: (1.25, 0.8) y = ax 2 + bx + c 0.8 = (−1.26)(1.25)2 + (0.1391)(1.25) + 0.62 0.8= −1.174
Through the calculation, we could see the number is not match, this indicates that the line does not pass through the point of the hoop. This means, the second shot is actually not successful, the ball will not go into the hoop. From the line, we could see the other end of the parabola is not passing inside the hoop. 4. I revise the position of the parabola because at first it didn't fit the position. When we move the parabola, the equation will automatically change. Finally the equation that I got from both first second shot are: a. First shot y = −0.69x 2 +1.47x + 0.88 b. Second Shot y = −0.9x 2 +1.33x + 0.63
First shot
Second shot Limitation: Uncertainty is the main problem when we try to find the equation manually, because we could not see where exactly the point is. Further more, we only use 1 decimal place point, which makes it more inaccurate. Conclusion: From this investigation now I know the role of different variables in quadratic formula equation, through using slider. In vertex form; a determines whether the parabola is open up or close; h determines the horizontal translation of the parabola (against the y-axis; right or left) and k determines the vertical translation (against the x-axis up or down)