Cassandra EmelianchikMicah Cox

Brett Griffin

With three seconds left on the clock of our football game, we are held at the 38 yard line. We can either wait and take a 55 yard kick, or make a Hail Mary pass and hope to gain some yardage. We decide to do some quick thinking and use the equation y = -1/14x^2 + 23/7x – 45/14 to represent our kicker’s best kick. Once we graph our equation, it will tell us where the ball starts on the field, where it will land, how high it will go, the path it will take, and consequently, whether we will make the field goal or not.

y = -1/14x^2 + 23/7x – 45/14x = -b/2ax = (-23/7)/2(-1/14)x = -3.428571429/-0.1428571429x = 23y = -1/14(23)^2 + 23/7(23) –45/14y = 34.57142857

1. Plug in the variables2. Solve3. Plug x into the equation4. Solve for y

To start off, we need to solve our equation for the vertex which can tell us how high the ball will go, the vertex on our graph, and the axis of symmetry. We first use the equation x = -b/2a to find the x value of our vertex, or the highest point, of our kick.

So our vertex is (23, 34.57142857), the graph has a maximum of 34.57142857, and our axis of symmetry is 23. The axis of symmetry is the turning point of the ball in the air.

To find out where the ball will start and land on the field, we need to factor our equation.

Now we know that the ball starts at the 45 yard line and lands 1 yard short of the field goal.

y = -1/14x^2 + 23/7x –45/14

0 = -1/14x^2 + 23/7x –45/14

-14(0 = -1/14x^2 + 23/7x –45/14)

0 = x^2 – 46x + 450 = (x – 45)(x – 1)0 = x – 45 0 = x – 1x = 45 x = 1

1. Set the equation equal to zero

2. Get rid of the negative a and the coefficient of x^2

3. Factor the equation4. Set the x values equal

to zero and solve for x

X Y

21 34.28571429

22 34.5

23 34.57142857

24 34.5

25 34.28571429

Now that we have our vertex and intercepts, we can find other points on our graph.

1. Start by making a table with x and y values.2. Have two numbers both above and below the x value of the vertex 3. Plug in two of the x values into your original equation. (make sure that

these values are either one unit above or below the x value, and two units above or below the x value; never two values the same amount away from the vertex)

4. Solve for y with each different value5. Plug what you get for y into the spot on your table that corresponds with

the x value you plugged in6. Now you have points that lay on your kick’s path to plot on your graph.

Now that we have all of the information that we need, we can graph it. We start by first plotting the points, and then connecting them. On our graph we have the axis of symmetry running through the vertex at (23, 34.57142857).

VertexMaximumAxis of symmetryRoots

•The y-axis represents the position of the field goal along the x-axis (so at x = 0 is the field goal)•From the 10 yard mark to the 0 mark on the x-axis of the graph is the end zone

To get a more precise graph, we plugged our information into the graphing calculator.

The kicker starts at 45 yards on the opposing team’s side. He kicks the ball, which peaks at 34.57142857 feet in the air, at 23 yards on the field. From there it goes back down and lands 9 yards into the end zone, or 1 yard away from the field goal. Even if the kicker were to get that one last yard, his kick would not have had enough height to make the field goal.

Based off of our graph, we would make a Hail Mary pass in the last three seconds of the game in hope of gaining yards. We would not take the kick at the 38 yard line because our kicker’s kick would land 1 yard short of the field goal. To make the field goal, we would need to gain at least 14 yards. This is because on our graph, the y-intercept is -3.214285714, and as the field goal is 10 feet tall, we would need to gain 14 yards. That extra yardage would be the minimum amount of yards that we would need to gain to get the kick’s height high enough to get the field goal. Thus, we will make a Hail Mary pass in hopes of gaining at least 14 yards.