1 Free Throw Shot Semester #1 ProjectAngela Huebert Mr. Eastman Period 3

2 1 2.75 ft 12.25 ft 10 ft 4 ft 6 ft 15 ft To find the actual measurements of the scale diagram, I used the given information that the distance between Rhee Bounder and the basket is 15ft to set up proportions for each section. If 6cm is the scale of 15ft, then I can set up an equation such as 6𝑐𝑚 15𝑓𝑡 = 𝑦 𝑥 where y is the measurement in centimeters from the scale diagram and you solve for x, which is the actual distance in feet. 6𝑐𝑚 15𝑓𝑡 = 4𝑐𝑚 𝑥𝑓𝑡 6x = 60 X=10 The distance between the basket and the floor in the diagram is 4cm, so I plugged it into the proportion and solved to get 10ft. The other sections were solved the same way.

3 1 To find the quadratic function that models the relationship between the horizontal distance before the basket and the vertical distance above and below the basket, which is located at the origin of the graph, I took the point (-12.25, 0) because it is one of the zeroes. I put it into factored form and then multiplied it out to get standard form. I used the sliders on Desmos to create a graph in which I plugged in all of the values I already had as well as the equation in standard form to find the estimated value of a and determined that it was less than -0.1, but not by much. The rounded value of a that the slider showed was I also added another point (D) to the graph to complete the parabola, and I found it by taking the distance from point A to the line of symmetry (the line down the center of the parabola, the average between the x-values on each “leg” of the parabola that have the same y-value and applying it to the opposite side of the line of symmetry. The y-value remained the same. x = x = 0 x = 0 x = 0 (x ) (x) 𝑥 𝑥+0 The points and equation that I plugged into Desmos to find the value of a. A (-15,-4) B (-12.25,0) C (0,0) D (2.75, -4)

4 Key Features in Relation to Problem2 Key Features in Relation to Problem Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot A (-15, -4) B (-12.25,0) E (-6.125, 3.638) D (2.75, -4) C (0,0) Point A: Rhee Bounder stood 15ft away from the base of the basket, and that is where he released the ball to begin the throw. With his height of 6ft factored in, he stood 4ft below the height at which the hoop was located. Point B: one of the zeroes (where the line crosses the x-axis). After Rhee threw the ball, this was the point in which it passed an invisible line that extends from the basket. The second photo was taken. Point C: one of the zeroes (where the line crosses the x-axis), the origin, and y-intercept. This was the point at which the ball passed through the center of the hoop, which is located at the origin of the graph. The third photo was taken. Point D: This was a point that I added to the parabola (reflecting point A across the dashed line, or line of symmetry) to make it even. The ball wouldn’t have realistically reached this point, as it would have dropped almost straight down after going into the hoop. Point E: vertex (highest point on the graph). This is the highest point that the ball reached on Rhee Bounder’s throw. After this point, the ball began its descent towards the hoop. It is located on the line of symmetry. Points A, B, and C were points at which Foe Tagrafer took pictures.

5 2 Vertex (Point E) Free Throw Shot Vertical Distance (feet)Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot As the vertex, Point E lies on the line of symmetry, which is the line that basically divides the parabola in half down the y-axis. I knew that the line of symmetry is located at x= because that is the point between the two zeroes: After finding the x-value, I only had to find the y-value, so I plugged it into the standard form equation and solved it out using both a=-0.1 and a= f(-6.125) = -0.1 ((-6.125) (-6.125)) f(-6.125) = -0.1 ( ) f(-6.125) = -0.1 ( ) f(-6.125) = round: 3.752 f(-6.125) = ((-6.125) (-6.125)) f(-6.125) = ( ) f(-6.125) = ( ) f(-6.125) = round: 3.638 The equation where I used a= was more accurate than the first equation I used. A (-15, -4) B (-12.25,0) E (-6.125, 3.638) D (2.75, -4) C (0,0)

6 2 Zeroes (Points B and C) Free Throw Shot Vertical Distance (feet)Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot First I attempted to plug a=-0.1 into the standard form equation. f(-12.25) = -0.1 ((-12.25) (-12.25)) f(-12.25) = -0.1 ( ) f(-12.25) = -0.1 (0) f(-12.25) = 0 f(0) = -0.1 ( (0)) f(0) = -0.1 (0+0) f(0) = -0.1 (0) f(0) = 0 It worked for these two points, but that’s because no matter what a is, it’s going to be multiplied by 0, which would equal 0. So I plugged in a more specific a= for the other points as well. f(-12.25) = ((-12.25) (12.25)) f(-12.25) = ( ) f(-12.25) = (0) f(0) = ( (0)) f(0) = (0 + 0) f(0) = (0) A (-15, -4) B (-12.25,0) D (2.75, -4) E (-6.125, 3.638) C (0,0)

7 2 Y-Intercept (Point C) Free Throw Shot Vertical Distance (feet)Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot Point C is also one of the zeroes, so I just took the equation that I used to find those. I could do this because the y-intercept is found by plugging in 0 as the x-value when solving the equation. I also began with a=-0.1 and then changed it to a= f(0) = -0.1 ( (0)) f(0) = -0.1 (0 + 0) f(0) = -0.1 (0) f(0) = 0 f(0) = ( (0)) f(0) = (0 + 0) f(0) = (0) E (-6.125, 3.638) A (-15, -4) B (-12.25,0) C (0,0) D (2.75, -4)

