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INTRODUCTION

“This lab will have you take a look at the parabolic path to try to determine the maximum altitude the plane reaches. First, you will work with data given about the parabola to come up with a quadratic model for the flight. Then you will work to find the maximum value of the model. Now for the data:

WRITING A 3 BY 3 SYSTEM OF EQUATIONS FOR A, B, AND C. Using the equation provided in the packet, h=at^2

+ bt +c, I created my first equation 23645 = 4a + 2b +c.

For my second equation, I plugged in 32015 = a(20)^2 + b(20) +c. After multiplying everything out, I got the equation 32015 = 400a +20b + c.

For the third system of equations, I plugged in 33715 = a(40)^2 + b(40) + c, and then after multiplying everything out, I got the equation 33715 = 1600a +40b +c.

SOLVING THE THREE SYSTEMS OF EQUATIONS To solve a system of three equations, you must first choose two

equations and choose which variable you want to eliminate. I chose to first eliminate variable ‘c’ from the equations. I chose the equation 23645 = 4a + 2b + c and the equation 32105 = 400a + 20b + c. I multiplied the first equation by a -1, so that the variable ‘c’ would cancel from both equations. After adding those two equations together, I was left with 8460 = 396a + 18b.

Next, I chose the equations 32105 = 400a +20b + c and 33715 = 1600a + 40b +c. I once again multiplied the first equation by a -1 to eliminate the variable c from both equations. After adding the two equations together, I was left with 1610 = 1200a + 20b.

My third step was to solve for each variable to see what they would equal. I used elimination on the equations that I had simplified earlier. I multiplied the equation 8460 = 396a + 18b by 20, and then multiplied the equation 1610 = 1200a + 20b by -18, so that the variable ‘b’ would cancel. I divided to get the variable ‘a’ alone, and found that a = =10.25.

SOLVING THE THREE SYSTEMS OF EQUATIONS (PT 2)To solve for variable ‘b’, I plugged variable ‘a’ (-10.25) into an

equation that I had solved for early that just had the variables ‘a’ and ‘b’ in it.

1610 = 1200(-10.25) + 20b

1610 = =12300 + 20b

20b = 13910

b = 695.5

Next, I am solving for the last variable, ‘c’.

33715 = 1600(-10.25) +40(695.5) + c

33715 = -16400 + 27820 + c

c = 22295

FORM A QUADRATIC MODEL OF THE DATA

Using the equation provided, h = at^2 + bt + c , I created the equation h = -10.25t^2 +695.5t + 22295.

FIND THE MAXIMUM VALUE OF THE QUADRATIC FUNCTIONTo find the maximum value of a function, you must put the

equation in vertex form.

H – 22295 = -10.25t^2 +695.5t subtract over the number without a variable

H – 22295 = -10.25(t^2 – 67.85 + 1150.91) factor out ‘a’ and complete the square

H – 34091.83 = -10.25(t – 33.93)^2 multiply 1150.91 by -22295 and subtract to the left side

H = -10.25(t – 33.93)^2 + 34091.83 balance the equation by adding over 34091.83 so that H is by itself.

The max. value of a function is the vertex, so the max. value for this equation is (33.93, 34091.83).

SKETCH THE PARABOLA

REFLECTIVE WRITING

“Did this project change the way you think about how math can be applied to the real world? Write one paragraph stating what ideas changed and why. If this project did not change the way you think, write how this project gave further evidence to support your existing opinion about applying math. Be specific.”

Response: Yes, this project did change the way that I think how math can be applied to the real world. With zero gravity ride, it is imperative that the parabola has the right curve. If it is too steep, there can be damage done to the passengers. If it is not curved enough, then you will not be able to feel the zero gravity effect.