8 2 Points A and D Free Throw Shot Vertical Distance (feet)Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot These two points are just more points on the parabola; they are not key features, and they were only used to find the value of a and double check my equations. Point A was given to me in the problem, but I added Point D to make it easier to visualize the situation. I could plot Point D because I knew that it will fall on the same y-value (-4) as Point A did. To find the x-value (2.75), I found the distance between Point A and the line of symmetry (8.875) and then applied that to the other side of the line (reflection). a=-0.1 f(-15) = -0.1 ((-15) (-15)) f(-15) = -0.1 ( ) f(-15) = -0.1 (41.25) f(-15) = f(2.75) = -0.1 ((2.75) (2.75)) f(2.75) = -0.1 ( ) f(2.75) = -0.1 (41.25) f(2.75) = a= f(-15) = ((-15) (-15)) f(-15) = ( ) f(-15) = (41.25) f(-15) = f(2.75) = ((2.75) (2.75)) f(2.75) = ( ) f(2.75) = (41.25) f(2.75) = E (-6.125, 3.638) A (-15, -4) B (-12.25,0) C (0,0) D (2.75, -4)

9 3 To Find a To Find b I began with the standard form that I got from the zeroes: From there, I plugged in point A(-15,-4) and solved it out (below) until I got a decimal, which was the same one as what I had found by using Desmos, but it was repeating. I then found the fraction form of it so that I could easily find the value of b. Once again, I began with the factored form of This time, instead of plugging in a point from the parabola, I plugged in a and distributed it throughout the equation. The value of b ended up being another repeating decimal, so I went back to distribute a to the but this time I turned the into a fraction and then multiplied it by the fraction form of a, which gave me b as a fraction.

10 Desmos Sliders vs. Algebraically (decimal vs. fractional value of a)3 Desmos Sliders vs. Algebraically (decimal vs. fractional value of a) In this graph, I decided to show the two different values of a with the two answers that I got when I solved it in two ways. As you can see, there is very little difference, if any, between the two. This is because the decimal form of a that I had found on the computer was just the rounded decimal version of the fraction that I found when I solved it algebraically the second time. Though there are two slightly different a values, they both fall on the same line and cross through the correct points. The blue line that is visibly different from the others represents the a value that I found using the Desmos sliders first, which was -0.1, which was not accurate. Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot ORANGE DASHES a = BLACK DASHES a = -16/165 BLUE LINE a = -0.1

11 4 (Option #1) Stepped 2 Feet Back Stepped 1 Foot CloserThe ball will not make it into the basket at all; the farther distance (but with the same throw) means that the ball will fall short of both the hoop and the backboard. The ball would only go in the basket if it managed to hit the square on the backboard and drop in. Otherwise, it would not make it into the basket because of his changed position. In the graphs, the domain (set by the hoop’s diameter, which is 1.5ft) represents how far forward or backwards the ball can travel and still make it through the hoop. The range (set by the height of the square on the backboard, which is 1.5ft) represents the area that the ball can contact the back-board at and still fall into the basket. -0.75 < x < 0.75 0 < y < 1.5 Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot Horizontal Distance (feet) Vertical Distance (feet) Free Throw Shot

12 Maximum Distance to Move Forwards and Backwards4 (Option #1) Maximum Distance to Move Forwards and Backwards Rhee Bounder can only move 0.75ft back or 2.18ft forwards from his starting position 15ft away from the base of the basket and still make it into the hoop. To find these distances, I used Desmos sliders again. I created a “box” by limiting the domain and range to < x < 0.75 and 0 < y < 1.5 which allowed me to see the area that Rhee had to throw the ball into for it to make it directly into the hoop or bounce off of the backboard and drop in. The lowest point on the graph the “ball” could pass through – if he stepped back – is (-0.75,0) and the highest point the “ball” could pass through – if he stepped forward – would be (0.75,1.5).

13 Maximum Distance to Move Forwards and Backwards4 (Option #1) Maximum Distance to Move Forwards and Backwards In Desmos, I graphed the original throw and points with the equation, and then took the factored form of it ( ). I know that moving forward or backwards on the x-axis would mean adding or subtracting the number that I was supposed to solve for, so I used the number n and used the factored form to add n and find the distance he could move back. The equation was plugged into Desmos and then I used a slider to adjust the value of n until I was able to find an equation that formed a parabola that passed through the point (-0.75,0). To solve for n when he moves forward, I did the same thing as I did for his move back, only n was subtracted instead of added, so the equation became , and repeated the process with the point (0.75,1.5)

14 4 (Option #1) Free Throw Shot Vertical Distance (feet)(0.75, 1.5) (-0.75, 0) Vertical Distance (feet) (-15.75, -4) (-15, -4) (-12.28, -4) domain of Rhee’s total movement: -15.75ft < x < ft maximum backwards original maximum forwards Horizontal Distance (feet)